Binomial Theorem

17 JEE Main questions · Maths

MathsJEE Main 21 Jan 2026, Shift 1·Q16·MCQ·Moderate
Open
Coefficient of a term
If the coefficient of $x$ in the expansion of $(ax^2 + bx + c)(1 - 2x)^{26}$ is $-56$ and the coefficients of $x^2$ and $x^3$ are both zero, then $a + b + c$ is equal to
1. (1)1300
2. (2)1500
3. (3)1403
4. (4)1483
MathsJEE Main 21 Jan 2026, Shift 2·Q24·Numerical·Moderate
Open
Products of binomial coefficients
If $\left(\frac{1}{{}^{15}C_0} + \frac{1}{{}^{15}C_1}\right)\left(\frac{1}{{}^{15}C_1} + \frac{1}{{}^{15}C_2}\right)\cdots\left(\frac{1}{{}^{15}C_{12}} + \frac{1}{{}^{15}C_{13}}\right) = \frac{\alpha^{13}}{{}^{14}C_0\,{}^{14}C_1\cdots{}^{14}C_{12}},$ then $30\alpha$ is equal to .
MathsJEE Main 22 Jan 2026, Shift 1·Q16·MCQ·Tough
Open
Coefficient in sum of binomial expansions
The coefficient of $x^{48}$ in $(1 + x) + 2(1 + x)^2 + 3(1 + x)^3 + \ldots + 100(1 + x)^{100}$ is equal to:
1. (1) $100\cdot{}^{100}C_{49} - {}^{100}C_{50}$
2. (2) ${}^{100}C_{50} + {}^{101}C_{49}$
3. (3) $100\cdot{}^{100}C_{49} - {}^{100}C_{48}$
4. (4) $100\cdot{}^{101}C_{49} - {}^{101}C_{50}$
We corrected a printing error in the original paper
Source printed option (4) as “ $100\cdot{}^{101}C_{49} - {}^{100}C_{50}$ ” (then no option is correct); corrected to the intended “ $100\cdot{}^{101}C_{49} - {}^{101}C_{50}$ ” (the correct coefficient and the source solution's result; printed key (4)).
MathsJEE Main 22 Jan 2026, Shift 2·Q20·MCQ·Tough
Open
Sums of binomial coefficients
Let $C_r$ denote the coefficient of $x^r$ in the binomial expansion of $(1 + x)^n$ , $n \in \mathbb{N}$ , $0 \le r \le n$ . If $P_n = C_0 - C_1 + \frac{2^2}{3}C_2 - \frac{2^3}{4}C_3 + \cdots + \frac{(-2)^n}{n+1}C_n,$ then the value of $\displaystyle\sum_{n=1}^{25}\frac{1}{P_{2n}}$ equals.
1. (1)580
2. (2)525
3. (3)650
4. (4)675
MathsJEE Main 23 Jan 2026, Shift 1·Q12·MCQ·Moderate
Open
Sum of binomial coefficients
The value of $\dfrac{{}^{100}C_{50}}{51} + \dfrac{{}^{100}C_{51}}{52} + \ldots + \dfrac{{}^{100}C_{100}}{101}$ is:
1. (1) $\dfrac{2^{101}}{100}$
2. (2) $\dfrac{2^{100}}{100}$
3. (3) $\dfrac{2^{101}}{101}$
4. (4) $\dfrac{2^{100}}{101}$
MathsJEE Main 23 Jan 2026, Shift 1·Q14·MCQ·Moderate
Open
Coefficients in binomial expansion
The sum of all possible values of $n \in \mathbb{N}$ , so that the coefficients of $x$ , $x^2$ and $x^3$ in the expansion of $(1 + x^2)^2(1 + x)^n$ , are in arithmetic progression is:
1. (1)3
2. (2)7
3. (3)12
4. (4)9
MathsJEE Main 24 Jan 2026, Shift 1·Q14·MCQ·Moderate
Open
Sum of binomial coefficients
Let $S = \dfrac{1}{25!} + \dfrac{1}{3!\,23!} + \dfrac{1}{5!\,21!} + \ldots$ up to 13 terms. If $13S = \dfrac{2^k}{n!}$ , $k \in \mathbb{N}$ , then $n + k$ is equal to
1. (1)51
2. (2)52
3. (3)49
4. (4)50
MathsJEE Main 28 Jan 2026, Shift 2·Q1·MCQ·Moderate
Open
Divisibility and integral part
Given below two statements :
Statement I : $25^{13} + 20^{13} + 8^{13} + 3^{13}$ is divisible by 7.
Statement II : The integral part of $\left(7 + 4\sqrt3\right)^{25}$ is an odd number.
In the light of the above statements, choose the correct answer from the options given below :
1. (1)Both Statement I and Statement II are false.
2. (2)Both Statement I and Statement II are true.
3. (3)Statement I is false but Statement II is true.
4. (4)Statement I is true but Statement II is false.
MathsJEE Main 28 Jan 2026, Shift 2·Q2·MCQ·Moderate
Open
Sum of binomial coefficients
The sum of the coefficients of $x^{499}$ and $x^{500}$ in $(1 + x)^{1000} + x(1 + x)^{999} + x^2(1 + x)^{998} + \cdots + x^{1000}$ is
1. (1) ${}^{1001}C_{501}$
2. (2) ${}^{1002}C_{500}$
3. (3) ${}^{1002}C_{501}$
4. (4) ${}^{1000}C_{501}$
MathsJEE Main 2 Apr 2026, Shift 1·Q7·MCQ·Moderate
Open
Properties of binomial coefficients
The number of elements in the set $S = \left\{(r, k) : k \in \mathbb{Z} \text{ and } {}^{36}C_{r+1} = \dfrac{6\left({}^{35}C_r\right)}{(k^2 - 3)}\right\},$ is
1. (1)2
2. (2)4
3. (3)8
4. (4)16
MathsJEE Main 4 Apr 2026, Shift 1·Q9·MCQ·Tough
Open
Sum of binomial coefficients
Let the smallest value of $k \in \mathbb{N}$ , for which the coefficient of $x^3$ in $(1 + x)^3 + (1 + x)^4 + (1 + x)^5 + \ldots + (1 + x)^{99} + (1 + kx)^{100}$ , $x \neq 0$ , is $\left(43n + \dfrac{101}{4}\right)\left({}^{100}C_3\right)$ for some $n \in \mathbb{N}$ , be $p$ . Then the value of $p + n$ is:
1. (1)10
2. (2)11
3. (3)12
4. (4)13
MathsJEE Main 4 Apr 2026, Shift 2·Q9·MCQ·Easy–moderate
Open
Term independent of x
In the expansion of $\left(9x - \dfrac{1}{3\sqrt{x}}\right)^{18}$ , $x > 0$ , if the term independent of $x$ is $(221)k$ . then $k$ is equal to:
1. (1)84
2. (2)78
3. (3)168
4. (4)198
MathsJEE Main 5 Apr 2026, Shift 1·Q23·Numerical·Moderate
Open
Coefficient of a term
If the sum of the coefficients of $x^7$ and $x^{14}$ in the expansion of $\left(\dfrac{1}{x^3} - x^4\right)^n$ , $x \neq 0$ , is zero, then the value of $n$ is .
MathsJEE Main 5 Apr 2026, Shift 2·Q7·MCQ·Easy–moderate
Open
Coefficient of a term
The coefficient of $x^2$ in the expansion of $\left(2x^2 + \dfrac{1}{x}\right)^{10}$ , $x \neq 0$ , is:
1. (1)3240
2. (2)3360
3. (3)3480
4. (4)3600
We corrected a printing error in the original paper
Answer corrected; the source key is (3) — the general term is $\binom{10}{r}2^{10-r}x^{20-3r}$ , so $r = 6$ and the coefficient is $\binom{10}{6}2^4 = 3360$ , option (2).
MathsJEE Main 6 Apr 2026, Shift 1·Q8·MCQ·Easy–moderate
Open
Middle term of binomial expansion
If the coefficients of the middle terms in the binomial expansions of $(1 + \alpha x)^{26}$ and $(1 - \alpha x)^{28}$ , $\alpha \neq 0$ , are equal, then the value of $\alpha$ is:
1. (1)1
2. (2) $\dfrac{14}{13}$
3. (3) $\dfrac{27}{7}$
4. (4) $\dfrac{7}{27}$
MathsJEE Main 6 Apr 2026, Shift 2·Q22·Numerical·Moderate
Open
Binomial expansion coefficients
If $\left(1 - x^3\right)^{10} = \displaystyle\sum_{r=0}^{10} a_r x^r (1 - x)^{30-2r}$ , then $\dfrac{9a_9}{a_{10}}$ is equal to .
MathsJEE Main 8 Apr 2026, Shift 2·Q8·MCQ·Tough
Open
Binomial coefficient sums by integration
If $26\left(\frac{2^3}{3}\left({}^{12}C_2\right) + \frac{2^5}{5}\left({}^{12}C_4\right) + \frac{2^7}{7}\left({}^{12}C_6\right) + \ldots + \frac{2^{13}}{13}\left({}^{12}C_{12}\right)\right) = 3^{13} - \alpha,$ then $\alpha$ is equal to:
1. (1)45
2. (2)48
3. (3)51
4. (4)54
We corrected a printing error in the original paper
Source printed the first term as “ $\frac{2^3}{3}({}^{13}C_2)$ ”; corrected to the intended ${}^{12}C_2$ (all other terms are ${}^{12}C_k$ ; with ${}^{13}C_2$ the value is $\alpha = -781$ , no option).