Binomial Theorem Questions with Solutions

13 questions1 easy10 medium2 hard

The binomial theorem gives the expansion of (a + b)^n for a positive integer n as the sum of terms nCr a^(n-r) b^r. This topic covers the general term T(r+1) = nCr a^(n-r) b^r, finding a specific term or the coefficient of a power of x, the term independent of x, the middle term(s), and properties of binomial coefficients. Almost every question is solved using the general term. Write T(r+1), simplify the power of x, and set it equal to the power you need (or to zero for the term independent of x) to find r. Remember that the expansion has n + 1 terms, so there is one middle term when n is even and two when n is odd. The sum of all binomial coefficients is 2^n. Practise the binomial theorem questions below, each with a step-by-step solution.

All Questions Easy Medium Hard

Q1.
The coefficient of $x^{10}$ in the expansion of $(x^{2} + \frac{1}{x})^{12} + (x + \frac{1}{x ^{2}})^{12}$ is:
A. 0B. 12C. 66D. 112
mediumAsked in NIMCET 2026View solution ›
Q2.
The coefficient of $x^{n}$ in the expansion of $(1 + x)^{2 n}$ and $(1 + x)^{2 n - 1}$ are in the ratio
A. $1 : 2$ B. $1 : 3$ C. $3 : 1$ D. $2 : 1$
mediumAsked in MAH MCA CET 2025View solution ›
Q3.
The coefficient of $x^{n}$ in the expansion of $(1 + x)^{2 n}$ and $(1 + x)^{2 n - 1}$ are in the ratio
A. $1 : 2$ B. $1 : 3$ C. $3 : 1$ D. $2 : 1$
mediumAsked in MAH MCA CET 2025View solution ›
Q4.
The coefficient of $x^{n}$ in the expansion of $(1 - 9 x + 20 x^{2})^{- 1}$ is
A. $5^{n + 1} - 4^{n + 1}$ B. $2 n + 1$ C. $20^{n + 10}$ D. $9^{2 n}$
mediumAsked in MAH MCA CET 2024View solution ›
Q5.
The coefficient of $x^{50}$ in $(1 + x)^{1000} + 2 x (1 + x)^{999} + 3 x^{2} (1 + x)^{998} + \dots + 1001 x^{1000}$ is
A. $^{1005} C_{50}$ B. $^{1005} C_{48}$ C. $^{1002} C_{50}$ D. $^{1002} C_{51}$
hardAsked in NIMCET 2023View solution ›
Q6.
If $\sum_{K = 0}^{2 n} (- 1)^{K} (^{2 n} C_{K})^{2} = A$ , then $\sum_{K = 0}^{2 n} (- 1)^{K} (K - 2 n) (^{2 n} C_{K})^{2}$ is equal to
A. $n A$ B. $- n A$ C. $0$ D. $A$
hardAsked in NIMCET 2011View solution ›
Q7.
If $(1 + x)^{n} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{n} x^{n}$ , then $(1 + \frac{a _{1}}{a _{0}}) (1 + \frac{a _{2}}{a _{1}}) (1 + \frac{a _{3}}{a _{2}}) \dots (1 + \frac{a _{n}}{a _{n - 1}})$ is equal to
A. $\frac{n ^{n}}{n !}$ B. $\frac{( n + 1 ) ^{n}}{n !}$ C. $\frac{n ^{n + 1}}{( n + 1 ) !}$ D. $\frac{( n - 1 ) ^{n}}{n !}$
mediumAsked in NIMCET 2010View solution ›
Q8.
If $(1 + x - 2 x^{2})^{6} = 1 + a_{1} x + a_{2} x^{2} + \dots + a_{12} x^{12}$ , then the value of $a_{2} + a_{4} + a_{6} + \dots + a_{12}$ is:
A. 1024B. 64C. 32D. 31
mediumAsked in NIMCET 2009 +1View solution ›
Q9.
In the expression $(x + 1) (x + 4) (x + 9) (x + 16) \dots (x + 400)$ , the coefficient of $x^{19}$ is
A. $2870$ B. $210$ C. $4001$ D. $1900$
easyAsked in NIMCET 2008View solution ›
Q10.
If x is so small that $x^{2}$ and higher powers of x can be neglected, then $\frac{( 9 + 2 x ) ^{1/2} ( 3 + 4 x )}{( 1 - x ) ^{1/5}}$ is approximately equal to
A. $9 + \frac{74}{15} x$ B. $9 + \frac{74}{5} x$ C. $3 + \frac{74}{15} x$ D. $3 + \frac{74}{5} x$
mediumAsked in NIMCET 2014View solution ›
Q11.
If $(1 - x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{2 n} x^{2 n}$ , then $a_{0} + a_{2} + a_{4} + \dots + a_{2 n}$ is
A. $\frac{3 ^{n} + 1}{2}$ B. $\frac{3 ^{n} - 1}{2}$ C. $\frac{1 - 3 ^{n}}{2}$ D. $3^{n} + \frac{1}{2}$
mediumAsked in NIMCET 2017View solution ›
Q12.
The coefficient of $x^{n}$ in the expansion of $(1 - 2 x + 3 x^{2} - 4 x^{3} + \dots t o \infty)^{- n}$ is
A. $\frac{( 2 n ) !}{n ! ( n - 1 ) !}$ B. $\frac{( 2 n ) !}{[ ( n - 1 ) ! ] ^{2}}$ C. $\frac{( 2 n ) !}{( n ! ) ^{2}}$ D. None of these
mediumAsked in NIMCET 2018View solution ›
Q13.
If a and b are the greatest values of $^{2 n} C_{r}$ and $^{2 n - 1} C_{r}$ respectively, then
A. a = 2bB. b = 2aC. a = bD. $a^{2} = 2 b^{2}$
mediumAsked in NIMCET 2019View solution ›