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JEE • MATHEMATICS • BINOMIAL THEOREM

Binomial theorem practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 91-120 of 951 questions

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91 Medium

91 of 951

Application +4 | -1

\left(x-2+\frac{1}{x}\right)^{30}=n_0 x^{30}+n_1 x^{29}+\ldots \ldots+n_{29} x+n_{30}+n_{31} x^{-1}+\ldots \ldots+n_{60} x^{-30}

and

C=n_0+n_1+n_2+\ldots \ldots+n_{60}

. Find the value of

(a+b)

C-n_{30}=-\left(\begin{array}{l}a \\ b\end{array}\right)

\left[{\left(\begin{array}{l}n \\ r\end{array}\right) \text{ denotes }{ }^n C_r}\right]

View answer and explanation

Correct option: A

Option A (90) is correct because the sum of coefficients C is 0, and C - n_30 = - (a choose b), leading to a = 60, b = 30, and a+b = 90.

92 Medium

92 of 951

Application +4 | -1

The interval in which

x

must lie so that the numerically greatest term in the expansion of

{\left( {1 - x} \right)^{21}}

has the numerically greatest coefficient, is

View answer and explanation

Correct option: B

\left( {\frac{5}{6},\frac{6}{5}} \right)

Option B

\left( {\frac{5}{6},\frac{6}{5}} \right)

is correct because it represents the interval where the numerically greatest term has the numerically greatest coefficient.

93 Medium

93 of 951

Application +4 | -1

\frac{1}{\left(\begin{array}{c}101 \\ 0\end{array}\right)}-\frac{1}{\left(\begin{array}{c}101 \\ 1\end{array}\right)}+\frac{1}{\left(\begin{array}{c}101 \\ 2\end{array}\right)}-\frac{1}{\left(\begin{array}{c}101 \\ 3\end{array}\right)}+\ldots+\frac{(-1)^{101}}{\left(\begin{array}{l}101 \\ 101\end{array}\right)}=

View answer and explanation

Correct option: D

Option D (0) is correct because the terms cancel out in pairs due to the symmetry of binomial coefficients and alternating signs.

94 Hard

94 of 951

Application +4 | -1

{S}_k=\sum\limits_{{r=1}}^n r^k

, then

\sum\limits_{{r=1}}^{k+1}\left(\begin{array}{c}k+1 \\ r\end{array}\right) {S}_{k+1 - r}=

View answer and explanation

Correct option: A

(n+1)^{k+1}-1

Option A (

(n+1)^{k+1}-1

) is correct because the given summation represents the expansion of

(n+1)^{k+1}

minus the term for

r=0

which is 1.

95 Easy

95 of 951

Analyzing +4 | -1

The value of the sum of the series

3.{\;^n}{C_0} - 8.{\;^n}{C_1} + 13\;{\;^n}{C_2} - 18.{\;^n}{C_3} + \; \ldots

upto

\left( {n + 1} \right)

terms is

View answer and explanation

Correct option: A

Option A (0) is correct because the sum of the series is zero due to the alternating signs and the properties of binomial coefficients.

96 Medium

96 of 951

Analyzing +4 | -1

{\left( {1 + x} \right)^n} = {C_0} + {C_1}x + {C_2}{x^2} + ... + {C_n}{x^n}

, then the value of

\sum \limits_{0 \le r < s \le n} \sum \left( {r + s} \right)\left( {{C_r} + {C_s}} \right)

View answer and explanation

Correct option: A

{n^2}{.2^n}

Option A is correct because the value of the summation is found to be n² * 2^n by using properties of binomial coefficients and summations.

97 Medium

97 of 951

Application +4 | -1

S_n=\sum_{r=0}^n \frac{1}{{ }^n C_r}

and

t_n=\sum_{r=0}^n \frac{r}{{ }^n C_r}

, then

\frac{t_n}{S_n}

is equal to-

View answer and explanation

Correct option: A

\frac{1}{2} \mathbf{n}

Option A (

\frac{1}{2}n

) is correct because by using the property

{ }^n C_r = { }^n C_{n-r}

and summing the series, we can find the ratio

\frac{t_n}{S_n}

98 Medium

98 of 951

Application +4 | -1

The sum of coefficients in the expansion of

\left(1+x-3 y^2\right)^{2163}

View answer and explanation

Correct option: B

-1

Option B (-1) is correct because the sum of coefficients is found by substituting x=1 and y=1 into the expression, resulting in (1+1-3)⁽²¹⁶³⁾ = (-1)⁽²¹⁶³⁾ = -1.

99 Medium

99 of 951

Application +4 | -1

The remainder when

4^{101}

is divided by 101 is

View answer and explanation

Correct option: A

Option A (4) is correct because, according to Fermat's Little Theorem, a^(p-1) is congruent to 1 mod p, where p is a prime number.

100 Hard

100 of 951

Analyzing +4 | -1

Let

n

be an even integer and

k=\frac{3 n}{2}

. Then, the value of

\sum\limits_{{r=1}}^k(-3)^{r-1}{ }\,\,^{3 n} C_{2 r-1}

View answer and explanation

Correct option: A

Option A (0) is correct because the given summation can be simplified using complex numbers and De Moivre's theorem, which results in the sum being equal to 0.

101 Easy

101 of 951

Application +4 | -1

The coefficient of

{x^{ - 10}}

{\left( {{x^2} - \frac{1}{{{x^3}}}} \right)^{10}},

View answer and explanation

Correct option: B

210

Option B (210) is correct because the coefficient of

{x^{ - 10}}

can be found by determining the correct term in the binomial expansion.

102 Medium

102 of 951

Application +4 | -1

The fractional part of

\frac{2^{4 n}}{15}

View answer and explanation

Correct option: A

\frac{1}{15}

Option A (

\frac{{1}}{{15}}

) is correct because the fractional part of

\frac{{2^{{4n}}}}{{15}}

\frac{{1}}{{15}}

, which is found using binomial expansion.

103 Medium

103 of 951

Application +4 | -1

The coefficient of

{x^n}

in the expansion of

\frac{1}{{\left( {1 - x} \right)\left( {3 - x} \right)}},

View answer and explanation

Correct option: A

\frac{{{3^{n + 1}} - 1}}{{{{2.3}^{n + 1}}}}

Option A (

\frac{{{3^{n + 1}} - 1}}{{{{2.3}^{n + 1}}}}

) is correct because it is the coefficient of

{x^n}

obtained by partial fraction decomposition and binomial expansion.

104 Medium

104 of 951

Analyzing +4 | -1

{\left( {1 + x + {x^2}} \right)^n} = \sum \limits_{r = 0}^{2n} {a_r}{x^r}

then,

{a_1} - 2{a_2} + 3{a_3} \ldots - 2n{a_{2n}}\;

is equal to

View answer and explanation

Correct option: B

- n

The correct answer is B (-n) because by differentiating the given equation and substituting x = -1, we can find the value of the expression.

105 Medium

105 of 951

Application +4 | -1

Given that positive integers

r>1, n>2

and the coefficient of

(3r)

th and

(r+2)

nd terms in the expansion of

(1+x)^{2 n}

are equal, then

n=

View answer and explanation

Correct option: A

n=2 r

Option A is correct because equating the coefficients and using the properties of binomial coefficients leads to n = 2r.

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106 Easy

106 of 951

Application +4 | -1

The 6th term in the expansion of

{\left( {2{x^2} - \frac{1}{{3{x^2}}}} \right)^{10}}

View answer and explanation

Correct option: B

- \frac{{896}}{{27}}

Option B (

- \frac{{896}}{{27}}

) is correct because it is the 6th term in the expansion of the given binomial expression, calculated using the binomial theorem.

107 Medium

107 of 951

Application +4 | -1

a, b, p \in {N}

and

p

is prime, then

(a+b)^p-a^p-b^p

is divisible by

View answer and explanation

Correct option: A

p

Option A (

p

) is correct because

(a+b)^p-a^p-b^p

can be expressed as a sum of terms, each of which is divisible by

p

due to the properties of binomial coefficients and prime numbers.

108 Easy

108 of 951

Application +4 | -1

1.2 .3+2.3 .4+3.4 .5+\ldots

upto

n

terms is equal to

View answer and explanation

Correct option: C

\frac{1}{4} n(n+1)(n+2)(n+3)

Option C (

\frac{1}{4} n(n+1)(n+2)(n+3)

) is correct because it represents the sum of the series 1.2.3 + 2.3.4 + 3.4.5 + ... up to n terms.

109 Easy

109 of 951

Application +4 | -1

If coefficient of 9th, 10th and 11th terms of

(1+x)^n

are in A.P., then value of

n

can be

View answer and explanation

Correct option: C

Option C (23) is correct because when n=23, the coefficients of the 9th, 10th, and 11th terms of (1+x)^n are in arithmetic progression.

110 Medium

110 of 951

Analyzing +4 | -1

Coefficient of

x^{301}

(1+x)^{500}+x(1+x)^{499}+x^2(1+x)^{498}+\cdots+x^{500}

is equal to

View answer and explanation

Correct option: C

{ }^{501} C_{301}

Option C (

{ }^{501} C_{301}

) is correct because the coefficient of

x^{301}

can be simplified using combinatorial identities to

{ }^{501} C_{301}

111 Easy

111 of 951

Application +4 | -1

The coefficient of

x^{10}

in the expansion of

\left(1+x^2-x^3\right)^8

, is

View answer and explanation

Correct option: B

476

Option B (476) is correct because the coefficient of

x^{10}

in the expansion of

(1+x^2-x^3)^8

can be found by considering the relevant terms in the multinomial expansion.

112 Easy

112 of 951

Application +4 | -1

It the number of terms in the expansion of

{\left( {1 - \frac{2}{x} + \frac{4}{{{x^2}}}} \right)^n},\,\,x \ne 0

, is 28, then the sum of the coefficients of all the terms in this expansion is

View answer and explanation

Correct option: D

729

The correct answer is D (729) because the sum of the coefficients is found by substituting x = 1 into the expansion.

113 Medium

113 of 951

Application +4 | -1

(1+x)^{100}=c_0+c_1 x+c_2 x^2+\ldots+c_{100} x^{100}

, then

2^1 \frac{c_0}{1}+2^2 \frac{c_1}{2}+2^3 \frac{c_2}{3}+\ldots+2^{101} \frac{c_{100}}{101}=

View answer and explanation

Correct option: B

\frac{3^{101}-1}{101}

Option B is the correct answer because the given expression can be evaluated using integration of the binomial expansion.

114 Easy

114 of 951

Understanding +4 | -1

The largest coefficient in the expansion of

{\left( {1 + x} \right)^{24}}

View answer and explanation

Correct option: C

{\;^{24}}{C_{12}}

Option C (

{\;^{24}}{C_{12}}

) is correct because in the expansion of

{\left( {1 + x} \right)^{24}}

, the largest coefficient is the middle term, which is the 13th term, and its coefficient is

{\;^{24}}{C_{12}}

115 Medium

115 of 951

Analyzing +4 | -1

Sum of the first 20 terms of the series

\frac{1}{(2)(4)}+\frac{(1)(3)}{(2)(4)(6)}+\frac{(1)(3)(5)}{(2)(4)(6)(8)}+\ldots

View answer and explanation

Correct option: C

\frac{1}{2}-\frac{1}{2^{42}}\left({ }^{42} C_{21}\right)

Option C,

\frac{{1}}{{2}}-\frac{{1}}{{2^{42}}}\left({ }^{{42}} C_{{21}}\right)

, is correct because the series can be expressed as a telescoping sum, and the sum of the first 20 terms is equal to

b_1 - b_{21}

116 Medium

116 of 951

Application +4 | -1

The value of

{ }^{n-1} C_0 \cdot{ }^n C_1+{ }^{n-1} C_1 \cdot{ }^n C_2+{ }^{n-1} C_2 \cdot{ }^n C_3 +\ldots+{ }^{n-1} C_{n-1} \cdot{ }^n C_n \text { is equal to }

View answer and explanation

Correct option: B

{ }^{2 n-1} C_{n-1}

Option B (

{ }^{2 n-1} C_{n-1}

) is correct because the sum represents the coefficient of x^(n-1) in the product of two binomial expansions.

117 Medium

117 of 951

Application +4 | -1

If terms independent of

x

in the expansion of

\left(3 x-\frac{1}{x}\right)^{20}

and

\left(x+\frac{\sqrt[3]{3^{10}}}{x}\right)^{18}

are

A

and

B

respectively then

\left(\frac{9}{38} A+B\right)

equals:

View answer and explanation

Correct option: B

3^{10} \cdot { }^{19} C_9

Option B is correct because after calculating A and B, substituting into the expression yields

3^{{10}} \cdot {{}^{{19}}}C_9

118 Easy

118 of 951

Application +4 | -1

(1+p x)^n=1+24 x+2642 x^2+\ldots

then

View answer and explanation

Correct option: B

p=2, n=12

Option B (p=2, n=12) is correct. The values of p and n are determined by comparing the coefficients of x and x² in the binomial expansion.

119 Easy

119 of 951

Application +4 | -1

The coefficient of

x^n

in the expansion of

(1-x)^{-2}

View answer and explanation

Correct option: B

(n+1)

Option B (

(n+1)

) is correct because the coefficient of

x^n

in the expansion of

(1-x)^{-2}

is indeed

n+1

120 Easy

120 of 951

Application +4 | -1

In the expansion of

(x+a)^n

, if the sum of odd terms be

A

and the sum of even terms be

B

, then value of

\left(x^2-a^2\right)^n

View answer and explanation

Correct option: D

A^2-B^2

Option D (

A^2-B^2

) is correct because it directly results from multiplying (A+B) and (A-B), which represent (x+a)^n and (x-a)^n respectively.

Revision notes and key takeaways

Review BINOMIAL THEOREM in your textbook or class notes, then use this set to check your understanding.

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