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Binomial Theorem and Its Simple Applications — JEE Previous Year Questions

Every Binomial Theorem and Its Simple Applications question asked in JEE Main and JEE Advanced across the last 186 papers — 211 questions, each with its correct answer. Free to read, no account needed.

Questions

211

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158/186

Appearance rate

85%

All 211 Binomial Theorem and Its Simple Applications questions

Most recent papers first.

Q1·MathematicsMultiple correctJEE Advanced 2026
Let R\mathbb{R}R denote the set of all real numbers. Consider the polynomial function f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R defined by f(x)=d10dx10((x2−1)10)f(x) = \dfrac{d^{10}}{dx^{10}}\left((x^{2} - 1)^{10}\right)f(x)=dx10d10​((x2−1)10), for all x∈Rx \in \mathbb{R}x∈R. Here d10dx10((x2−1)10)\dfrac{d^{10}}{dx^{10}}\left((x^{2} - 1)^{10}\right)dx10d10​((x2−1)10) is the 10th order derivative of the function (x2−1)10(x^{2} - 1)^{10}(x2−1)10. Then which of the following statements is (are) TRUE ?
  1. (A)The coefficient of x8x^{8}x8 in the polynomial f(x)f(x)f(x) is (−10)(18!8!)(-10)\left(\dfrac{18!}{8!}\right)(−10)(8!18!​)
  2. (B)The value of f(1)+f(−1)f(1) + f(-1)f(1)+f(−1) is equal to 10! 21110!\, 2^{11}10!211
  3. (C)The degree of the polynomial f(x)f(x)f(x) is 10
  4. (D)The constant term of the polynomial f(x)f(x)f(x) is −(10!5!)-\left(\dfrac{10!}{5!}\right)−(5!10!​)

Correct answer: (A), (B), (C)

Step-by-step solution →
Q2·MathematicsSingle correctJEE Main 2026
If 26(233(122)+255(124)+277(126)+…+21313(1212))=313−α26\left(\frac{2^3}{3}\binom{12}{2} + \frac{2^5}{5}\binom{12}{4} + \frac{2^7}{7}\binom{12}{6} + \ldots + \frac{2^{13}}{13}\binom{12}{12}\right) = 3^{13} - \alpha26(323​(212​)+525​(412​)+727​(612​)+…+13213​(1212​))=313−α, then α\alphaα is equal to:
  1. (A)45
  2. (B)48
  3. (C)51
  4. (D)54

Correct answer: (C)

Step-by-step solution →
Q3·MathematicsNumericalJEE Main 2026
If (1−x3)10=∑r=010arxr(1−x)30−2r(1 - x^{3})^{10} = \sum_{r=0}^{10} a_{r} x^{r} (1 - x)^{30-2r}(1−x3)10=∑r=010​ar​xr(1−x)30−2r, then 9a9a10\frac{9a_{9}}{a_{10}}a10​9a9​​ is equal to __________.

Correct answer: 30

Step-by-step solution →
Q4·MathematicsSingle correctJEE Main 2026
If the coefficients of the middle terms in the binomial expansions of (1 + αx)26^{26}26 and (1 − αx)28^{28}28, α ≠ 0, are equal, then the value of α is:
  1. (A)1
  2. (B)1413\frac{14}{13}1314​
  3. (C)277\frac{27}{7}727​
  4. (D)727\frac{7}{27}277​

Correct answer: (D)

Step-by-step solution →
Q5·MathematicsSingle correctJEE Main 2026
The coefficient of x2x^2x2 in the expansion of (2x2+1x)10\left( 2x^2 + \frac{1}{x} \right)^{10}(2x2+x1​)10, x≠0x \neq 0x=0, is :
  1. (A)3240
  2. (B)3360
  3. (C)3480
  4. (D)3600

Correct answer: (B)

Step-by-step solution →
Q6·MathematicsNumericalJEE Main 2026
If the sum of the coefficients of x7x^7x7 and x14x^{14}x14 in the expansion of (1x3−x4)n\left(\frac{1}{x^3} - x^4\right)^n(x31​−x4)n, x≠0x \neq 0x=0, is zero, then the value of nnn is __________.

Correct answer: 21

Step-by-step solution →
Q7·MathematicsSingle correctJEE Main 2026
In the expansion of (9x−13x)18\left(9x - \frac{1}{3\sqrt{x}}\right)^{18}(9x−3x​1​)18, x>0x > 0x>0, if the term independent of xxx is (221)k(221)k(221)k, then kkk is equal to:
  1. (A)84
  2. (B)78
  3. (C)168
  4. (D)198

Correct answer: (A)

Step-by-step solution →
Q8·MathematicsSingle correctJEE Main 2026
Let the smallest value of k∈Nk \in \mathbb{N}k∈N, for which the coefficient of x3x^3x3 in (1+x)3+(1+x)4+(1+x)5+…+(1+x)99+(1+kx)100(1 + x)^3 + (1 + x)^4 + (1 + x)^5 + \ldots + (1 + x)^{99} + (1 + kx)^{100}(1+x)3+(1+x)4+(1+x)5+…+(1+x)99+(1+kx)100, x≠0x \neq 0x=0, is (43n+1014)(100C3)\left(43n + \frac{101}{4}\right)\left({}^{100}C_3\right)(43n+4101​)(100C3​) for some n∈Nn \in \mathbb{N}n∈N, be ppp. Then the value of p+np + np+n is:
  1. (A)101010
  2. (B)111111
  3. (C)121212
  4. (D)131313

Correct answer: (B)

Step-by-step solution →
Q9·MathematicsSingle correctJEE Main 2026
The number of elements in the set S={(r,k):k∈Z and 36Cr+1=6(35Cr)(k2−3)}S=\left\{(r,k):k\in\mathbb{Z}\ \text{and}\ {}^{36}C_{r+1}=\frac{6\left({}^{35}C_{r}\right)}{(k^{2}-3)}\right\}S={(r,k):k∈Z and 36Cr+1​=(k2−3)6(35Cr​)​}, is :
  1. (A)2
  2. (B)4
  3. (C)8
  4. (D)16

Correct answer: (B)

Step-by-step solution →
Q10·MathematicsSingle correctJEE Main 2026
If for 3≤r≤303 \leq r \leq 303≤r≤30, (3030−r)+3(3031−r)+3(3032−r)+(3033−r)=(mr)\binom{30}{30-r} + 3\binom{30}{31-r} + 3\binom{30}{32-r} + \binom{30}{33-r} = \binom{m}{r}(30−r30​)+3(31−r30​)+3(32−r30​)+(33−r30​)=(rm​), then mmm equals:
  1. (A)313131
  2. (B)323232
  3. (C)333333
  4. (D)343434

Correct answer: (C)

Step-by-step solution →
Q11·MathematicsSingle correctJEE Main 2026
The sum of the coefficients of x499x^{499}x499 and x500x^{500}x500 in (1+x)1000+x(1+x)999+x2(1+x)998+……+x1000(1 + x)^{1000} + x(1 + x)^{999} + x^{2}(1 + x)^{998} + \ldots\ldots + x^{1000}(1+x)1000+x(1+x)999+x2(1+x)998+……+x1000 is
  1. (A)1001C501^{1001}C_{501}1001C501​
  2. (B)1002C500^{1002}C_{500}1002C500​
  3. (C)1002C501^{1002}C_{501}1002C501​
  4. (D)1000C501^{1000}C_{501}1000C501​

Correct answer: (B)

Step-by-step solution →
Q12·MathematicsSingle correctJEE Main 2026
Given below two statements : Statement I : 2513+2013+813+31325^{13} + 20^{13} + 8^{13} + 3^{13}2513+2013+813+313 is divisible by 7. Statement II : The integral part of (7+43)25\left(7+4\sqrt{3}\right)^{25}(7+43​)25 is an odd number. In the light of the above statements, choose the correct answer from the options given below :
  1. (A)Both Statement I and Statement II are false.
  2. (B)Both Statement I and Statement II are true.
  3. (C)Statement I is false but Statement II is true.
  4. (D)Statement I is true but Statement II is false.

Correct answer: (B)

Step-by-step solution →
Q13·MathematicsSingle correctJEE Main 2026
Let S =125!+13!23!+15!21!+...= \frac{1}{25!} + \frac{1}{3!23!} + \frac{1}{5!21!} + ...=25!1​+3!23!1​+5!21!1​+... up to 13 terms. If 13S=2kn!13\mathrm{S} = \frac{2^{k}}{n!}13S=n!2k​, k∈N, then n + k is equal to
  1. (A)51
  2. (B)52
  3. (C)49
  4. (D)50

Correct answer: (C)

Step-by-step solution →
Q14·MathematicsSingle correctJEE Main 2026
The value of 100C5051+100C5152+…+100C100101\frac{^{100}C_{50}}{51} + \frac{^{100}C_{51}}{52} + \ldots + \frac{^{100}C_{100}}{101}51100C50​​+52100C51​​+…+101100C100​​ is :
  1. (A)2101100\frac{2^{101}}{100}1002101​
  2. (B)2100100\frac{2^{100}}{100}1002100​
  3. (C)2101101\frac{2^{101}}{101}1012101​
  4. (D)2100101\frac{2^{100}}{101}1012100​

Correct answer: (D)

Step-by-step solution →
Q15·MathematicsSingle correctJEE Main 2026
The sum of all possible values of n∈Nn \in Nn∈N, so that the coefficients of xxx, x2x^{2}x2 and x3x^{3}x3 in the expansion of (1+x2)2(1+x)n(1 + x^{2})^{2} (1 + x)^{n}(1+x2)2(1+x)n, are in arithmetic progression is :
  1. (A)3
  2. (B)7
  3. (C)12
  4. (D)9

Correct answer: (D)

Step-by-step solution →
Q16·MathematicsSingle correctJEE Main 2026
Let CrC_rCr​ denote the coefficient of xrx^rxr in the binomial expansion of (1+x)n(1 + x)^n(1+x)n, n∈Nn \in \mathbb{N}n∈N, 0≤r≤n0 \le r \le n0≤r≤n. If Pn=C0−C1+223C2−234C3+.....+(−2)nn+1CnP_n = C_0 - C_1 + \frac{2^2}{3}C_2 - \frac{2^3}{4}C_3 + ..... + \frac{(-2)^n}{n+1}C_nPn​=C0​−C1​+322​C2​−423​C3​+.....+n+1(−2)n​Cn​, then the value of ∑n=1251P2n\sum_{n=1}^{25} \frac{1}{P_{2n}}∑n=125​P2n​1​ equals.
  1. (A)580
  2. (B)525
  3. (C)650
  4. (D)675

Correct answer: (D)

Step-by-step solution →
Q17·MathematicsNumericalJEE Main 2026
If (115C0+115C1)(115C1+115C2)⋯(115C12+115C13)=α1314C0 14C1⋯14C12\left(\frac{1}{{}^{15}C_{0}} + \frac{1}{{}^{15}C_{1}}\right)\left(\frac{1}{{}^{15}C_{1}} + \frac{1}{{}^{15}C_{2}}\right)\cdots\left(\frac{1}{{}^{15}C_{12}} + \frac{1}{{}^{15}C_{13}}\right) = \frac{\alpha^{13}}{{}^{14}C_{0}\,{}^{14}C_{1}\cdots{}^{14}C_{12}}(15C0​1​+15C1​1​)(15C1​1​+15C2​1​)⋯(15C12​1​+15C13​1​)=14C0​14C1​⋯14C12​α13​, then 30α30\alpha30α is equal to _______.

Correct answer: 32

Step-by-step solution →
Q18·MathematicsSingle correctJEE Main 2026
If the coefficient of x in the expansion of (ax2+bx+c)(1−2x)26(ax^{2}+bx+c)(1-2x)^{26}(ax2+bx+c)(1−2x)26 is −56-56−56 and the coefficients of x2x^{2}x2 and x3x^{3}x3 are both zero, then a + b + c is equal to
  1. (A)1300
  2. (B)1500
  3. (C)1403
  4. (D)1483

Correct answer: (C)

Step-by-step solution →
Q19·MathematicsNumericalJEE Advanced 2025
Let a0,a1,……,a23a_0, a_1, \ldots\ldots, a_{23}a0​,a1​,……,a23​ be real numbers such that (1+25x)23=∑i=023aixi\left(1 + \frac{2}{5}x\right)^{23} = \sum_{i=0}^{23} a_i x^i(1+52​x)23=∑i=023​ai​xi for every real number x. Let ara_rar​ be the largest among the numbers aja_jaj​ for 0≤j≤230 \leq j \leq 230≤j≤23. Then the value of r is ______

Correct answer: 6

Step-by-step solution →
Q20·MathematicsIntegerJEE Main 2025
The product of the last two digits of (1919)1919(1919)^{1919}(1919)1919 is ___

Correct answer: 63

Step-by-step solution →
Q21·MathematicsSingle correctJEE Main 2025
The number of integral terms in the expansion of (51/2+71/8)1016\left(5^{1/2}+7^{1/8}\right)^{1016}(51/2+71/8)1016 is:
  1. (A)127127127
  2. (B)130130130
  3. (C)129129129
  4. (D)128128128

Correct answer: (D)

Step-by-step solution →
Q22·MathematicsSingle correctJEE Main 2025
The remainder when ((64)(64))(64)\left((64)^{(64)}\right)^{(64)}((64)(64))(64) is divided by 7 is equal to
  1. (A)4
  2. (B)1
  3. (C)3
  4. (D)6

Correct answer: (B)

Step-by-step solution →
Q23·MathematicsIntegerJEE Main 2025
The sum of the series 2×1×20C4−3×2×20C5+4×3×20C6−5×4×20C7+⋯+18×17×20C202\times 1\times{}^{20}C_4-3\times 2\times{}^{20}C_5+4\times 3\times{}^{20}C_6-5\times 4\times{}^{20}C_7+\dots+18\times 17\times{}^{20}C_{20}2×1×20C4​−3×2×20C5​+4×3×20C6​−5×4×20C7​+⋯+18×17×20C20​, is equal to __________.

Correct answer: 34

Step-by-step solution →
Q24·MathematicsSingle correctJEE Main 2025
If 12(151)+22(152)+32(153)+⋯+152(1515)=2m⋅3n⋅5k1^2\binom{15}{1}+2^2\binom{15}{2}+3^2\binom{15}{3}+\dots+15^2\binom{15}{15}=2^m\cdot3^n\cdot5^k12(115​)+22(215​)+32(315​)+⋯+152(1515​)=2m⋅3n⋅5k, where m,n,k∈Nm,n,k\in\mathbb{N}m,n,k∈N, then m+n+km+n+km+n+k is equal to:
  1. (A)19
  2. (B)21
  3. (C)18
  4. (D)15

Correct answer: (A)

Step-by-step solution →
Q25·MathematicsSingle correctJEE Main 2025
For an integer n≥2n\ge2n≥2, if the arithmetic mean of all coefficients in the binomial expansion of (x+y)2n−3(x+y)^{2n-3}(x+y)2n−3 is 16, then the distance of the point P(2n−1,n2−4n)P(2n-1, n^2-4n)P(2n−1,n2−4n) from the line x+y=8x+y=8x+y=8 is:
  1. (A)2\sqrt{2}2​
  2. (B)222\sqrt{2}22​
  3. (C)525\sqrt{2}52​
  4. (D)323\sqrt{2}32​

Correct answer: (D)

Step-by-step solution →
Q26·MathematicsSingle correctJEE Main 2025
In the expansion of (23+133)n\left(\sqrt[3]{2}+\dfrac{1}{\sqrt[3]{3}}\right)^n(32​+33​1​)n, n∈Nn\in\mathbb{N}n∈N, if the ratio of 15th15^{\text{th}}15th term from the beginning to the 15th15^{\text{th}}15th term from the end is 16\dfrac{1}{6}61​, then the value of nC3{}^nC_3nC3​ is:
  1. (A)4060
  2. (B)1040
  3. (C)2300
  4. (D)4960

Correct answer: (C)

Step-by-step solution →
Q27·MathematicsSingle correctJEE Main 2025
The sum of all rational terms in the expansion of (2+3)8\left(2+\sqrt3\right)^8(2+3​)8 is:
  1. (A)16923
  2. (B)3763
  3. (C)33845
  4. (D)18817

Correct answer: (D)

Step-by-step solution →
Q28·MathematicsSingle correctJEE Main 2025
If ∑r=19r+32r 9Cr=α(32)9−β\displaystyle\sum_{r=1}^{9}\dfrac{r+3}{2^r}\,{}^9C_r=\alpha\left(\dfrac{3}{2}\right)^9-\betar=1∑9​2rr+3​9Cr​=α(23​)9−β, α,β∈N\alpha,\beta\in\mathbb{N}α,β∈N, then (α+β)2(\alpha+\beta)^2(α+β)2 is equal to:
  1. (A)27
  2. (B)9
  3. (C)81
  4. (D)18

Correct answer: (C)

Step-by-step solution →
Q29·MathematicsIntegerJEE Main 2025
Let (1+x+x2)10=a0+a1x+a2x2+…+a20x20(1+x+x^2)^{10}=a_0+a_1x+a_2x^2+\ldots+a_{20}x^{20}(1+x+x2)10=a0​+a1​x+a2​x2+…+a20​x20. If (a1+a3+a5+…+a19)−11a2=121k(a_1+a_3+a_5+\ldots+a_{19})-11a_2=121k(a1​+a3​+a5​+…+a19​)−11a2​=121k, then kkk is equal to __________.

Correct answer: 239

Step-by-step solution →
Q30·MathematicsSingle correctJEE Main 2025
If ∑r=010(10r+1−110r)⋅11Cr+1=α11−11111010\displaystyle\sum_{r=0}^{10}\left(\dfrac{10^{r+1}-1}{10^r}\right)\cdot {}^{11}C_{r+1}=\dfrac{\alpha^{11}-11^{11}}{10^{10}}r=0∑10​(10r10r+1−1​)⋅11Cr+1​=1010α11−1111​, then α\alphaα is equal to:
  1. (A)15
  2. (B)11
  3. (C)24
  4. (D)20

Correct answer: (D)

Step-by-step solution →
Q31·MathematicsSingle correctJEE Main 2025
The term independent of xxx in the expansion of ((x+1)x2/3+1−x1/3−(x+1)x−x1/2)10\left(\dfrac{(x+1)}{x^{2/3}+1-x^{1/3}}-\dfrac{(x+1)}{x-x^{1/2}}\right)^{10}(x2/3+1−x1/3(x+1)​−x−x1/2(x+1)​)10, x>1x>1x>1 is:
  1. (A)210
  2. (B)150
  3. (C)240
  4. (D)120

Correct answer: (A)

Step-by-step solution →
Q32·MathematicsSingle correctJEE Main 2025
The remainder, when 71037^{103}7103 is divided by 23, is equal to:
  1. (A)14
  2. (B)9
  3. (C)17
  4. (D)6

Correct answer: (A)

Step-by-step solution →
Q33·MathematicsSingle correctJEE Main 2025
The least value of n for which the number of integral terms in the Binomial expansion of (73+1112)n\left(\sqrt[3]{7}+\sqrt[12]{11}\right)^n(37​+1211​)n is 183, is:
  1. (A)2184
  2. (B)2148
  3. (C)2172
  4. (D)2196

Correct answer: (A)

Step-by-step solution →
Q34·MathematicsSingle correctJEE Main 2025
Let the coefficients of three consecutive terms TrT_rTr​, Tr+1T_{r+1}Tr+1​ and Tr+2T_{r+2}Tr+2​ in the binomial expansion of (a+b)12(a+b)^{12}(a+b)12 be in a G.P. and let p be the number of all possible values of r. Let q be the sum of all rational terms in the binomial expansion of (34+43)12\left(\sqrt[4]{3}+\sqrt[3]{4}\right)^{12}(43​+34​)12. Then p+q is equal to:
  1. (A)283
  2. (B)295
  3. (C)287
  4. (D)299

Correct answer: (A)

Step-by-step solution →
Q35·MathematicsIntegerJEE Main 2025
If α=1+∑r=16(−3)r−1 12C2r−1\alpha=1+\sum_{r=1}^{6}(-3)^{r-1}\,^{12}C_{2r-1}α=1+∑r=16​(−3)r−112C2r−1​, then the distance of the point (12,3)(12, \sqrt{3})(12,3​) from the line αx−3y+1=0\alpha x-\sqrt{3}y+1=0αx−3​y+1=0 is ____.

Correct answer: 5

Step-by-step solution →
Q36·MathematicsSingle correctJEE Main 2025
For some n≠10n\ne 10n=10, let the coefficients of the 5th5^{th}5th, 6th6^{th}6th and 7th7^{th}7th terms in the binomial expansion of (1+x)n+4(1+x)^{n+4}(1+x)n+4 be in A.P. Then the largest coefficient in the expansion of (1+x)n+4(1+x)^{n+4}(1+x)n+4 is :
  1. (A)707070
  2. (B)353535
  3. (C)202020
  4. (D)101010

Correct answer: (B)

Step-by-step solution →
Q37·MathematicsSingle correctJEE Main 2025
Suppose A and B are the coefficients of 30th30^{\text{th}}30th and 12th12^{\text{th}}12th terms respectively in the binomial expansion of (1+x)2n−1(1+x)^{2n-1}(1+x)2n−1. If 2A=5B2A=5B2A=5B, then nnn is equal to:
  1. (A)22
  2. (B)21
  3. (C)20
  4. (D)19

Correct answer: (B)

Step-by-step solution →
Q38·MathematicsSingle correctJEE Main 2025
If in the expansion of (1+x)p(1−x)q(1+x)^p(1-x)^q(1+x)p(1−x)q, the coefficients of xxx and x2x^2x2 are 111 and −2-2−2, respectively, then p2+q2p^2+q^2p2+q2 is equal to :
  1. (A)8
  2. (B)18
  3. (C)13
  4. (D)20

Correct answer: (C)

Step-by-step solution →
Q39·MathematicsIntegerJEE Main 2025
The sum of all rational terms in the expansion of (1+21/3+31/2)6(1+2^{1/3}+3^{1/2})^6(1+21/3+31/2)6 is equal to _______

Correct answer: 612

Step-by-step solution →
Q40·MathematicsIntegerJEE Main 2025
If ∑r=130r2(30Cr)230Cr−1=α×229\displaystyle\sum_{r=1}^{30}\dfrac{r^2\left(^{30}C_r\right)^2}{^{30}C_{r-1}}=\alpha\times2^{29}r=1∑30​30Cr−1​r2(30Cr​)2​=α×229, then α\alphaα is equal to ______.

Correct answer: 465

Step-by-step solution →
Q41·MathematicsSingle correctJEE Main 2025
Let α,β,γ\alpha,\beta,\gammaα,β,γ and δ\deltaδ be the coefficients of x7,x5,x3x^7,x^5,x^3x7,x5,x3 and xxx respectively in the expansion of (x+x3−1)5+(x−x3−1)5\left(x+\sqrt{x^3-1}\right)^5+\left(x-\sqrt{x^3-1}\right)^5(x+x3−1​)5+(x−x3−1​)5, x>1x>1x>1. If uuu and vvv satisfy the equations αu+βv=18\alpha u+\beta v=18αu+βv=18, γu+δv=20\gamma u+\delta v=20γu+δv=20, then u+vu+vu+v equals:
  1. (A)5
  2. (B)4
  3. (C)3
  4. (D)8

Correct answer: (A)

Step-by-step solution →
Q42·MathematicsNumericalJEE Main 2024
The remainder when 4282024428^{2024}4282024 is divided by 212121 is ________.

Correct answer: 1

Step-by-step solution →
Q43·MathematicsSingle correctJEE Main 2024
The coefficient of x70x^{70}x70 in x2(1+x)98+x3(1+x)97+x4(1+x)96+…+x54(1+x)46x^2(1 + x)^{98} + x^3(1 + x)^{97} + x^4(1 + x)^{96} + \ldots + x^{54}(1 + x)^{46}x2(1+x)98+x3(1+x)97+x4(1+x)96+…+x54(1+x)46 is 99Cp−46Cq{}^{99}C_p - {}^{46}C_q99Cp​−46Cq​. Then a possible value to p+qp + qp+q is:
  1. (A)55
  2. (B)61
  3. (C)68
  4. (D)83

Correct answer: (D)

Step-by-step solution →
Q44·MathematicsSingle correctJEE Main 2024
The sum of the coefficient of x2/3x^{2/3}x2/3 and x−2/5x^{-2/5}x−2/5 in the binomial expansion of (x2/3+12x−2/5)9\left(x^{2/3}+\dfrac{1}{2}x^{-2/5}\right)^{9}(x2/3+21​x−2/5)9 is:
  1. (A)214\dfrac{21}{4}421​
  2. (B)6916\dfrac{69}{16}1669​
  3. (C)6316\dfrac{63}{16}1663​
  4. (D)194\dfrac{19}{4}419​

Correct answer: (A)

Step-by-step solution →
Q45·MathematicsNumericalJEE Main 2024
Let α=∑r=0n(4r2+2r+1) nCr\alpha=\displaystyle\sum_{r=0}^{n}(4r^2+2r+1)\,{}^nC_rα=r=0∑n​(4r2+2r+1)nCr​ and β=∑r=0nnCrr+1\beta=\displaystyle\sum_{r=0}^{n}\dfrac{{}^nC_r}{r+1}β=r=0∑n​r+1nCr​​. If 140<2αβ<281140<\dfrac{2\alpha}{\beta}<281140<β2α​<281, then the value of nnn is ___

Correct answer: 5

Step-by-step solution →
Q46·MathematicsSingle correctJEE Main 2024
If the term independent of xxx in the expansion of (a x2+12x3)10\left(\sqrt{a}\,x^2+\dfrac{1}{2x^3}\right)^{10}(a​x2+2x31​)10 is 105105105, then a2a^2a2 is equal to
  1. (A)444
  2. (B)999
  3. (C)666
  4. (D)222

Correct answer: (A)

Step-by-step solution →
Q47·MathematicsSingle correctJEE Main 2024
Let 0≤r≤n0 \le r \le n0≤r≤n. If n+1Cr+1:nCr:n−1Cr−1=55:35:21^{n+1}C_{r+1} : {}^{n}C_{r} : {}^{n-1}C_{r-1} = 55 : 35 : 21n+1Cr+1​:nCr​:n−1Cr−1​=55:35:21, then 2n+5r2n + 5r2n+5r is equal to:
  1. (A)60
  2. (B)62
  3. (C)50
  4. (D)55

Correct answer: (C)

Step-by-step solution →
Q48·MathematicsNumericalJEE Main 2024
If S(x)=(1+x)+2(1+x)2+3(1+x)3+⋯+60(1+x)60S(x)=(1+x)+2(1+x)^{2}+3(1+x)^{3}+\dots+60(1+x)^{60}S(x)=(1+x)+2(1+x)2+3(1+x)3+⋯+60(1+x)60, x≠0x\neq 0x=0, and (60)2S(60)=a(b)b+b(60)^{2}S(60)=a(b)^{b}+b(60)2S(60)=a(b)b+b, where a,b∈Na,b\in Na,b∈N, then (a+b)(a+b)(a+b) equal to ______.

Correct answer: 3660

Step-by-step solution →
Q49·MathematicsNumericalJEE Main 2024
If the second, third and fourth terms in the expansion of (x+y)n(x+y)^{n}(x+y)n are 135, 30 and 103\dfrac{10}{3}310​, respectively, then 6(n3+x2+y)6(n^{3}+x^{2}+y)6(n3+x2+y) is equal to _______.

Correct answer: 806

Step-by-step solution →
Q50·MathematicsSingle correctJEE Main 2024
If the constant term in the expansion of (35x+2x53)12\left(\dfrac{\sqrt[5]{3}}{x}+\dfrac{2x}{\sqrt[3]{5}}\right)^{12}(x53​​+35​2x​)12, x≠0x\ne 0x=0, is α×28×35\alpha\times 2^8\times\sqrt[5]{3}α×28×53​, then 25α25\alpha25α is equal to:
  1. (A)639
  2. (B)724
  3. (C)693
  4. (D)742

Correct answer: (C)

Step-by-step solution →
Q51·MathematicsNumericalJEE Main 2024
If the constant term in the expansion of (1+2x−3x3)(32x2−13x)9\left(1 + 2x - 3x^3\right)\left(\dfrac{3}{2}x^2 - \dfrac{1}{3x}\right)^9(1+2x−3x3)(23​x2−3x1​)9 is ppp, then 108p108p108p is equal to ___.

Correct answer: 54

Step-by-step solution →
Q52·MathematicsSingle correctJEE Main 2024
If the coefficients of x4x^4x4, x5x^5x5 and x6x^6x6 in the expansion of (1+x)n(1+x)^n(1+x)n are in the arithmetic progression, then the maximum value of nnn is
  1. (A)14
  2. (B)21
  3. (C)28
  4. (D)7

Correct answer: (A)

Step-by-step solution →
Q53·MathematicsSingle correctJEE Main 2024
The sum of all rational terms in the expansion of (21/5+51/3)15\left(2^{1/5}+5^{1/3}\right)^{15}(21/5+51/3)15 is equal to:
  1. (A)313331333133
  2. (B)633633633
  3. (C)931931931
  4. (D)613161316131

Correct answer: (A)

Step-by-step solution →
Q54·MathematicsNumericalJEE Main 2024
If the coefficient of x30x^{30}x30 in the expansion of (1+1x)6(1+x2)7(1−x3)8\left(1+\dfrac1x\right)^6(1+x^2)^7(1-x^3)^8(1+x1​)6(1+x2)7(1−x3)8; x≠0x\ne0x=0 is ∣α∣|\alpha|∣α∣, then ∣α∣|\alpha|∣α∣ equals ___

Correct answer: 678

Step-by-step solution →
Q55·MathematicsSingle correctJEE Main 2024
Let mmm and nnn be the coefficients of seventh and thirteenth terms respectively in the expansion of (13x1/3+12x2/3)18\left(\dfrac{1}{3}x^{1/3}+\dfrac{1}{2x^{2/3}}\right)^{18}(31​x1/3+2x2/31​)18. Then (nm)1/3\left(\dfrac{n}{m}\right)^{1/3}(mn​)1/3 is:
  1. (A)49\dfrac{4}{9}94​
  2. (B)19\dfrac{1}{9}91​
  3. (C)14\dfrac{1}{4}41​
  4. (D)94\dfrac{9}{4}49​

Correct answer: (D)

Step-by-step solution →
Q56·MathematicsNumericalJEE Main 2024
In the expansion of (1+x)(1−x2)(1+3x+3x2+1x3)5(1+x)\left(1-x^2\right)\left(1+\dfrac{3}{x}+\dfrac{3}{x^2}+\dfrac{1}{x^3}\right)^5(1+x)(1−x2)(1+x3​+x23​+x31​)5, x≠0x\ne 0x=0, the sum of the coefficient of x3x^3x3 and x−13x^{-13}x−13 is equal to ______.

Correct answer: 118

Step-by-step solution →
Q57·MathematicsNumericalJEE Main 2024
Let the coefficient of xrx^rxr in the expansion of (x+3)n−1+(x+3)n−2(x+2)+(x+3)n−3(x+2)2+…+(x+2)n−1(x+3)^{n-1}+(x+3)^{n-2}(x+2)+(x+3)^{n-3}(x+2)^2+\ldots+(x+2)^{n-1}(x+3)n−1+(x+3)n−2(x+2)+(x+3)n−3(x+2)2+…+(x+2)n−1 be ara_rar​. If ∑r=0nar=βn−αn\displaystyle\sum_{r=0}^{n}a_r=\beta^n-\alpha^nr=0∑n​ar​=βn−αn, β,α∈N\beta,\alpha\in Nβ,α∈N, then the value of β2+α2\beta^2+\alpha^2β2+α2 equals ______.

Correct answer: 25

Step-by-step solution →
Q58·MathematicsNumericalJEE Main 2024
Let α=∑k=0n(nCk)2k+1\alpha=\displaystyle\sum_{k=0}^{n}\dfrac{\left({}^nC_k\right)^2}{k+1}α=k=0∑n​k+1(nCk​)2​ and β=∑k=0n−1nCk nCk+1k+2\beta=\displaystyle\sum_{k=0}^{n-1}\dfrac{{}^nC_k\,{}^nC_{k+1}}{k+2}β=k=0∑n−1​k+2nCk​nCk+1​​. If 5α=6β5\alpha=6\beta5α=6β, then nnn equals ______.

Correct answer: 10

Step-by-step solution →
Q59·MathematicsNumericalJEE Main 2024
Number of integral terms in the expansion of {7(1/2)+11(1/6)}824\left\{7^{(1/2)}+11^{(1/6)}\right\}^{824}{7(1/2)+11(1/6)}824 is equal to ___

Correct answer: 138

Step-by-step solution →
Q60·MathematicsNumericalJEE Main 2024
If 11C12+11C23+⋯+11C910=nm\dfrac{^{11}C_1}{2}+\dfrac{^{11}C_2}{3}+\cdots+\dfrac{^{11}C_9}{10}=\dfrac{n}{m}211C1​​+311C2​​+⋯+1011C9​​=mn​ with gcd⁡(n,m)=1\gcd(n,m)=1gcd(n,m)=1, then n+mn+mn+m is equal to ______.

Correct answer: 2041

Step-by-step solution →
Q61·MathematicsNumericalJEE Main 2024
Remainder when 64323264^{32^{32}}643232 is divided by 999 is equal to ___.

Correct answer: 1

Step-by-step solution →
Q62·MathematicsNumericalJEE Main 2024
The coefficient of x2012x^{2012}x2012 in the expansion of (1−x)2008(1+x+x2)2007(1-x)^{2008}(1+x+x^2)^{2007}(1−x)2008(1+x+x2)2007 is equal to __________.

Correct answer: 0

Step-by-step solution →
Q63·MathematicsSingle correctJEE Main 2024
If AAA denotes the sum of all the coefficients in the expansion of (1−3x+10x2)n(1-3x+10x^2)^n(1−3x+10x2)n and BBB denotes the sum of all the coefficients in the expansion of (1+x2)n(1+x^2)^n(1+x2)n, then:
  1. (A)A=B3A=B^3A=B3
  2. (B)3A=B3A=B3A=B
  3. (C)B=A3B=A^3B=A3
  4. (D)A=3BA=3BA=3B

Correct answer: (A)

Step-by-step solution →
Q64·MathematicsNumericalJEE Advanced 2023
Let a and b be two nonzero real numbers. If the coefficient of x5x^5x5 in the expansion of (ax2+7027bx)4\left(ax^2 + \frac{70}{27bx}\right)^4(ax2+27bx70​)4 is equal to the coefficient of x−5x^{-5}x−5 in the expansion of (ax−1bx2)7\left(ax - \frac{1}{bx^2}\right)^7(ax−bx21​)7 , then the value of 2b is

Correct answer: 3

Step-by-step solution →
Q65·MathematicsSingle correctJEE Main 2023
Let (a+bx+cx2)10=∑i=020pixi(a+bx+cx^2)^{10}=\sum_{i=0}^{20}p_i x^i(a+bx+cx2)10=∑i=020​pi​xi, a,b,c∈Na,b,c\in\mathbb{N}a,b,c∈N. If p1=20p_1=20p1​=20 and p2=210p_2=210p2​=210, then 2(a+b+c)2(a+b+c)2(a+b+c) is equal to
  1. (A)8
  2. (B)12
  3. (C)15
  4. (D)6

Correct answer: (B)

Step-by-step solution →
Q66·MathematicsNumericalJEE Main 2023
Let α\alphaα be the constant term in the binomial expansion of (x−6x3/2)n, n≤15\left(\sqrt x-\dfrac{6}{x^{3/2}}\right)^{n},\,n\le 15(x​−x3/26​)n,n≤15. If the sum of the coefficients of the remaining terms in the expansion is 649649649 and the coefficient of x−nx^{-n}x−n is λα\lambda\alphaλα, then λ\lambdaλ is equal to _____.

Correct answer: 36

Step-by-step solution →
Q67·MathematicsNumericalJEE Main 2023
The remainder, when 71037^{103}7103 is divided by 17, is

Correct answer: 12

Step-by-step solution →
Q68·MathematicsSingle correctJEE Main 2023
The coefficient of x5x^5x5 in the expansion of (2x3−13x2)5\left(2x^3 - \dfrac{1}{3x^2}\right)^5(2x3−3x21​)5 is
  1. (A)8
  2. (B)9
  3. (C)809\dfrac{80}{9}980​
  4. (D)263\dfrac{26}{3}326​

Correct answer: (C)

Step-by-step solution →
Q69·MathematicsSingle correctJEE Main 2023
Fractional part of the number 4202215\dfrac{4^{2022}}{15}1542022​ is equal to:
  1. (A)415\dfrac{4}{15}154​
  2. (B)115\dfrac{1}{15}151​
  3. (C)1415\dfrac{14}{15}1514​
  4. (D)815\dfrac{8}{15}158​

Correct answer: (B)

Step-by-step solution →
Q70·MathematicsSingle correctJEE Main 2023
If 1n+1 nCn+1n nCn−1+⋯+12 nC1+nC0=102310\dfrac{1}{n+1}\,{}^nC_n+\dfrac{1}{n}\,{}^nC_{n-1}+\cdots+\dfrac{1}{2}\,{}^nC_1+{}^nC_0=\dfrac{1023}{10}n+11​nCn​+n1​nCn−1​+⋯+21​nC1​+nC0​=101023​ then n is equal to
  1. (A)6
  2. (B)9
  3. (C)8
  4. (D)7

Correct answer: (B)

Step-by-step solution →
Q71·MathematicsSingle correctJEE Main 2023
The sum of the coefficients of the first 50 terms in the binomial expansion of (1−x)100(1-x)^{100}(1−x)100, is equal to
  1. (A)−100C50-{}^{100}C_{50}−100C50​
  2. (B)99C49{}^{99}C_{49}99C49​
  3. (C)−99C49-{}^{99}C_{49}−99C49​
  4. (D)100C50{}^{100}C_{50}100C50​

Correct answer: (C)

Step-by-step solution →
Q72·MathematicsSingle correctJEE Main 2023
The sum of the coefficients of three consecutive terms in the binomial expansion of (1+x)n+2(1+x)^{n+2}(1+x)n+2, which are in the ratio 1:3:51:3:51:3:5, is equal to
  1. (A)25
  2. (B)63
  3. (C)41
  4. (D)92

Correct answer: (B)

Step-by-step solution →
Q73·MathematicsNumericalJEE Main 2023
The mean of the coefficients of x,x2,…,x7x, x^2, \ldots, x^7x,x2,…,x7 in the binomial expansion of (2+x)9(2 + x)^9(2+x)9 is _______ .

Correct answer: 2736

Step-by-step solution →
Q74·MathematicsSingle correctJEE Main 2023
If the 1011th1011^{\text{th}}1011th term from the end in the binomial expansion of (4x5−52x)2022\left(\dfrac{4x}{5}-\dfrac{5}{2x}\right)^{2022}(54x​−2x5​)2022 is 102410241024 times the 1011th1011^{\text{th}}1011th term from the beginning, then ∣x∣|x|∣x∣ is equal to
  1. (A)121212
  2. (B)888
  3. (C)101010
  4. (D)151515

Correct answer: (C)

Step-by-step solution →
Q75·MathematicsNumericalJEE Main 2023
The number of integral terms in the expansion of (312+514)680\left(3^{\frac{1}{2}} + 5^{\frac{1}{4}}\right)^{680}(321​+541​)680 is equal to _______ .

Correct answer: 171

Step-by-step solution →
Q76·MathematicsSingle correctJEE Main 2023
If (22)2022+(2022)22(22)^{2022}+(2022)^{22}(22)2022+(2022)22 leaves the remainder α\alphaα when divided by 333 and β\betaβ when divided by 777, then (α2+β2)(\alpha^2+\beta^2)(α2+β2) is equal to:
  1. (A)101010
  2. (B)555
  3. (C)202020
  4. (D)131313

Correct answer: (B)

Step-by-step solution →
Q77·MathematicsSingle correctJEE Main 2023
If the coefficient of x7x^7x7 in (ax−1bx2)13\left(ax-\dfrac{1}{bx^2}\right)^{13}(ax−bx21​)13 and the coefficient of x−5x^{-5}x−5 in (ax+1bx2)13\left(ax+\dfrac{1}{bx^2}\right)^{13}(ax+bx21​)13 are equal, then a4b4a^4 b^4a4b4 is equal to:
  1. (A)444444
  2. (B)222222
  3. (C)111111
  4. (D)333333

Correct answer: (B)

Step-by-step solution →
Q78·MathematicsSingle correctJEE Main 2023
If the coefficients of xxx and x2x^2x2 in (1+x)p(1−x)q(1+x)^p(1-x)^q(1+x)p(1−x)q are 444 and −5-5−5 respectively, then 2p+3q2p+3q2p+3q is equal to:
  1. (A)636363
  2. (B)696969
  3. (C)666666
  4. (D)606060

Correct answer: (A)

Step-by-step solution →
Q79·MathematicsNumericalJEE Main 2023
The coefficient of x7x^7x7 in (1−x+2x3)10\left(1-x+2x^3\right)^{10}(1−x+2x3)10 is _________.

Correct answer: 960

Step-by-step solution →
Q80·MathematicsSingle correctJEE Main 2023
The absolute difference of the coefficients of x10x^{10}x10 and x7x^{7}x7 in the expansion of (2x2+12x)11\left(2x^{2}+\dfrac{1}{2x}\right)^{11}(2x2+2x1​)11 is equal to
  1. (A)123−1212^{3}-12123−12
  2. (B)113−1111^{3}-11113−11
  3. (C)103−1010^{3}-10103−10
  4. (D)133−1313^{3}-13133−13

Correct answer: (A)

Step-by-step solution →
Q81·MathematicsNumericalJEE Main 2023
Let [t][t][t] denotes the greatest integer ≤t\le t≤t. If the constant term in the expansion of (3x2−12x5)7\left(3x^{2}-\frac{1}{2x^{5}}\right)^{7}(3x2−2x51​)7 is α\alphaα, then [α][\alpha][α] is equal to

Correct answer: 1275

Step-by-step solution →
Q82·MathematicsSingle correctJEE Main 2023
25190−19190−8190+219025^{190}-19^{190}-8^{190}+2^{190}25190−19190−8190+2190 is divisible by
  1. (A)34 but not by 14
  2. (B)both 14 and 34
  3. (C)neither 14 nor 34
  4. (D)14 but not by 34

Correct answer: (A)

Step-by-step solution →
Q83·MathematicsSingle correctJEE Main 2023
If the coefficients of the three consecutive terms in the expansion of (1+x)n(1+x)^{n}(1+x)n are in the ratio 1:5:201:5:201:5:20, then the coefficient of the fourth term is
  1. (A)3654
  2. (B)1827
  3. (C)5481
  4. (D)2436

Correct answer: (A)

Step-by-step solution →
Q84·MathematicsNumericalJEE Main 2023
The coefficient of x18x^{18}x18 in the expansion of (x4−1x3)15\left(x^{4}-\dfrac{1}{x^{3}}\right)^{15}(x4−x31​)15 is _____.

Correct answer: 5005

Step-by-step solution →
Q85·MathematicsSingle correctJEE Main 2023
If the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of (24+134)n\left(\sqrt[4]{2}+\dfrac{1}{\sqrt[4]{3}}\right)^{n}(42​+43​1​)n is 6:1\sqrt6:16​:1, then the third term from the beginning is:
  1. (A)60260\sqrt2602​
  2. (B)60360\sqrt3603​
  3. (C)30230\sqrt2302​
  4. (D)30330\sqrt3303​

Correct answer: (B)

Step-by-step solution →
Q86·MathematicsSingle correctJEE Main 2023
Among the statements: (S1):20232022−19992022(S1):2023^{2022}-1999^{2022}(S1):20232022−19992022 is divisible by 8. (S2):13(13)n−11n−13(S2):13(13)^n-11n-13(S2):13(13)n−11n−13 is divisible by 144 for infinitely many n∈Nn\in \mathbb{N}n∈N.
  1. (A)both (S1) and (S2) are incorrect
  2. (B)only (S2) is correct
  3. (C)both (S1) and (S2) are correct
  4. (D)only (S1) is correct

Correct answer: (C)

Step-by-step solution →
Q87·MathematicsSingle correctJEE Main 2023
If the coefficients of x7x^7x7 in (ax2+12bx)11\left(ax^2+\frac{1}{2bx}\right)^{11}(ax2+2bx1​)11 and x−7x^{-7}x−7 in (ax−13bx2)11\left(ax-\frac{1}{3bx^2}\right)^{11}(ax−3bx21​)11 are equal, then
  1. (A)64ab=24364ab=24364ab=243
  2. (B)729ab=32729ab=32729ab=32
  3. (C)243ab=64243ab=64243ab=64
  4. (D)32ab=72932ab=72932ab=729

Correct answer: (B)

Step-by-step solution →
Q88·MathematicsSingle correctJEE Main 2023
If 2nC3:nC3=10:1^{2n}C_3:^nC_3=10:12nC3​:nC3​=10:1, then the ratio (n2+3n):(n2−3n+4)(n^2+3n):(n^2-3n+4)(n2+3n):(n2−3n+4) is:
  1. (A)35:1635:1635:16
  2. (B)65:3765:3765:37
  3. (C)27:1127:1127:11
  4. (D)2:12:12:1

Correct answer: (D)

Step-by-step solution →
Q89·MathematicsNumericalJEE Main 2023
If the term without x in the expansion of (x23+αx3)22\left(x^{\frac23}+\dfrac{\alpha}{x^3}\right)^{22}(x32​+x3α​)22 is 7315, then ∣α∣|\alpha|∣α∣ is equal to

Correct answer: 1

Step-by-step solution →
Q90·MathematicsSingle correctJEE Main 2023
The value of 11! 50!+13! 48!+15! 46!+⋯+149! 2!+151! 1!\frac{1}{1!\,50!}+\frac{1}{3!\,48!}+\frac{1}{5!\,46!}+\cdots+\frac{1}{49!\,2!}+\frac{1}{51!\,1!}1!50!1​+3!48!1​+5!46!1​+⋯+49!2!1​+51!1!1​ is
  1. (A)25051!\frac{2^{50}}{51!}51!250​
  2. (B)25150!\frac{2^{51}}{50!}50!251​
  3. (C)25050!\frac{2^{50}}{50!}50!250​
  4. (D)25151!\frac{2^{51}}{51!}51!251​

Correct answer: (A)

Step-by-step solution →
Q91·MathematicsNumericalJEE Main 2023
Let the sixth term in the binomial expansion of (2log⁡2(10−3x)+2(x−2)log⁡235)m\left(\sqrt{2^{\log_2(10-3^x)}}+\sqrt[5]{2^{(x-2)\log_2 3}}\right)^m(2log2​(10−3x)​+52(x−2)log2​3​)m, in the increasing powers of 2(x−2)log⁡232^{(x-2)\log_2 3}2(x−2)log2​3, be 21. If the binomial coefficients of the second, third and fourth terms in the expansion are respectively the first, third and fifth terms of an A.P., then the sum of the squares of all possible values of x is

Correct answer: 4

Step-by-step solution →
Q92·MathematicsNumericalJEE Main 2023
The remainder, when 19200+2320019^{200}+23^{200}19200+23200 is divided by 494949, is _______ .

Correct answer: 29

Step-by-step solution →
Q93·MathematicsNumericalJEE Main 2023
Let α>0\alpha > 0α>0, be the smallest number such that the expansion of (x23+2x3)30\left(x^{\frac{2}{3}} + \frac{2}{x^3}\right)^{30}(x32​+x32​)30 has a term βx−α, β∈N\beta x^{-\alpha},\ \beta \in \mathbb{N}βx−α, β∈N. Then α\alphaα is equal to

Correct answer: 2

Step-by-step solution →
Q94·MathematicsNumericalJEE Main 2023
If the constant term in the binomial expansion of (x5/22−4xl)9\left(\dfrac{x^{5/2}}{2}-\dfrac{4}{x^{l}}\right)^{9}(2x5/2​−xl4​)9 is −84-84−84 and the coefficient of x−3lx^{-3l}x−3l is 2αβ2^{\alpha}\beta2αβ, where β<0\beta<0β<0 is an odd number, then ∣αl−β∣|\alpha l-\beta|∣αl−β∣ is equal to

Correct answer: 98

Step-by-step solution →
Q95·MathematicsNumericalJEE Main 2023
The coefficient of x−6x^{-6}x−6, in the expansion of (4x5+52x2)9\left(\dfrac{4x}{5}+\dfrac{5}{2x^2}\right)^{9}(54x​+2x25​)9, is

Correct answer: 5040

Step-by-step solution →
Q96·MathematicsNumericalJEE Main 2023
The remainder on dividing 5995^{99}599 by 11 is :

Correct answer: 9

Step-by-step solution →
Q97·MathematicsNumericalJEE Main 2023
The 50th50^{th}50th root of a number xxx is 121212 and the 50th50^{th}50th root of another number yyy is 181818. Then the remainder obtained on dividing (x+y)(x + y)(x+y) by 252525 is _______.

Correct answer: 23

Step-by-step solution →
Q98·MathematicsSingle correctJEE Main 2023
If the coefficient of x15x^{15}x15 in the expansion of (ax3+1bx1/3)15\left(ax^3+\dfrac{1}{bx^{1/3}}\right)^{15}(ax3+bx1/31​)15 is equal to the coefficient of x−15x^{-15}x−15 in the expansion of (ax1/3−1bx3)15\left(ax^{1/3}-\dfrac{1}{bx^3}\right)^{15}(ax1/3−bx31​)15, where aaa and bbb are positive real numbers, then for each such ordered pair (a,b)(a,b)(a,b):
  1. (A)a=3a=3a=3
  2. (B)ab=1ab=1ab=1
  3. (C)a=ba=ba=b
  4. (D)a=3ba=3ba=3b

Correct answer: (B)

Step-by-step solution →
Q99·MathematicsSingle correctJEE Main 2023
x=(83+13)13x=(8\sqrt{3} + 13)^{13}x=(83​+13)13 and y=(72+9)9y=(7\sqrt{2} + 9)^{9}y=(72​+9)9. If [t][t][t] denotes the greatest integer ≤t\le t≤t, then:
  1. (A)[x][x][x] is odd but [y][y][y] is even
  2. (B)[x]+[y][x] + [y][x]+[y] is even
  3. (C)[x][x][x] and [y][y][y] are both odd
  4. (D)[x][x][x] is even but [y][y][y] is odd

Correct answer: (B)

Step-by-step solution →
Q100·MathematicsSingle correctJEE Main 2023
The coefficient of x301x^{301}x301 in (1+x)500+x(1+x)499+x2(1+x)498+⋯+x500(1+x)^{500}+x(1+x)^{499}+x^2(1+x)^{498}+\cdots+x^{500}(1+x)500+x(1+x)499+x2(1+x)498+⋯+x500 is:
  1. (A)500C300^{500}C_{300}500C300​
  2. (B)501C200^{501}C_{200}501C200​
  3. (C)501C302^{501}C_{302}501C302​
  4. (D)500C301^{500}C_{301}500C301​

Correct answer: (B)

Step-by-step solution →
Q101·MathematicsSingle correctJEE Main 2023
Let KKK be the sum of the coefficients of the odd powers of xxx in the expansion of (1+x)99(1+x)^{99}(1+x)99. Let aaa be the middle term in the expansion of (2+12)200\left(2+\dfrac{1}{\sqrt2}\right)^{200}(2+2​1​)200. If 200C99⋅Ka=2ℓ⋅mn\dfrac{{}^{200}C_{99}\cdot K}{a}=\dfrac{2^{\ell}\cdot m}{n}a200C99​⋅K​=n2ℓ⋅m​, where mmm and nnn are odd numbers, then the ordered pair (ℓ,n)(\ell,n)(ℓ,n) is equal to:
  1. (A)(50,51)(50,51)(50,51)
  2. (B)(50,101)(50,101)(50,101)
  3. (C)(51,99)(51,99)(51,99)
  4. (D)(51,101)(51,101)(51,101)

Correct answer: (B)

Step-by-step solution →
Q102·MathematicsNumericalJEE Main 2023
Let the coefficients of three consecutive terms in the binomial expansion of (1+2x)n(1 + 2x)^n(1+2x)n be in the ratio 2:5:82 : 5 : 82:5:8. Then the coefficient of the term, which is in the middle of these three terms, is ________ .

Correct answer: 1120

Step-by-step solution →
Q103·MathematicsNumericalJEE Main 2023
If the co-efficient of x9x^9x9 in (αx3+1βx)11\left(\alpha x^3 + \dfrac{1}{\beta x}\right)^{11}(αx3+βx1​)11 and the co-efficient of x−9x^{-9}x−9 in (αx−1βx3)11\left(\alpha x - \dfrac{1}{\beta x^3}\right)^{11}(αx−βx31​)11 are equal, then (αβ)2(\alpha\beta)^2(αβ)2 is equal to ________ .

Correct answer: 1

Step-by-step solution →
Q104·MathematicsSingle correctJEE Main 2023
If ara_rar​ is the coefficient of x10−rx^{10-r}x10−r in the Binomial expansion of (1+x)10(1+x)^{10}(1+x)10, then ∑r=110r3(arar−1)2\displaystyle\sum_{r=1}^{10}r^3\left(\dfrac{a_r}{a_{r-1}}\right)^2r=1∑10​r3(ar−1​ar​​)2 is equal to:
  1. (A)544554455445
  2. (B)302530253025
  3. (C)489548954895
  4. (D)121012101210

Correct answer: (D)

Step-by-step solution →
Q105·MathematicsSingle correctJEE Main 2023
∑k=0651−kC3\displaystyle\sum_{k=0}^{6} {}^{51-k}C_3k=0∑6​51−kC3​ is equal to
  1. (A)51C4−45C4{}^{51}C_4 - {}^{45}C_451C4​−45C4​
  2. (B)52C3−45C3{}^{52}C_3 - {}^{45}C_352C3​−45C3​
  3. (C)52C4−45C4{}^{52}C_4 - {}^{45}C_452C4​−45C4​
  4. (D)51C3−45C3{}^{51}C_3 - {}^{45}C_351C3​−45C3​

Correct answer: (C)

Step-by-step solution →
Q106·MathematicsNumericalJEE Main 2023
The remainder when (2023)2023(2023)^{2023}(2023)2023 is divided by 353535 is

Correct answer: 7

Step-by-step solution →
Q107·MathematicsNumericalJEE Main 2023
The constant term in the expansion of (2x+1x7+3x2)5\left(2x+\dfrac{1}{x^7}+3x^2\right)^5(2x+x71​+3x2)5 is _______.

Correct answer: 1080

Step-by-step solution →
Q108·MathematicsSingle correctJEE Main 2023
If (30C1)2+2(30C2)2+3(30C3)2+⋯+30(30C30)2=α 60!(30!)2\left(^{30}C_1\right)^2+2\left(^{30}C_2\right)^2+3\left(^{30}C_3\right)^2+\cdots+30\left(^{30}C_{30}\right)^2=\dfrac{\alpha\,60!}{(30!)^2}(30C1​)2+2(30C2​)2+3(30C3​)2+⋯+30(30C30​)2=(30!)2α60!​ then α\alphaα is equal to:
  1. (A)303030
  2. (B)101010
  3. (C)606060
  4. (D)151515

Correct answer: (D)

Step-by-step solution →
Q109·MathematicsNumericalJEE Main 2023
Let the sum of the coefficients of the first three terms in the expansion of (x−3x2)n, x≠0, n∈N\left(x-\dfrac{3}{x^2}\right)^n,\ x\neq0,\ n\in\mathbb{N}(x−x23​)n, x=0, n∈N, be 376376376. Then the coefficient of x4x^4x4 is

Correct answer: 405

Step-by-step solution →
Q110·MathematicsNumericalJEE Main 2023
Suppose ∑r=02023r2 2023Cr=2023×α×22022\sum\limits_{r=0}^{2023} r^2\,{}^{2023}C_r=2023\times\alpha\times2^{2022}r=0∑2023​r22023Cr​=2023×α×22022. Then the value of α\alphaα is _______ .

Correct answer: 1012

Step-by-step solution →
Q111·MathematicsSingle correctJEE Main 2023
The value of ∑r=02222Cr 23Cr\sum\limits_{r=0}^{22}{}^{22}C_r\,{}^{23}C_rr=0∑22​22Cr​23Cr​ is
  1. (A)44C23{}^{44}C_{23}44C23​
  2. (B)45C23{}^{45}C_{23}45C23​
  3. (C)44C22{}^{44}C_{22}44C22​
  4. (D)45C24{}^{45}C_{24}45C24​

Correct answer: (B)

Step-by-step solution →
Q112·MathematicsNumericalJEE Main 2022
Let the ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of (24+134)n\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^{n}(42​+43​1​)n, in the increasing powers of 134\frac{1}{\sqrt[4]{3}}43​1​ be 64:1\sqrt[4]{6} : 146​:1. If the sixth term from the beginning is α34\frac{\alpha}{\sqrt[4]{3}}43​α​, then α\alphaα is equal to __________.

Correct answer: 84

Step-by-step solution →
Q113·MathematicsNumericalJEE Main 2022
If ∑k=110K2(10CK)2=22000L\sum_{k=1}^{10} K^2 \left({}^{10}C_K\right)^2 = 22000L∑k=110​K2(10CK​)2=22000L, then L is equal to ___.

Correct answer: 221

Step-by-step solution →
Q114·MathematicsNumericalJEE Main 2022
If 1+(2+49C1+49C2+…+49C49)(50C2+50C4+…+50C50)1+ \left(2 + {}^{49}C_{1} + {}^{49}C_{2} + \ldots + {}^{49}C_{49}\right)\left({}^{50}C_{2} + {}^{50}C_{4} + \ldots + {}^{50}C_{50}\right)1+(2+49C1​+49C2​+…+49C49​)(50C2​+50C4​+…+50C50​) is equal to 2n.m2^{n}.m2n.m, where m is odd, then n+mn + mn+m is equal to _____

Correct answer: 99

Step-by-step solution →
Q115·MathematicsNumericalJEE Main 2022
Let the coefficients of the middle terms in the expansion of (16+βx)4\left(\frac{1}{\sqrt{6}}+\beta x\right)^{4}(6​1​+βx)4, (1−3βx)2\left(1-3\beta x\right)^{2}(1−3βx)2 and (1−β2x)6\left(1-\frac{\beta}{2}x\right)^{6}(1−2β​x)6, β>0\beta>0β>0, respectively form the first three terms of an A.P. If d is the common difference of this A.P., then 50−2dβ250-\frac{2d}{\beta^{2}}50−β22d​ is equal to _____

Correct answer: 57

Step-by-step solution →
Q116·MathematicsSingle correctJEE Main 2022
The remainder when 72022+320227^{2022} + 3^{2022}72022+32022 is divided by 5 is:
  1. (A)0
  2. (B)2
  3. (C)3
  4. (D)4

Correct answer: (C)

Step-by-step solution →
Q117·MathematicsNumericalJEE Main 2022
Let for the 9th9^{\text{th}}9th term in the binomial expansion of (3+6x)n(3 + 6x)^{n}(3+6x)n, in the increasing powers of 6x6x6x, to be the greatest for x=32x = \frac{3}{2}x=23​, the least value of n is n0n_0n0​. If k is the ratio of the coefficient of x6x^{6}x6 to the coefficient of x3x^{3}x3, then k+n0k + n_0k+n0​ is equal to:

Correct answer: 24

Step-by-step solution →
Q118·MathematicsSingle correctJEE Main 2022
The remainder when (2021)2022+(2022)2021(2021)^{2022} + (2022)^{2021}(2021)2022+(2022)2021 is divided by 7 is
  1. (A)0
  2. (B)1
  3. (C)2
  4. (D)6

Correct answer: (A)

Step-by-step solution →
Q119·MathematicsNumericalJEE Main 2022
If the coefficients of x and x2^{2}2 in the expansion of (1 + x)p^{p}p(1 − x)q^{q}q, p, q ≤ 15, are −3 and−5 respectively, then the coefficient of x3^{3}3 is equal to______________.

Correct answer: 23

Step-by-step solution →
Q120·MathematicsSingle correctJEE Main 2022
∑i,j=0i≠jnnCi  nCj\sum\limits_{\substack{i,j=0 \\ i\neq j}}^{n}{}^{n}C_{i}\;{}^{n}C_{j}i,j=0i=j​∑n​nCi​nCj​ is equal to
  1. (A)22n−2nCn2^{2n}-{}^{2n}C_{n}22n−2nCn​
  2. (B)22n−1−2n−1Cn−12^{2n-1}-{}^{2n-1}C_{n-1}22n−1−2n−1Cn−1​
  3. (C)22n−12 2nCn2^{2n}-\frac{1}{2}\,{}^{2n}C_{n}22n−21​2nCn​
  4. (D)2n−1+2n−1Cn2^{n-1}+{}^{2n-1}C_{n}2n−1+2n−1Cn​

Correct answer: (B)

Step-by-step solution →
Q121·MathematicsNumericalJEE Main 2022
If the maximum value of the term independent of t in the expansion of (t2x15+(1−x)110t)15\left( t^{2}x^{\frac{1}{5}} + \frac{(1-x)^{\frac{1}{10}}}{t} \right)^{15}(t2x51​+t(1−x)101​​)15, x≥0x \ge 0x≥0, is K, then 8K is equal to ________.

Correct answer: 6006

Step-by-step solution →
Q122·MathematicsNumericalJEE Main 2022
Let the coefficients of x−1x^{-1}x−1 and x−3x^{-3}x−3 in the expansion of (2x15−1x15)15\left(2x^{\frac{1}{5}} - \frac{1}{x^{\frac{1}{5}}}\right)^{15}(2x51​−x51​1​)15, x>0x > 0x>0, be mmm and nnn respectively. If r is a positive integer such mn2=15Cr.2rmn^2 = {}^{15}C_r . 2^rmn2=15Cr​.2r, then the value of r is equal to____.

Correct answer: 5

Step-by-step solution →
Q123·MathematicsSingle correctJEE Main 2022
If the constant term in the expansion of (3x3−2x2+5x5)10\left(3x^3 - 2x^2 + \frac{5}{x^5}\right)^{10}(3x3−2x2+x55​)10 is 2k⋅l2^k \cdot l2k⋅l, where lll is an odd integer, then the value of k is equal to :
  1. (A)6
  2. (B)7
  3. (C)8
  4. (D)9

Correct answer: (D)

Step-by-step solution →
Q124·MathematicsSingle correctJEE Main 2022
Let n≥5n \geq 5n≥5 be an integer. If 9n−8n−1=64α9^n - 8n - 1 = 64\alpha9n−8n−1=64α and 6n−5n−1=25β6^n - 5n - 1 = 25\beta6n−5n−1=25β, then α−β\alpha - \betaα−β is equal to:
  1. (A)1+nC2(8−5)+nC3(82−52)+...+nCn(8n−1−5n−1)1 + {}^nC_2 (8-5) + {}^nC_3 (8^2 - 5^2) + ... + {}^nC_n (8^{n-1} - 5^{n-1})1+nC2​(8−5)+nC3​(82−52)+...+nCn​(8n−1−5n−1)
  2. (B)1+nC3(8−5)+nC4(82−52)+...+nCn(8n−2−5n−2)1 + {}^nC_3 (8-5) + {}^nC_4 (8^2 - 5^2) + ... + {}^nC_n (8^{n-2} - 5^{n-2})1+nC3​(8−5)+nC4​(82−52)+...+nCn​(8n−2−5n−2)
  3. (C)nC3(8−5)+nC4(82−52)+...+nCn(8n−2−5n−2){}^nC_3 (8-5) + {}^nC_4 (8^2 - 5^2) + ... + {}^nC_n (8^{n-2} - 5^{n-2})nC3​(8−5)+nC4​(82−52)+...+nCn​(8n−2−5n−2)
  4. (D)nC4(8−5)+nC5(82−52)+...+nCn(8n−3−5n−3){}^nC_4 (8-5) + {}^nC_5 (8^2 - 5^2) + ... + {}^nC_n (8^{n-3} - 5^{n-3})nC4​(8−5)+nC5​(82−52)+...+nCn​(8n−3−5n−3)

Correct answer: (C)

Step-by-step solution →
Q125·MathematicsSingle correctJEE Main 2022
If ∑k=131(31Ck)(31Ck−1)−∑k=130(30Ck)(30Ck−1)=α (60!)(30!)(31!),\sum_{k=1}^{31}\left({}^{31}\mathrm{C}_{k}\right)\left({}^{31}\mathrm{C}_{k-1}\right)-\sum_{k=1}^{30}\left({}^{30}\mathrm{C}_{k}\right)\left({}^{30}\mathrm{C}_{k-1}\right)=\frac{\alpha\,(60!)}{(30!)(31!)},∑k=131​(31Ck​)(31Ck−1​)−∑k=130​(30Ck​)(30Ck−1​)=(30!)(31!)α(60!)​, Where α∈R,\alpha \in \mathrm{R},α∈R, then the value of 16α16\alpha16α is equal to
  1. (A)1411
  2. (B)1320
  3. (C)1615
  4. (D)1855

Correct answer: (A)

Step-by-step solution →
Q126·MathematicsNumericalJEE Main 2022
The number of positive integers k such that the constant term in the binomial expansion of (2x3+3xk)12\left(2x^3 + \frac{3}{x^k}\right)^{12}(2x3+xk3​)12, x≠0x \neq 0x=0 is 28⋅ℓ,2^8 \cdot \ell,28⋅ℓ, where ℓ\ellℓ is an odd integer, is ______.

Correct answer: 2

Step-by-step solution →
Q127·MathematicsSingle correctJEE Main 2022
The term independent of x in the expression of (1−x2+3x3)(52x3−15x2)11\left(1 - x^{2} + 3x^{3}\right)\left(\frac{5}{2}x^{3} - \frac{1}{5x^{2}}\right)^{11}(1−x2+3x3)(25​x3−5x21​)11, x≠0x \neq 0x=0 is
  1. (A)740\frac{7}{40}407​
  2. (B)33200\frac{33}{200}20033​
  3. (C)39200\frac{39}{200}20039​
  4. (D)1150\frac{11}{50}5011​

Correct answer: (B)

Step-by-step solution →
Q128·MathematicsNumericalJEE Main 2022
If the sum of the coefficients of all the positive powers of x, in the binomial expansion of (xn+2x5)7\left(x^{n}+\dfrac{2}{x^{5}}\right)^{7}(xn+x52​)7 is 939, then the sum of all the possible integral values of n is :

Correct answer: 57

Step-by-step solution →
Q129·MathematicsNumericalJEE Main 2022
If the coefficient of x10x^{10}x10 in the binomial expansion of (x514+5x13)60\left(\frac{\sqrt{x}}{5^{\frac{1}{4}}}+\frac{\sqrt{5}}{x^{\frac{1}{3}}}\right)^{60}(541​x​​+x31​5​​)60 is 5kl5^{k}l5kl, where lll, k ∈\in∈ N and lll is co-prime to 5, then k is equal to _____.

Correct answer: 5

Step-by-step solution →
Q130·MathematicsNumericalJEE Main 2022
If (40C0)+(41C1)+(42C2)+...+(60C20)=mn 60C20\left(^{40}C_{0}\right)+\left(^{41}C_{1}\right)+\left(^{42}C_{2}\right)+...+\left(^{60}C_{20}\right)=\frac{m}{n}\,^{60}C_{20}(40C0​)+(41C1​)+(42C2​)+...+(60C20​)=nm​60C20​, m and n are coprime, then m+nm+nm+n is equal to _____ .

Correct answer: 102

Step-by-step solution →
Q131·MathematicsSingle correctJEE Main 2022
The remainder when (2021)2023(2021)^{2023}(2021)2023 is divided by 7 is :
  1. (A)1
  2. (B)2
  3. (C)5
  4. (D)6

Correct answer: (C)

Step-by-step solution →
Q132·MathematicsNumericalJEE Main 2022
Let CrC_rCr​ denote the binomial coefficient of xrx^rxr in the expansion of (1+x)10(1 + x)^{10}(1+x)10. If α, β ∈ R. C1+3⋅2C2+5⋅3C3+…C_1 + 3\cdot 2C_2 + 5\cdot 3C_3 + \ldotsC1​+3⋅2C2​+5⋅3C3​+… upto 10 terms =α×2112β−1(C0+C12+C23+… upto 10 terms)= \frac{\alpha \times 2^{11}}{2^{\beta} - 1}\left(C_0 + \frac{C_1}{2} + \frac{C_2}{3} + \ldots\ \text{upto 10 terms}\right)=2β−1α×211​(C0​+2C1​​+3C2​​+… upto 10 terms) then the value of α + β is equal to

Correct answer: 286

Step-by-step solution →
Q133·MathematicsSingle correctJEE Main 2022
The coefficient of x101x^{101}x101 in the expression (5+x)500+x(5+x)499+x2(5+x)498+.....x500(5+x)^{500} + x(5+x)^{499} + x^2(5+x)^{498} + .....x^{500}(5+x)500+x(5+x)499+x2(5+x)498+.....x500, x>0x > 0x>0, is
  1. (A)501C101(5)399^{501}C_{101}(5)^{399}501C101​(5)399
  2. (B)501C101(5)400^{501}C_{101}(5)^{400}501C101​(5)400
  3. (C)501C100(5)400^{501}C_{100}(5)^{400}501C100​(5)400
  4. (D)500C101(5)399^{500}C_{101}(5)^{399}500C101​(5)399

Correct answer: (A)

Step-by-step solution →
Q134·MathematicsNumericalJEE Main 2022
If the sum of the coefficients of all the positive even powers of x in the binomial expansion of (2x3+3x)10\left(2x^{3} + \frac{3}{x}\right)^{10}(2x3+x3​)10 is 510−β⋅395^{10} - \beta \cdot 3^{9}510−β⋅39, then β\betaβ is equal to ______

Correct answer: 83

Step-by-step solution →
Q135·MathematicsSingle correctJEE Main 2022
The remainder when 320223^{2022}32022 is divided by 5 is
  1. (A)1
  2. (B)2
  3. (C)3
  4. (D)4

Correct answer: (D)

Step-by-step solution →
Q136·MathematicsNumericalJEE Main 2022
The remainder on dividing 1+3+32+33+...+320211 + 3 + 3^{2} + 3^{3} + ... + 3^{2021}1+3+32+33+...+32021 by 50 is ______.

Correct answer: 4

Step-by-step solution →
Q137·MathematicsNumericalJEE Main 2021
If the sum of the coefficients in the expansion of (x+y)n(x + y)^{n}(x+y)n is 4096, then the greatest coefficient in the expansion is ________ .

Correct answer: 924

Step-by-step solution →
Q138·MathematicsNumericalJEE Main 2021
If the coefficient of a7b8a^{7}b^{8}a7b8 in the expansion of (a+2b+4ab)10(a + 2b + 4ab)^{10}(a+2b+4ab)10 is K.216K.2^{16}K.216, then K is equal to ______.

Correct answer: 315

Step-by-step solution →
Q139·MathematicsNumericalJEE Main 2021
If (3644)k\left( \frac{3^{6}}{4^{4}} \right) k(4436​)k is the term, independent of x, in the binomial expansion of (x4−12x2)12\left( \frac{x}{4} - \frac{12}{x^{2}} \right)^{12}(4x​−x212​)12, then k is equal to ______.

Correct answer: 55

Step-by-step solution →
Q140·MathematicsSingle correctJEE Main 2021
∑k=020(20Ck)2\sum_{k=0}^{20} \left({}^{20}C_k\right)^2∑k=020​(20Ck​)2 is equal to :
  1. (A)40C21{}^{40}C_{21}40C21​
  2. (B)40C19{}^{40}C_{19}40C19​
  3. (C)40C20{}^{40}C_{20}40C20​
  4. (D)41C20{}^{41}C_{20}41C20​

Correct answer: (C)

Step-by-step solution →
Q141·MathematicsNumericalJEE Main 2021
3×722+2×1022−443 \times 7^{22} + 2 \times 10^{22} - 443×722+2×1022−44 when divided by 18 leaves the remainder __________ .

Correct answer: 15

Step-by-step solution →
Q142·MathematicsSingle correctJEE Main 2021
If 20Cr^{20}C_r20Cr​ is the co-efficient of xrx^rxr in the expansion of (1+x)20(1 + x)^{20}(1+x)20, then the value of ∑r=020r2 20Cr\sum_{r=0}^{20} r^2\, ^{20}C_r∑r=020​r220Cr​ is equal to :
  1. (A)420×219420 \times 2^{19}420×219
  2. (B)380×219380 \times 2^{19}380×219
  3. (C)380×218380 \times 2^{18}380×218
  4. (D)420×218420 \times 2^{18}420×218

Correct answer: (D)

Step-by-step solution →
Q143·MathematicsNumericalJEE Main 2021
Let (nk)\begin{pmatrix} n \\ k \end{pmatrix}(nk​) denotes nCk^{n}C_{k}nCk​ and [nk]={(nk),if 0≤k≤n0,otherwise\begin{bmatrix} n \\ k \end{bmatrix} = \begin{cases} \begin{pmatrix} n \\ k \end{pmatrix}, & \text{if } 0 \le k \le n \\ 0, & \text{otherwise} \end{cases}[nk​]=⎩⎨⎧​(nk​),0,​if 0≤k≤notherwise​ If Ak=∑i=09(9i)[1212−k+i]+∑i=08(8i)[1313−k+i]A_{k} = \sum_{i=0}^{9} \begin{pmatrix} 9 \\ i \end{pmatrix} \begin{bmatrix} 12 \\ 12-k+i \end{bmatrix} + \sum_{i=0}^{8} \begin{pmatrix} 8 \\ i \end{pmatrix} \begin{bmatrix} 13 \\ 13-k+i \end{bmatrix}Ak​=∑i=09​(9i​)[1212−k+i​]+∑i=08​(8i​)[1313−k+i​] and A4−A3=190 pA_{4} - A_{3} = 190\,pA4​−A3​=190p, then p is equal to :

Correct answer: 49

Step-by-step solution →
Q144·MathematicsSingle correctJEE Main 2021
If the coefficients of x7x^7x7 in (x2+1bx)11\left(x^2 + \frac{1}{bx}\right)^{11}(x2+bx1​)11 and x−7x^{-7}x−7 in (x−1bx2)11\left(x - \frac{1}{bx^2}\right)^{11}(x−bx21​)11, b≠0b \ne 0b=0 are equal, then the value of b is equal to :
  1. (A)−1-1−1
  2. (B)1
  3. (C)2
  4. (D)−2-2−2

Correct answer: (B)

Step-by-step solution →
Q145·MathematicsSingle correctJEE Main 2021
A possible value of 'x', for which the ninth term in the expansion of {3log⁡325x−1+7+3(−18)log⁡3(5x−1+1)}10\left\{ 3^{\log_3 \sqrt{25^{x-1} + 7}} + 3^{\left( -\frac{1}{8} \right) \log_3 \left( 5^{x-1} + 1 \right)} \right\}^{10}{3log3​25x−1+7​+3(−81​)log3​(5x−1+1)}10 in the increasing powers of 3(−18)log⁡3(5x−1+1)3^{\left( -\frac{1}{8} \right) \log_3 \left( 5^{x-1} + 1 \right)}3(−81​)log3​(5x−1+1) is equal to 180, is :
  1. (A)1
  2. (B)2
  3. (C)0
  4. (D)−1

Correct answer: (A)

Step-by-step solution →
Q146·MathematicsNumericalJEE Main 2021
Let n∈Nn \in Nn∈N and [x] denote the greatest integer less than or equal to x. If the sum of (n+1) terms nC0,3.nC1,5.nC2,7.nC3,.....^{n}C_{0}, 3.^{n}C_{1}, 5.^{n}C_{2}, 7.^{n}C_{3},.....nC0​,3.nC1​,5.nC2​,7.nC3​,..... is equal to 2100.1012^{100}.1012100.101, then 2[n−12]2\left[\dfrac{n-1}{2}\right]2[2n−1​] is equal to

Correct answer: 98

Step-by-step solution →
Q147·MathematicsSingle correctJEE Main 2021
If the greatest value of the term independent of 'x' in the expansion of (xsin⁡α+acos⁡αx)10\left(x\sin\alpha + a\frac{\cos\alpha}{x}\right)^{10}(xsinα+axcosα​)10 is 10!(5!)2\frac{10!}{\left(5!\right)^2}(5!)210!​, then the value of 'a' is equal to :
  1. (A)-1
  2. (B)2
  3. (C)-2
  4. (D)1

Correct answer: (B)

Step-by-step solution →
Q148·MathematicsNumericalJEE Main 2021
The ratio of the coefficient of the middle term in the expansion of (1+x)20\left( 1+x \right)^{20}(1+x)20 and the sum of the coefficients of two middle terms in expansion of (1+x)19\left( 1+x \right)^{19}(1+x)19 is.......

Correct answer: 1

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Q149·MathematicsNumericalJEE Main 2021
If the co-efficients of x7x^{7}x7 and x8x^{8}x8 in the expansion of (2+x3)n\left(2 + \dfrac{x}{3}\right)^{n}(2+3x​)n are equal, then the value of n is equal to .........

Correct answer: 55

Step-by-step solution →
Q150·MathematicsSingle correctJEE Main 2021
If b is very small as compared to the value of a, so that the cube and other higher powers of ba\frac{b}{a}ab​ can be neglected in the identity 1a−b+1a−2b+1a−3b+....+1a−nb=αn+βn2+γn3\frac{1}{a-b}+\frac{1}{a-2b}+\frac{1}{a-3b}+....+\frac{1}{a-nb}=\alpha n+\beta n^{2}+\gamma n^{3}a−b1​+a−2b1​+a−3b1​+....+a−nb1​=αn+βn2+γn3,
  1. (A)a+b3a2\frac{a+b}{3a^{2}}3a2a+b​
  2. (B)a2+b3a3\frac{a^{2}+b}{3a^{3}}3a3a2+b​
  3. (C)b23a3\frac{b^{2}}{3a^{3}}3a3b2​
  4. (D)a+b23a3\frac{a+b^{2}}{3a^{3}}3a3a+b2​

Correct answer: (C)

Step-by-step solution →
Q151·MathematicsSingle correctJEE Main 2021
The sum of all those terms which are rational numbers in the expansion of (213+314)12\left(2^{\frac{1}{3}} + 3^{\frac{1}{4}}\right)^{12}(231​+341​)12 is :
  1. (A)89
  2. (B)35
  3. (C)27
  4. (D)43

Correct answer: (D)

Step-by-step solution →
Q152·MathematicsNumericalJEE Main 2021
the term independent of 'x' in the expression of (x+1x2/3−x1/3+1−x−1x−x1/2)10\left( \frac{x+1}{x^{2/3} - x^{1/3} + 1} - \frac{x-1}{x - x^{1/2}} \right)^{10}(x2/3−x1/3+1x+1​−x−x1/2x−1​)10, where x≠0,1x \neq 0, 1x=0,1 is equal to.......

Correct answer: 210

Step-by-step solution →
Q153·MathematicsNumericalJEE Main 2021
If the constant term, in binomial expansion of (2xr+1x2)10\left(2x^{r} + \dfrac{1}{x^{2}}\right)^{10}(2xr+x21​)10 is 180, then r is equal to.......

Correct answer: 8

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Q154·MathematicsNumericalJEE Main 2021
The number of elements in the set {n∈{1,2,3,....100}|(11)n>(10)n+(9)n}\left\{n \in \{1,2,3,.... 100\} \middle| (11)^n > (10)^n + (9)^n\right\}{n∈{1,2,3,....100}∣(11)n>(10)n+(9)n} is..........

Correct answer: 96

Step-by-step solution →
Q155·MathematicsSingle correctJEE Main 2021
For the natural numbers m,nm,nm,n, if (1−y)m(1+y)n=1+a1y+a2y2+....+am+nym+n(1-y)^{m}(1+y)^{n}=1+a_{1}y+a_{2}y^{2}+....+a_{m+n}y^{m+n}(1−y)m(1+y)n=1+a1​y+a2​y2+....+am+n​ym+n and a1=a2=10,a_{1}=a_{2}=10,a1​=a2​=10, then the value of (m+n)(m+n)(m+n) is equal to :
  1. (A)646464
  2. (B)888888
  3. (C)808080
  4. (D)100100100

Correct answer: (C)

Step-by-step solution →
Q156·MathematicsNumericalJEE Main 2021
The number of rational terms in the binomial expansion of (414+516)120\left(4^{\frac{1}{4}} + 5^{\frac{1}{6}}\right)^{120}(441​+561​)120 is.....

Correct answer: 21

Step-by-step solution →
Q157·MathematicsSingle correctJEE Main 2021
The coefficient of x256x^{256}x256 in the expansion of (1−x)101(x2+x+1)100(1-x)^{101}\left(x^2+x+1\right)^{100}(1−x)101(x2+x+1)100 is :
  1. (A)−100C15-{}^{100}C_{15}−100C15​
  2. (B)100C16{}^{100}C_{16}100C16​
  3. (C)−100C16-{}^{100}C_{16}−100C16​
  4. (D)100C15{}^{100}C_{15}100C15​

Correct answer: (D)

Step-by-step solution →
Q158·MathematicsNumericalJEE Main 2021
The term independent of x in the expansion of [x+1x2/3−x1/3+1−x−1x−x1/2]10\left[\frac{x+1}{x^{2/3} - x^{1/3} + 1} - \frac{x-1}{x - x^{1/2}}\right]^{10}[x2/3−x1/3+1x+1​−x−x1/2x−1​]10, x ≠ 1, is equal to ________ .

Correct answer: 210

Step-by-step solution →
Q159·MathematicsSingle correctJEE Main 2021
Let (1+x+2x2)20=a0+a1x+a2x2+...+a40x40(1 + x + 2x^{2})^{20} = a_{0} + a_{1}x + a_{2}x^{2} + ... + a_{40}x^{40}(1+x+2x2)20=a0​+a1​x+a2​x2+...+a40​x40. then a1+a3+a5+...+a37a_{1} + a_{3} + a_{5} + ... + a_{37}a1​+a3​+a5​+...+a37​ is equal to
  1. (A)220(220−21)2^{20}(2^{20} - 21)220(220−21)
  2. (B)219(220−21)2^{19}(2^{20} - 21)219(220−21)
  3. (C)219(220+21)2^{19}(2^{20} + 21)219(220+21)
  4. (D)220(220+21)2^{20}(2^{20} + 21)220(220+21)

Correct answer: (B)

Step-by-step solution →
Q160·MathematicsSingle correctJEE Main 2021
If the fourth term in the expansion of (x+xlog⁡2x)7\left(x + x^{\log_2 x}\right)^7(x+xlog2​x)7 is 4480, then the value of x where x∈Nx \in Nx∈N is equal to :
  1. (A)2
  2. (B)4
  3. (C)3
  4. (D)1

Correct answer: (A)

Step-by-step solution →
Q161·MathematicsNumericalJEE Main 2021
If (2021)3762(2021)^{3762}(2021)3762 is divided by 17, then the remainder is _______ .

Correct answer: 4

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Q162·MathematicsNumericalJEE Main 2021
Let the coefficients of third, fourth and fifth terms in the expansion of (x+ax2)n\left(x+\frac{a}{x^{2}}\right)^{n}(x+x2a​)n, x≠0x \neq 0x=0, be in the ratio 12:8:312 : 8 : 312:8:3. Then the term independent of xxx in the expansion, is equal to ________.

Correct answer: 4

Step-by-step solution →
Q163·MathematicsSingle correctJEE Main 2021
The value of ∑r=06(6Cr⋅6C6−r)\sum_{r=0}^{6}\left({}^{6}C_{r}\cdot{}^{6}C_{6-r}\right)∑r=06​(6Cr​⋅6C6−r​) is equal to :
  1. (A)112411241124
  2. (B)132413241324
  3. (C)102410241024
  4. (D)924924924

Correct answer: (D)

Step-by-step solution →
Q164·MathematicsSingle correctJEE Main 2021
If n is the number of irrational terms in the expansion of (31/4+51/8)60(3^{1/4} + 5^{1/8})^{60}(31/4+51/8)60, then (n – 1) is divisible by :
  1. (A)26
  2. (B)30
  3. (C)8
  4. (D)7

Correct answer: (A)

Step-by-step solution →
Q165·MathematicsNumericalJEE Main 2021
Let n be a positive integer. Let A=∑k=0n(−1)k nCk[(12)k+(34)k+(78)k+(1516)k+(3132)k]A=\sum\limits_{k=0}^{n}(-1)^{k}\ {}^{n}C_{k}\left[\left(\frac{1}{2}\right)^{k}+\left(\frac{3}{4}\right)^{k}+\left(\frac{7}{8}\right)^{k}+\left(\frac{15}{16}\right)^{k}+\left(\frac{31}{32}\right)^{k}\right]A=k=0∑n​(−1)k nCk​[(21​)k+(43​)k+(87​)k+(1615​)k+(3231​)k] If 63A=1−123063A = 1-\frac{1}{2^{30}}63A=1−2301​, then n is equal to _______.

Correct answer: 6

Step-by-step solution →
Q166·MathematicsSingle correctJEE Main 2021
The maximum value of the term independent of ‘t’ in the expansion of (tx15+(1−x)110t)10\left( tx^{\frac{1}{5}} + \frac{(1-x)^{\frac{1}{10}}}{t} \right)^{10}(tx51​+t(1−x)101​​)10 where x ∈ (0,1) is:
  1. (A)10!3(5!)2\frac{10!}{\sqrt{3}(5!)^{2}}3​(5!)210!​
  2. (B)2.10!3(5!)2\frac{2.10!}{3(5!)^{2}}3(5!)22.10!​
  3. (C)10!3(5!)2\frac{10!}{3(5!)^{2}}3(5!)210!​
  4. (D)2.10!33(5!)2\frac{2.10!}{3\sqrt{3}(5!)^{2}}33​(5!)22.10!​

Correct answer: (D)

Step-by-step solution →
Q167·MathematicsNumericalJEE Main 2021
Let m, n ∈ N and gcd (2, n) = 1. If 30(300)+29(301)+....+2(3028)+1(3029)=n⋅2m30\binom{30}{0}+29\binom{30}{1}+....+2\binom{30}{28}+1\binom{30}{29}=n\cdot 2^{m}30(030​)+29(130​)+....+2(2830​)+1(2930​)=n⋅2m, then n + m is equal to _______.

Correct answer: 45

Step-by-step solution →
Q168·MathematicsNumericalJEE Main 2021
The total number of two digit numbers ‘n’, such that 3n^{n}n+7n^{n}n is a multiple of 10, is ______.

Correct answer: 45

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Q169·MathematicsNumericalJEE Main 2021
If the remainder when x is divided by 4 is 3, then the remainder when (2020+x)2022^{2022}2022is divided by 8 is ______.

Correct answer: 1

Step-by-step solution →
Q170·MathematicsSingle correctJEE Main 2021
If n≥2n\ge 2n≥2 is a positive integer, then the sum of the series n+1C2+2(2C2+3C2+4C2+....+nC2)^{n+1}C_{2}+2\left(^{2}C_{2}+^{3}C_{2}+^{4}C_{2}+....+^{n}C_{2}\right)n+1C2​+2(2C2​+3C2​+4C2​+....+nC2​) is :
  1. (A)n(n+1)2(n+2)12\frac{n(n+1)^{2}(n+2)}{12}12n(n+1)2(n+2)​
  2. (B)n(n−1)(2n+1)6\frac{n(n-1)(2n+1)}{6}6n(n−1)(2n+1)​
  3. (C)n(n+1)(2n+1)6\frac{n(n+1)(2n+1)}{6}6n(n+1)(2n+1)​
  4. (D)n(2n+1)(3n+1)6\frac{n(2n+1)(3n+1)}{6}6n(2n+1)(3n+1)​

Correct answer: (C)

Step-by-step solution →
Q171·MathematicsNumericalJEE Main 2021
For integers n and r, let (nr)={nCr,if n≥r≥00,otherwise\binom{n}{r}=\begin{cases}{}^{n}C_{r}, & \text{if } n\ge r\ge 0 \\ 0, & \text{otherwise}\end{cases}(rn​)={nCr​,0,​if n≥r≥0otherwise​ The maximum value of k for which the sum ∑i=0k(10i)(15k−i)+∑i=0k+1(12i)(13k+1−i)\sum_{i=0}^{k}\binom{10}{i}\binom{15}{k-i}+\sum_{i=0}^{k+1}\binom{12}{i}\binom{13}{k+1-i}∑i=0k​(i10​)(k−i15​)+∑i=0k+1​(i12​)(k+1−i13​) exists, is equal to________.

Correct answer: 12

Step-by-step solution →
Q172·MathematicsSingle correctJEE Main 2021
The value of −15C1+2.15C2−3.15C3+....−15.15C15+14C1+14C3+14C5+....+14C11-{}^{15}C_1 + 2.{}^{15}C_2 - 3.{}^{15}C_3 + .... -15.{}^{15}C_{15} + {}^{14}C_1 + {}^{14}C_3 + {}^{14}C_5 + ....+ {}^{14}C_{11}−15C1​+2.15C2​−3.15C3​+....−15.15C15​+14C1​+14C3​+14C5​+....+14C11​ is:
  1. (A)2142^{14}214
  2. (B)213−132^{13} - 13213−13
  3. (C)216−12^{16} - 1216−1
  4. (D)213−142^{13} - 14213−14

Correct answer: (D)

Step-by-step solution →
Q173·MathematicsMultiple correctJEE Advanced 2020
For non-negative integers sss and rrr, let (sr)={s!r!(s−r)!if r≤s,0if r>s.\binom{s}{r} = \begin{cases} \frac{s!}{r!(s-r)!} & \text{if } r \le s, \\ 0 & \text{if } r > s. \end{cases}(rs​)={r!(s−r)!s!​0​if r≤s,if r>s.​ For positive integers mmm and nnn, let g(m,n)=∑p=0m+nf(m,n,p)(n+pp)g(m,n) = \sum_{p=0}^{m+n} \frac{f(m,n,p)}{\binom{n+p}{p}}g(m,n)=∑p=0m+n​(pn+p​)f(m,n,p)​ where for any nonnegative integer ppp, f(m,n,p)=∑i=0p(mi)(n+ip)(p+np−i)f(m,n,p) = \sum_{i=0}^{p} \binom{m}{i} \binom{n+i}{p} \binom{p+n}{p-i}f(m,n,p)=∑i=0p​(im​)(pn+i​)(p−ip+n​) Then which of the following statements is/are TRUE?
  1. (A)g(m,n)=g(n,m)g(m,n) = g(n,m)g(m,n)=g(n,m) for all positive integers m,nm, nm,n
  2. (B)g(m,n+1)=g(m+1,n)g(m,n+1) = g(m+1,n)g(m,n+1)=g(m+1,n) for all positive integers m,nm, nm,n
  3. (C)g(2m,2n)=2g(m,n)g(2m,2n) = 2g(m,n)g(2m,2n)=2g(m,n) for all positive integers m,nm, nm,n
  4. (D)g(2m,2n)=(g(m,n))2g(2m,2n) = (g(m,n))^{2}g(2m,2n)=(g(m,n))2 for all positive integers m,nm, nm,n

Correct answer: (A), (B), (D)

Step-by-step solution →
Q174·MathematicsSingle correctJEE Main 2020
If {p} denotes the fractional part of the number p, then {32008}\left\{\dfrac{3^{200}}{8}\right\}{83200​}, is equal to
  1. (A)58\dfrac{5}{8}85​
  2. (B)78\dfrac{7}{8}87​
  3. (C)38\dfrac{3}{8}83​
  4. (D)18\dfrac{1}{8}81​

Correct answer: (D)

Step-by-step solution →
Q175·MathematicsSingle correctJEE Main 2020
If the constant term in the binomial expansion of (x−kx2)10\left(\sqrt{x} - \frac{k}{x^2}\right)^{10}(x​−x2k​)10 is 405, then ∣k∣|k|∣k∣ equals:
  1. (A)999
  2. (B)111
  3. (C)333
  4. (D)222

Correct answer: (C)

Step-by-step solution →
Q176·MathematicsNumericalJEE Main 2020
The natural number m, for which the coefficient of x in the binomial expansion of (xm+1x2)22\left(x^{m}+\dfrac{1}{x^{2}}\right)^{22}(xm+x21​)22 is 1540, is ____

Correct answer: 13.00

Step-by-step solution →
Q177·MathematicsNumericalJEE Main 2020
The coefficient of x4x^4x4 in the expansion of (1+x+x2+x3)6(1 + x + x^2 + x^3)^6(1+x+x2+x3)6 in powers of x, is __________.

Correct answer: 120.00

Step-by-step solution →
Q178·MathematicsSingle correctJEE Main 2020
The value of ∑r=02050−rC6\sum_{r=0}^{20} {}^{50-r}C_{6}∑r=020​50−rC6​ is equal to:
  1. (A)50C6−30C6{}^{50}C_{6} - {}^{30}C_{6}50C6​−30C6​
  2. (B)51C7+30C7{}^{51}C_{7} + {}^{30}C_{7}51C7​+30C7​
  3. (C)51C7−30C7{}^{51}C_{7} - {}^{30}C_{7}51C7​−30C7​
  4. (D)50C7−30C7{}^{50}C_{7} - {}^{30}C_{7}50C7​−30C7​

Correct answer: (C)

Step-by-step solution →
Q179·MathematicsSingle correctJEE Main 2020
If for some positive integer n, the coefficients of three consecutive terms in the binomial expansion of (1+x)N+5(1 + x)^{N+5}(1+x)N+5 are in the ratio 5 : 10 : 14, then the largest coefficient in the expansion is:
  1. (A)252
  2. (B)330
  3. (C)792
  4. (D)462

Correct answer: (D)

Step-by-step solution →
Q180·MathematicsNumericalJEE Main 2020
Let (2x2+3x+4)10=∑r=020arxr(2x^{2} + 3x + 4)^{10} = \sum_{r=0}^{20} a_{r} x^{r}(2x2+3x+4)10=∑r=020​ar​xr. Then a7a13\frac{a_{7}}{a_{13}}a13​a7​​ is equal to __________.

Correct answer: 8

Step-by-step solution →
Q181·MathematicsSingle correctJEE Main 2020
If the term independent of x in the expansion of (32x2−13x)9\left( \frac{3}{2} x^2 - \frac{1}{3x} \right)^9(23​x2−3x1​)9 is k, then 18k is equal to:
  1. (A)9
  2. (B)5
  3. (C)7
  4. (D)11

Correct answer: (C)

Step-by-step solution →
Q182·MathematicsSingle correctJEE Main 2020
If the number of integral terms in the expansion of (312+518)n\left(3^{\frac{1}{2}} + 5^{\frac{1}{8}}\right)^{n}(321​+581​)n is exactly 33, then the least value of 'n' is
  1. (A)264
  2. (B)248
  3. (C)256
  4. (D)128

Correct answer: (C)

Step-by-step solution →
Q183·MathematicsNumericalJEE Main 2020
If Cr=25CrC_{r} = {}^{25}C_{r}Cr​=25Cr​ and C0+5.C1+9.C2+..........+(101).C25=225.kC_{0} + 5.C_{1} + 9.C_{2} + .......... + (101).C_{25} = 2^{25}.kC0​+5.C1​+9.C2​+..........+(101).C25​=225.k, then k is equal to ______

Correct answer: 51

Step-by-step solution →
Q184·MathematicsNumericalJEE Main 2020
The coefficient of x4x^{4}x4 in the expansion of (1+x+x2)10(1+x+x^{2})^{10}(1+x+x2)10 is ______.

Correct answer: 615

Step-by-step solution →
Q185·MathematicsSingle correctJEE Main 2020
If a, b and c are the greatest values of 19Cp^{19}C_{p}19Cp​, 20Cq^{20}C_{q}20Cq​ and 21Cr^{21}C_{r}21Cr​ respectively, then:
  1. (A)a10=b11=c42\dfrac{a}{10}=\dfrac{b}{11}=\dfrac{c}{42}10a​=11b​=42c​
  2. (B)a11=b22=c21\dfrac{a}{11}=\dfrac{b}{22}=\dfrac{c}{21}11a​=22b​=21c​
  3. (C)a11=b22=c42\dfrac{a}{11}=\dfrac{b}{22}=\dfrac{c}{42}11a​=22b​=42c​
  4. (D)a10=b11=c21\dfrac{a}{10}=\dfrac{b}{11}=\dfrac{c}{21}10a​=11b​=21c​

Correct answer: (C)

Step-by-step solution →
Q186·MathematicsSingle correctJEE Main 2020
If α\alphaα and β\betaβ be the coefficients of x4x^{4}x4 and x2x^{2}x2 respectively in the expansion of (x+x2−1)6+(x−x2−1)6\left(x+\sqrt{x^{2}-1}\right)^{6}+\left(x-\sqrt{x^{2}-1}\right)^{6}(x+x2−1​)6+(x−x2−1​)6, then:
  1. (A)α+β=−30\alpha+\beta=-30α+β=−30
  2. (B)α−β=60\alpha-\beta=60α−β=60
  3. (C)α−β=−132\alpha-\beta=-132α−β=−132
  4. (D)α+β=60\alpha+\beta=60α+β=60

Correct answer: (C)

Step-by-step solution →
Q187·MathematicsNumericalJEE Main 2020
If the sum of the coefficients of all even powers of x in the product (1+x+x2+......+x2n)(1−x+x2−x3+.....+x2n)(1+x+x^{2}+......+x^{2n})(1-x+x^{2}-x^{3}+.....+x^{2n})(1+x+x2+......+x2n)(1−x+x2−x3+.....+x2n) is 61, then n is equal to

Correct answer: 30

Step-by-step solution →
Q188·MathematicsNumericalJEE Advanced 2019
Suppose det⁡[∑k=0nk∑k=0nnCkk2∑k=0nnCkk∑k=0nnCk3k]=0\det \begin{bmatrix} \displaystyle\sum_{k=0}^{n} k & \displaystyle\sum_{k=0}^{n} {}^{n}C_k k^{2} \\ \displaystyle\sum_{k=0}^{n} {}^{n}C_k k & \displaystyle\sum_{k=0}^{n} {}^{n}C_k 3^{k} \end{bmatrix} = 0det​k=0∑n​kk=0∑n​nCk​k​k=0∑n​nCk​k2k=0∑n​nCk​3k​​=0 holds for some positive integer n. Then ∑k=0nnCkk+1\displaystyle\sum_{k=0}^{n} \dfrac{{}^{n}C_k}{k+1}k=0∑n​k+1nCk​​ equals. ____

Correct answer: 6.20

Step-by-step solution →
Q189·MathematicsSingle correctJEE Main 2019
The coefficient of x18x^{18}x18 in the product (1+x)(1−x)10(1+x+x2)9(1+x)(1-x)^{10}(1+x+x^{2})^{9}(1+x)(1−x)10(1+x+x2)9 is :
  1. (A)84
  2. (B)126
  3. (C)−126-126−126
  4. (D)−84-84−84

Correct answer: (A)

Step-by-step solution →
Q190·MathematicsSingle correctJEE Main 2019
The term independent of x in the expansion of (160−x881)⋅(2x2−3x2)6\left(\dfrac{1}{60} - \dfrac{x^{8}}{81}\right)\cdot\left(2x^{2} - \dfrac{3}{x^{2}}\right)^{6}(601​−81x8​)⋅(2x2−x23​)6 is equal to:
  1. (A)36
  2. (B)−36-36−36
  3. (C)−108-108−108
  4. (D)−72-72−72

Correct answer: (B)

Step-by-step solution →
Q191·MathematicsSingle correctJEE Main 2019
The smallest natural number n, such that the coefficient of x in the expansion of (x2+1x3)n\left(x^{2} + \dfrac{1}{x^{3}}\right)^{n}(x2+x31​)n is nC23^{n}C_{23}nC23​ is
  1. (A)38
  2. (B)58
  3. (C)23
  4. (D)35

Correct answer: (A)

Step-by-step solution →
Q192·MathematicsSingle correctJEE Main 2019
If the coefficients of x2^{2}2 and x3^{3}3 are both zero, in the expansion of the expression (1 + ax + bx2^{2}2) (1 − 3x)15^{15}15 in powers of x, then the ordered pair (a, b) is equal to
  1. (A)(−54, 315)
  2. (B)(28, 861)
  3. (C)(28, 315)
  4. (D)(−21, 714)

Correct answer: (C)

Step-by-step solution →
Q193·MathematicsSingle correctJEE Main 2019
If some three consecutive in the binomial expansion of (x+1)n(x+1)^{n}(x+1)n in powers of x are in the ratio 2:15:702 : 15 : 702:15:70, then the average of these three coefficient is:
  1. (A)964964964
  2. (B)625625625
  3. (C)227227227
  4. (D)232232232

Correct answer: (D)

Step-by-step solution →
Q194·MathematicsSingle correctJEE Main 2019
If the fourth term in the Binomial expansion of (2x+xlog⁡8x)6(x>0)\left(\dfrac{2}{x}+x^{\log_{8}x}\right)^{6}(x>0)(x2​+xlog8​x)6(x>0) is 20×8720\times 8^{7}20×87, then a value of x is:
  1. (A)838^{3}83
  2. (B)8−28^{-2}8−2
  3. (C)888
  4. (D)828^{2}82

Correct answer: (D)

Step-by-step solution →
Q195·MathematicsSingle correctJEE Main 2019
2⋅20C0+5⋅20C1+8⋅20C2+11⋅20C3+....+62⋅20C202\cdot {}^{20}C_{0}+5\cdot {}^{20}C_{1}+8\cdot {}^{20}C_{2}+11\cdot {}^{20}C_{3}+....+62\cdot {}^{20}C_{20}2⋅20C0​+5⋅20C1​+8⋅20C2​+11⋅20C3​+....+62⋅20C20​ is equal to
  1. (A)2232^{23}223
  2. (B)2262^{26}226
  3. (C)2242^{24}224
  4. (D)2252^{25}225

Correct answer: (D)

Step-by-step solution →
Q196·MathematicsSingle correctJEE Main 2019
The sum of the co-efficient of all even degree terms in x in the expansion of (x+x3−1)6+(x−x3−1)6,(x>1)\left(x+\sqrt{x^{3}-1}\right)^{6}+\left(x-\sqrt{x^{3}-1}\right)^{6},(x>1)(x+x3−1​)6+(x−x3−1​)6,(x>1) is equal to:
  1. (A)26
  2. (B)24
  3. (C)32
  4. (D)29

Correct answer: (B)

Step-by-step solution →
Q197·MathematicsSingle correctJEE Main 2019
If nC4,nC5^{n}C_{4}, {}^{n}C_{5}nC4​,nC5​ and nC6^{n}C_{6}nC6​ are in A.P., then n can be :
  1. (A)9
  2. (B)14
  3. (C)11
  4. (D)12

Correct answer: (B)

Step-by-step solution →
Q198·MathematicsSingle correctJEE Main 2019
The total number or irrational terms in the binomial expansion of (71/5−31/10)60\left(7^{1/5} - 3^{1/10}\right)^{60}(71/5−31/10)60 is :
  1. (A)55
  2. (B)49
  3. (C)48
  4. (D)54

Correct answer: (D)

Step-by-step solution →
Q199·MathematicsSingle correctJEE Main 2019
The value of r for which 20Cr 20C0+20Cr−1 20C1+20Cr−2 20C2+…+20C0 20Cr^{20}C_{r}\,^{20}C_{0}+^{20}C_{r-1}\,^{20}C_{1}+^{20}C_{r-2}\,^{20}C_{2}+\ldots+^{20}C_{0}\,^{20}C_{r}20Cr​20C0​+20Cr−1​20C1​+20Cr−2​20C2​+…+20C0​20Cr​ is maximum is:
  1. (A)151515
  2. (B)202020
  3. (C)111111
  4. (D)101010

Correct answer: (B)

Step-by-step solution →
Q200·MathematicsSingle correctJEE Main 2019
Let Sn=1+q+q2+…+qnS_{n}=1+q+q^{2}+\ldots+q^{n}Sn​=1+q+q2+…+qn and Tn=1+(q+12)+(q+12)2+…+(q+12)nT_{n}=1+\left(\dfrac{q+1}{2}\right)+\left(\dfrac{q+1}{2}\right)^{2}+\ldots+\left(\dfrac{q+1}{2}\right)^{n}Tn​=1+(2q+1​)+(2q+1​)2+…+(2q+1​)n where q is a real number and q≠1q\neq 1q=1. If 101C1+101C2⋅S1+…+101C101⋅S100=αT100{}^{101}C_{1}+{}^{101}C_{2}\cdot S_{1}+\ldots+{}^{101}C_{101}\cdot S_{100}=\alpha T_{100}101C1​+101C2​⋅S1​+…+101C101​⋅S100​=αT100​ then α is equal to:
  1. (A)2992^{99}299
  2. (B)202
  3. (C)200
  4. (D)21002^{100}2100

Correct answer: (D)

Step-by-step solution →
Q201·MathematicsSingle correctJEE Main 2019
The sum of the real values of x for which the middle term in the binomial expansion of (x33+3x)8\left(\frac{x^{3}}{3}+\frac{3}{x}\right)^{8}(3x3​+x3​)8 equals 5670 is:
  1. (A)0
  2. (B)6
  3. (C)4
  4. (D)8

Correct answer: (A)

Step-by-step solution →
Q202·MathematicsSingle correctJEE Main 2019
Let (x+10)50+(x−10)50=a0+a1x+a2x2+...+a50x50(x+10)^{50}+(x-10)^{50}=a_0+a_1x+a_2x^{2}+...+a_{50}x^{50}(x+10)50+(x−10)50=a0​+a1​x+a2​x2+...+a50​x50, for x∈Rx\in Rx∈R; then a2a0\frac{a_2}{a_0}a0​a2​​ is equal to:
  1. (A)12.50
  2. (B)12.00
  3. (C)12.25
  4. (D)12.75

Correct answer: (C)

Step-by-step solution →
Q203·MathematicsSingle correctJEE Main 2019
If ∑i=120(20Ci−120Ci+20Ci−1)3=k21\sum_{i=1}^{20}\left(\frac{^{20}C_{i-1}}{^{20}C_i+^{20}C_{i-1}}\right)^{3}=\frac{k}{21}∑i=120​(20Ci​+20Ci−1​20Ci−1​​)3=21k​, then k equals:
  1. (A)400
  2. (B)50
  3. (C)200
  4. (D)100

Correct answer: (D)

Step-by-step solution →
Q204·MathematicsSingle correctJEE Main 2019
If the third term in the binomial expansion of (1+xlog⁡2x)5\left(1+x^{\log_{2}x}\right)^{5}(1+xlog2​x)5 equals 2560, then a possible value of x is:
  1. (A)14\frac{1}{4}41​
  2. (B)424\sqrt{2}42​
  3. (C)18\frac{1}{8}81​
  4. (D)222\sqrt{2}22​

Correct answer: (A)

Step-by-step solution →
Q205·MathematicsSingle correctJEE Main 2019
If the fractional part of the number 240315\dfrac{2^{403}}{15}152403​ is k15\dfrac{k}{15}15k​, then k is equal to:
  1. (A)6
  2. (B)8
  3. (C)4
  4. (D)14

Correct answer: (B)

Step-by-step solution →
Q206·MathematicsSingle correctJEE Main 2019
The coefficient of t4t^{4}t4 in the expansion of (1−t61−t)3\left(\frac{1-t^{6}}{1-t}\right)^{3}(1−t1−t6​)3 is
  1. (A)12
  2. (B)15
  3. (C)10
  4. (D)14

Correct answer: (B)

Step-by-step solution →
Q207·MathematicsNumericalJEE Advanced 2018
Let X=(10C1)2+2(10C2)2+3(10C3)2+....+10(10C10)2X = ({}^{10}C_{1})^{2} + 2({}^{10}C_{2})^{2} + 3({}^{10}C_{3})^{2} + .... + 10({}^{10}C_{10})^{2}X=(10C1​)2+2(10C2​)2+3(10C3​)2+....+10(10C10​)2, where 10Cr{}^{10}C_{r}10Cr​, r∈{1,2,.....,10}r \in \{1, 2, ....., 10\}r∈{1,2,.....,10} denote binomial coefficients. Then the value of 11430X\frac{1}{1430}X14301​X is ______ .

Correct answer: 646

Step-by-step solution →
Q208·MathematicsIntegerJEE Advanced 2016
Let mmm be the smallest positive integer such that the coefficient of x2x^{2}x2 in the expansion of (1+x)2+(1+x)3+…+(1+x)49+(1+mx)50(1 + x)^{2} + (1 + x)^{3} + \ldots + (1 + x)^{49} + (1 + mx)^{50}(1+x)2+(1+x)3+…+(1+x)49+(1+mx)50 is (3n+1) 51C3(3n + 1)\,{}^{51}C_3(3n+1)51C3​ for some positive integer nnn. Then the value of nnn is

Correct answer: 5

Step-by-step solution →
Q209·MathematicsIntegerJEE Advanced 2015
The coefficient of x9x^{9}x9 in the expansion of (1+x)(1+x2)(1+x3)…(1+x100)(1 + x)(1 + x^{2})(1 + x^{3}) \ldots (1 + x^{100})(1+x)(1+x2)(1+x3)…(1+x100) is

Correct answer: 8

Step-by-step solution →
Q210·MathematicsSingle correctJEE Advanced 2014
Coefficient of x11x^{11}x11 in the expansion of (1+x2)4(1+x3)7(1+x4)12(1 + x^{2})^{4} (1 + x^{3})^{7} (1 + x^{4})^{12}(1+x2)4(1+x3)7(1+x4)12 is
  1. (A)1051
  2. (B)1106
  3. (C)1113
  4. (D)1120

Correct answer: (C)

Step-by-step solution →
Q211·MathematicsIntegerJEE Advanced 2013
The coefficients of three consecutive terms of (1+x)n+5(1+x)^{n+5}(1+x)n+5 are in the ratio 5:10:145:10:145:10:14. Then n = _______

Correct answer: 6

Step-by-step solution →

Binomial Theorem and Its Simple Applications — frequently asked

How many questions from Binomial Theorem and Its Simple Applications appear in JEE?

Binomial Theorem and Its Simple Applications has appeared in 158 of the last 186 JEE Main and JEE Advanced papers — about 85% of them — contributing 211 questions in total across those papers.

Is Binomial Theorem and Its Simple Applications an important chapter for JEE?

Judged by how often it is actually tested, it appears in roughly 85% of papers. Chapters above about 50% are effectively guaranteed to show up every session, so they repay thorough preparation; lower-frequency chapters are better treated as targeted revision.

Where do these Binomial Theorem and Its Simple Applications questions come from?

Every question is from an official JEE Main or JEE Advanced paper, transcribed from the original paper and tagged to this chapter. Answers follow the official answer key.

Other Mathematics chapters

  • Three Dimensional Geometry 344
  • Matrices and Determinants 342
  • Sets, Relations and Functions 313
  • Sequence and Series 287
  • Definite Integration 267
  • Vector Algebra 245
  • Differential Equations 240
  • Probability 226

All 26 Mathematics chapters →

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