Core Syllabus Module

Binomial Theorem

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Chapter Elite-20 Challenge Zone

Target Advanced

JEE Main 2025 (January)

LEVELJEE Advanced

Let $^{n} C_{r - 1} = 28$ , $^{n} C_{r} = 56$ and $^{n} C_{r + 1} = 70$ Let $A (4 cos t, 4 sin t)$ , $B (2 sin t, - 2 cos t)$ and $C (3 r - n, r^{2} - n - 1)$ be the vertices of a triangle ABC, where t is a parameter. If $(3 x - 1)^{2} + (3 y)^{2} = α .$ is the locus of the centroid of triangle ABC, then $α$ equals

JEE Main 2022 (28 July Shift 2)

LEVELJEE Advanced

Let the coefficients of the middle terms in the expansion of $(\frac{1}{6} + β x)^{4}, (1 - 3 β x)^{2}$ and $(1 - \frac{β}{2} x)^{6}, β > 0$ , respectively form the first three terms of an A.P. If $d$ is the common difference of this A.P., then $50 - \frac{2 d}{β ^{2}}$ is equal to ______.

JEE Advanced 1999

LEVELJEE Advanced

Let $n$ be any positive integer. Prove that $\sum_{k = 0}^{m} \frac{( k 2 n - k )}{( n 2 n - k )} (2 n - 4 k + 1) 2^{n - 2 k} = \frac{( m n )}{( n - m 2 n - 2 m )} 2^{n - 2 m}$ for each non-negative integer $m \leq n$ .

JEE Advanced 1993

LEVELJEE Advanced

Prove that $\sum_{r=1}^k (-3)^{r-1} ^{3n}C_{2r-1} = 0$ , where $k = (3 n) /2$ and $n$ is an even positive integer.

JEE Advanced 1989

LEVELJEE Advanced

Using mathematical induction, prove that $^{m} C_{0}^{n} C_{k} +^{m} C_{1}^{n} C_{k - 1} + .... +^{m} C_{k}^{n} C_{0} =^{(m + n)} C_{k}$ , where $m, n, k$ are positive integers, and $^{p} C_{q} = 0$ for $p < q$ .

JEE Main 2025 (January)

LEVELJEE Advanced

If $α = 1 + \sum_{r = 1}^{6} (- 3)^{r - 1}^{12} C_{2 r - 1}$ then the distance of the point $(12, 3)$ from the line $α x - 3 y + 1 = 0$ is

JEE Main 2020 - 4 Sep (Morning)

LEVELJEE Advanced

Let $(2 x^{2} + 3 x + 4)^{10} = \sum_{r = 0}^{20} a_{r} x^{r}$ . Then $\frac{a _{7}}{a _{13}}$ is equal to

JEE Main 2021 (18 March Shift 2)

LEVELJEE Main

The term independent of $x$ in the expansion of $[\frac{x + 1}{x ^{2/3} - x ^{1/3} + 1} - \frac{x - 1}{x - x ^{1/2}}]^{10}, x \neq = 1$ , is equal to ___

JEE Advanced 1991

LEVELJEE Main

Using induction or otherwise, prove that for any non-negative integers $m, n, r$ and $k$ , $\sum_{m = 0}^{k} (n - m) \frac{( r + m )!}{m !} = \frac{( r + k + 1 )!}{k !} [\frac{n}{r + 1} - \frac{k}{r + 2}]$

JEE Main 2019 (08 April Shift 2)

LEVELJEE Main

If the fourth term in the binomial expansion of $(1/ (1 0^{1 + l o g_{10} x}) + 12 x^{6})^{6}$ is equal to 200, and $x > 1$ , then the value of x is :

(A)

1 0^{3}

(B)

100

(C)

1 0^{4}

(D)

10

JEE Main 2021 (February)

LEVELJEE Main

The maximum value of the term independent of 't' in the expansion of $(t x^{1/5} + \frac{( 1 - x ) ^{1/10}}{t})^{10}$ where $x \in (0, 1)$ is:

(A)

\frac{10 !}{3 ( 5 ! ) ^{2}}

(B)

\frac{2.10 !}{3 ( 5 ! ) ^{2}}

(C)

\frac{10 !}{3 ( 5 ! ) ^{2}}

(D)

\frac{2.10 !}{3 3 ( 5 ! ) ^{2}}

JEE Advanced 2015

LEVELJEE Main

The coefficient of $x^{9}$ in the expansion of $(1 + x) (1 + x^{2}) (1 + x^{3}) \dots (1 + x^{100})$ is \dots.

JEE Main 2022 (28 June Shift 1)

LEVELJEE Main

If $\sum_{k = 1}^{31} (^{31} C_{k}) (^{31} C_{k - 1}) - \sum_{k = 1}^{30} (^{30} C_{k}) (^{30} C_{k - 1}) = \frac{α ( 60 !)}{( 30 !) ( 31 !)}$ , where $α \in R$ , then the value of $16 α$ is equal to

If $f (m, n) = \frac{( 1 - x ^{m} ) ( 1 - x ^{m - 1} ) \dots ( 1 - x ^{m - n + 1} )}{( 1 - x ) ( 1 - x ^{2} ) \dots ( 1 - x ^{n} )}$ where $m$ and $n$ are positive integers ( $n \leq m$ ), show that $f (m, n + 1) = f (m - 1, n + 1) + x^{m - n - 1} f (m - 1, n)$ .

JEE Advanced 1979

LEVELJEE Main

Given that $C_{1} + 2 C_{2} x + 3 C_{3} x^{2} + .... + 2 n C_{2 n} x^{2 n - 1} = 2 n (1 + x)^{2 n - 1}$ where $C_{r} = \frac{( 2 n )!}{r ! ( 2 n - r )!}, r = 0, 1, 2, ...., 2 n$ . Prove that $C_{1}^{2} - 2 C_{2}^{2} + 3 C_{3}^{2} - .... - 2 n C_{2 n}^{2} = (- 1)^{n} n^{n} C_{n}$ .

JEE Main 2025 (January)

LEVELJEE Main

Let the coefficients of three consecutive terms $T_{r}, T_{r + 1}$ and $T_{r + 2}$ in the binomial expansion of $(a + b)^{12}$ 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 $(43 + 34)^{12}$ . Then $p + q$ is equal to:

JEE Main 2022 (25 July Shift 1)

LEVELJEE Main

If the maximum value of the term independent of $t$ in the expansion of $(t^{2} x^{\frac{1}{5}} + \frac{( 1 - x ) ^{\frac{1}{10}}}{t})^{15}, x \geq 0$ , is $K$ , then $8 K$ is equal to ______.

JEE Advanced 1984

LEVELJEE Main

Given $s_{n} = 1 + q + q^{2} + .... + q^{n}$ ; $S_{n} = 1 + \frac{q + 1}{2} + (\frac{q + 1}{2})^{2} + .... + (\frac{q + 1}{2})^{n}, q \neq = 1$ Prove that $^{n + 1} C_{1} +^{n + 1} C_{2} s_{1} +^{n + 1} C_{3} s_{2} + .... +^{n + 1} C_{n + 1} s_{n} = 2^{n} S_{n}$

JEE Main 2025 (January)

LEVELJEE Main

The sum of all rational terms in the expansion of $(1 + 2^{1/3} + 3^{1/2})^{6}$ is equal to

JEE Main 2022 (27 July Shift 2)

LEVELJEE Main

Let for the $9^{t h}$ term in the binomial expansion of $(3 + 6 x)^{n}$ , in the increasing powers of $6 x$ , to be the greatest for $x = \frac{3}{2}$ , the least value of $n$ is $n_{0}$ . If $k$ is the ratio of the coefficient of $x^{6}$ to the coefficient of $x^{3}$ , then $k + n_{0}$ is equal to:

Micro-Topic Strategy Matrix

Binomial Expansion for Positive Integral Index

General Term and Middle Term

Properties of Binomial Coefficients

Greatest Term in Expansion

Multinomial Theorem

Binomial Theorem for any Index

Chapter Elite-20 Challenge Zone

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let n be any positive integer. Prove that ∑k=0m​(n2n−k​)(k2n−k​)​(2n−4k+1)2n−2k=(n−m2n−2m​)(mn​)​2n−2m for each non-negative integer m≤n.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Prove that \sum_{r=1}^k (-3)^{r-1} ^{3n}C_{2r-1} = 0, where k=(3n)/2 and n is an even positive integer.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Using mathematical induction, prove that mC0n​Ck​+mC1n​Ck−1​+....+mCkn​C0​=(m+n)Ck​, where m,n,k are positive integers, and pCq​=0 for p<q.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If α=1+∑r=16​(−3)r−112C2r−1​ then the distance of the point (12,3​) from the line αx−3​y+1=0 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let (2x2+3x+4)10=∑r=020​ar​xr. Then a13​a7​​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The term independent of x in the expansion of [x2/3−x1/3+1x+1​−x−x1/2x−1​]10,x=1, is equal to ___

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Using induction or otherwise, prove that for any non-negative integers m,n,r and k, ∑m=0k​(n−m)m!(r+m)!​=k!(r+k+1)!​[r+1n​−r+2k​]

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the fourth term in the binomial expansion of (1/(101+log10​x)​+12x6​)6 is equal to 200, and x>1, then the value of x is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The maximum value of the term independent of 't' in the expansion of (tx1/5+t(1−x)1/10​)10 where x∈(0,1) is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The coefficient of x9 in the expansion of (1+x)(1+x2)(1+x3)…(1+x100) is \dots.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∑k=131​(31Ck​)(31Ck−1​)−∑k=130​(30Ck​)(30Ck−1​)=(30!)(31!)α(60!)​, where α∈R, then the value of 16α is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the maximum value of the term independent of t in the expansion of (t2x51​+t(1−x)101​​)15,x≥0, is K, then 8K is equal to ______.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Given sn​=1+q+q2+....+qn; Sn​=1+2q+1​+(2q+1​)2+....+(2q+1​)n,q=1 Prove that n+1C1​+n+1C2​s1​+n+1C3​s2​+....+n+1Cn+1​sn​=2nSn​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of all rational terms in the expansion of (1+21/3+31/2)6 is equal to

Let $n$ be any positive integer. Prove that $\sum_{k = 0}^{m} \frac{( k 2 n - k )}{( n 2 n - k )} (2 n - 4 k + 1) 2^{n - 2 k} = \frac{( m n )}{( n - m 2 n - 2 m )} 2^{n - 2 m}$ for each non-negative integer $m \leq n$ .

Prove that $\sum_{r=1}^k (-3)^{r-1} ^{3n}C_{2r-1} = 0$ , where $k = (3 n) /2$ and $n$ is an even positive integer.

Using mathematical induction, prove that $^{m} C_{0}^{n} C_{k} +^{m} C_{1}^{n} C_{k - 1} + .... +^{m} C_{k}^{n} C_{0} =^{(m + n)} C_{k}$ , where $m, n, k$ are positive integers, and $^{p} C_{q} = 0$ for $p < q$ .

If $α = 1 + \sum_{r = 1}^{6} (- 3)^{r - 1}^{12} C_{2 r - 1}$ then the distance of the point $(12, 3)$ from the line $α x - 3 y + 1 = 0$ is

Let $(2 x^{2} + 3 x + 4)^{10} = \sum_{r = 0}^{20} a_{r} x^{r}$ . Then $\frac{a _{7}}{a _{13}}$ is equal to

The term independent of $x$ in the expansion of $[\frac{x + 1}{x ^{2/3} - x ^{1/3} + 1} - \frac{x - 1}{x - x ^{1/2}}]^{10}, x \neq = 1$ , is equal to ___

Using induction or otherwise, prove that for any non-negative integers $m, n, r$ and $k$ , $\sum_{m = 0}^{k} (n - m) \frac{( r + m )!}{m !} = \frac{( r + k + 1 )!}{k !} [\frac{n}{r + 1} - \frac{k}{r + 2}]$

If the fourth term in the binomial expansion of $(1/ (1 0^{1 + l o g_{10} x}) + 12 x^{6})^{6}$ is equal to 200, and $x > 1$ , then the value of x is :

The maximum value of the term independent of 't' in the expansion of $(t x^{1/5} + \frac{( 1 - x ) ^{1/10}}{t})^{10}$ where $x \in (0, 1)$ is:

The coefficient of $x^{9}$ in the expansion of $(1 + x) (1 + x^{2}) (1 + x^{3}) \dots (1 + x^{100})$ is \dots.

If $\sum_{k = 1}^{31} (^{31} C_{k}) (^{31} C_{k - 1}) - \sum_{k = 1}^{30} (^{30} C_{k}) (^{30} C_{k - 1}) = \frac{α ( 60 !)}{( 30 !) ( 31 !)}$ , where $α \in R$ , then the value of $16 α$ is equal to

If the maximum value of the term independent of $t$ in the expansion of $(t^{2} x^{\frac{1}{5}} + \frac{( 1 - x ) ^{\frac{1}{10}}}{t})^{15}, x \geq 0$ , is $K$ , then $8 K$ is equal to ______.

Given $s_{n} = 1 + q + q^{2} + .... + q^{n}$ ; $S_{n} = 1 + \frac{q + 1}{2} + (\frac{q + 1}{2})^{2} + .... + (\frac{q + 1}{2})^{n}, q \neq = 1$ Prove that $^{n + 1} C_{1} +^{n + 1} C_{2} s_{1} +^{n + 1} C_{3} s_{2} + .... +^{n + 1} C_{n + 1} s_{n} = 2^{n} S_{n}$

The sum of all rational terms in the expansion of $(1 + 2^{1/3} + 3^{1/2})^{6}$ is equal to