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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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631 Medium

631 of 951

Application +4 | -1

If sum of coefficients in the expression of

\left(\alpha x^2-2 x+1\right)^{35}

is equal to sum of coefficients in the expansion of

(x-\alpha y)^{35}

, then

\alpha

is equal to

View answer and explanation

Correct option: B

Option B (1) is correct because substituting x=1 into the polynomial and equating the coefficients leads to the solution α = 1.

632 Medium

632 of 951

Analyzing +4 | -1

\frac{{ }^n C_1}{{ }^n C_0}+2 \frac{{ }^n C_2}{{ }^n C_1}+3 \frac{{ }^n C_3}{{ }^n C_2}+\ldots+n \frac{{ }^n C_n}{{ }^n C_{n-1}}

is equal to

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because the given expression simplifies to the sum of the first n natural numbers, which is

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

633 Easy

633 of 951

Application +4 | -1

In the expansion of

{\left( {1 + x} \right)^{2n}}\left( {n \in N} \right),

the coefficients of

{\left( {p + 1} \right)^{th}}\,\mathrm{and}\,{\left( {p + 3} \right)^{th}}

terms are equal, then

View answer and explanation

Correct option: B

p = n - 1

Option B (

p = n - 1

) is correct because equating the coefficients of the (p+1)th and (p+3)th terms leads to the equation

p = n - 1

634 Easy

634 of 951

Analyzing +4 | -1

Sum of the infinite series

1 + \frac{1}{3}.\frac{1}{2} + \frac{2}{3}.\frac{5}{6}.\frac{1}{{{2^2}}} + \frac{2}{3}.\frac{5}{6}.\frac{8}{9}.\frac{1}{{{2^3}}} + \ldots \infty \;\mathrm{is}

View answer and explanation

Correct option: B

{4^{1/3}}

Option B (

{4^{1/3}}

) is correct because by comparing the given series with the binomial expansion of

(1+x)^n

, we can determine the values of

n

and

x

, and then find the sum.

635 Easy

635 of 951

Application +4 | -1

Let

f(n)=\sum\limits_{{k=1}}^n k^2 C_k^2

then

\frac{f(5)}{875}

View answer and explanation

Correct option: B

Option B (2) is correct because the value of the summation is calculated using combinatorial identities and simplification.

636 Hard

636 of 951

Application +4 | -1

For

n > 2

, let

C_r={ }^n C_r

and

f(n)=3 C_2+5 C_4+7 C_6+\ldots

, then

2^{-5} f(6)

equals

View answer and explanation

Correct option: D

Option D (4) is correct because the value of

2^{-5} f(6)

is calculated by using the given recurrence relation and binomial coefficients.

637 Easy

637 of 951

Application +4 | -1

If sum of the coefficient in the binomial expansion of

(1+2 x)^n

is 6561, then coefficient of the fourth term is:

View answer and explanation

Correct option: C

448

Option C (448) is correct because the coefficient of the fourth term in the expansion of (1+2x)⁸ is 448.

638 Easy

638 of 951

Application +4 | -1

In the expansion of

{\left( {{x^3} - \frac{1}{{{x^2}}}} \right)^n},n \in N,

if the sum of the coefficients of

{x^5}

and

{x^{10}}

is 0, then

n =

View answer and explanation

Correct option: C

Option C (15) is correct because setting the sum of coefficients of

{x^5}

and

{x^{10}}

to 0 in the expansion of

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

leads to the equation

n = 15

639 Easy

639 of 951

Application +4 | -1

(1+x)\left(1+x+x^2\right)\left(1+x+x^2+x^3\right) \ldots(1+x\left.+x^2+\ldots+x^n\right)=a_0+a_1 x+a_2 x^2+\ldots+a_m x^m

then the value of

a_1

View answer and explanation

Correct option: C

n

Option C (

n

) is correct because the coefficient of

x

in the given product is equivalent to the coefficient of

x

(1+x)^n

, which is

n

640 Medium

640 of 951

Analyzing +4 | -1

The sum of the series

\sum \limits_{r = 0}^n {\left( { - 1} \right)^r}{\;^n}{C_r}\left( {\frac{1}{{{2^r}}} + \frac{{{3^r}}}{{{2^{2r}}}} + \frac{{{7^r}}}{{{2^{3r}}}} + \frac{{{{15}^r}}}{{{2^{4r}}}} + ... + m\mathrm{\;terms}} \right)

View answer and explanation

Correct option: A

\frac{{{2^{mn}} - 1}}{{{2^{mn}}\left( {{2^n} - 1} \right)}}

Option A (

\frac{{{{2^{mn}} - 1}}}{{{{2^{mn}}\left( {{{2^n} - 1}} \right)}}}

) is the correct answer because it represents the sum of the given series after simplification using binomial theorem and geometric progression.

641 Medium

641 of 951

Application +4 | -1

The smallest positive integer greater than

(1+0.0001)^{10^3}

View answer and explanation

Correct option: C

Option C (2) is correct because the expression can be approximated using the binomial theorem, and the result is slightly greater than 1 but less than 2.

642 Easy

642 of 951

Application +4 | -1

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

, then

C_0^2 + C_1^2 + C_2^2 + C_3^2 + ... + C_n^2

is equal to

View answer and explanation

Correct option: B

\frac{{\left( {2n} \right)!}}{{n!n!}}

Option B (

\frac{{\left( {2n} \right)!}}{{n!n!}}

) is correct because it represents the sum of squares of binomial coefficients, which is equivalent to the middle term in the expansion of (1+x)⁽²ⁿ⁾.

643 Easy

643 of 951

Analyzing +4 | -1

Coefficient of

x^{10}

in the expansion of

\left(1+x^2\right)^{10}\left(x^2+2+\frac{1}{x^2}\right)^5

View answer and explanation

Correct option: C

Option C (0) is correct because after simplification, the coefficient of

x^{10}

in the expansion becomes the coefficient of

x^{15}

(1+x^2)^{15}

, which is 0.

644 Hard

644 of 951

Application +4 | -1

The coefficient of

{x^{53}}

in the following expansion

\sum \limits_{m = 0}^{100\;} {\;^{100}}{C_m}{\left( {x - 3} \right)^{100 - m}}{.2^m}

View answer and explanation

Correct option: C

- {\;^{100}}{C_{53}}

Option C (

- {\;^{100}}{C_{53}}

) is correct because the given sum simplifies to

(x-1)^{100}

, and the coefficient of

x^{53}

-{\;^{100}}{C_{53}}

645 Medium

645 of 951

Application +4 | -1

The sum of the numerical coefficients in the expansion of

\left(1+\frac{x}{3}+\frac{2 y}{3}\right)^{12}

View answer and explanation

Correct option: C

2^{12}

Option C (

2^{12}

) is correct because the sum of numerical coefficients is found by substituting x=1 and y=1 into the expression.

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SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs

646 Medium

646 of 951

Application +4 | -1

If the coefficient in the third term of the expansion of

\left(x^2+\frac{1}{4}\right)^n

when expanded in decreasing powers of

x

is 31, then

n

is equal to

View answer and explanation

Correct option: D

Option D (32) is correct because the third term in the binomial expansion is given by

{}^nC_2 (x^2)^{n-2} (1/4)^2

, and solving for n yields 32.

647 Easy

647 of 951

Application +4 | -1

If for

1 \leq \mathrm{m} \leq \mathrm{n}, \mathrm{f}(\mathrm{m}, \mathrm{n})={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{0}}-{ }^{\mathrm{n}} \mathrm{C}_1+{ }^{\mathrm{n}} \mathrm{C}_2-\ldots . .+(-1)^{\mathrm{m}-1} \cdot{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{m}-1}

, find

\mathrm{f}(7, 8)

View answer and explanation

Correct option: A

Option A (7) is correct because by simplifying the expression and using combinatorial identities, f(7, 8) is found to be 7.

648 Easy

648 of 951

Application +4 | -1

If the term independent of

x

in the expansion of

{\left( {\frac{3}{2}{x^2} - \frac{1}{{3x}}} \right)^9}\,\,\mathrm{is}\,\,\,k,\,\,\mathrm{then}\,18\,k

is equal to

View answer and explanation

Correct option: B

Option B (7) is correct because it is the value of 18k, where k is the term independent of x in the given binomial expansion.

649 Easy

649 of 951

Understanding +4 | -1

The expansion of

(8-3 x)^{3 / 2}

in terms of powers of

x

is valid only if

View answer and explanation

Correct option: B

|x| < \frac{8}{3}

Option B (

|x| < \frac{8}{3}

) is correct because the binomial expansion of

(1+x)^n

is valid only when

|x| < 1

. In this case, we need

|\frac{{3x}}{{8}}| < 1

650 Easy

650 of 951

Understanding +4 | -1

The number of terms in the expansion of

{\left( {a + b + c} \right)^n}

will be

View answer and explanation

Correct option: C

\frac{{\left( {n + 1} \right)\left( {n + 2} \right)}}{2}

Option C,

\frac{{\left( {n + 1} \right)\left( {n + 2} \right)}}{2}

, is the correct formula for calculating the number of terms in the expansion of

{\left( {a + b + c} \right)^n}

651 Easy

651 of 951

Application +4 | -1

The 7th term in

\left(\frac{1}{y}+y^2\right)^{10}

, when expanded in descending power of

y

, is

View answer and explanation

Correct option: C

210\, y^2

Option C (

210\, y^2

) is correct because it is the 7th term in the expansion.

652 Medium

652 of 951

Analyzing +4 | -1

{\left( {1 + x} \right)^{2n}} = {a_0} + {a_1}\;x + {a_2}\;{x^2} + \ldots + {a_{2n}}{x^{2n}},

then

{\left( {{a_0} - {a_2} + {a_4} - {a_6} + \ldots - {a_{2n}}} \right)^2} + {\left( {{a_1} - {a_3} + {a_5} - {a_7} + \ldots + {a_{2n - 1}}} \right)^2}

is equal to

View answer and explanation

Correct option: B

{4^n}

Option B (

{4^n}

) is correct because by substituting

x = i

and

x = -i

into the given expression and using complex number properties, we can simplify the expression to

{4^n}

653 Medium

653 of 951

Analyzing +4 | -1

Let

K

be the sum of the coefficients of the odd powers of

x

in the expansion of

(1+x)^{99}

. Let

a

be the middle term in the expansion of

\left(2+\frac{1}{\sqrt{2}}\right)^{200}

. If

\frac{{ }^{200} C_{99} K}{a}=\frac{2^{\prime} m}{n}

, where

m

and

n

are odd numbers, then the ordered pair

(I, n)

is equal to

View answer and explanation

Correct option: C

(50,101)

Option C (

(50,101)

) is correct because it is obtained by calculating the sum of coefficients of odd powers, the middle term, and then simplifying the given expression.

654 Easy

654 of 951

Analyzing +4 | -1

Let

{(1 + x + 2{x^2})^{20}} = {a_0} + {a_1}x + {a_2}{x^2} + ..... + {a_{40}}{x^{40}}

, then

{a_1} + {a_3} + {a_5} + .... + {a_{37}}

is equal to

View answer and explanation

Correct option: B

{2^{19}}({2^{20}} - 21)

Option B (

{2^{19}}({2^{20}} - 21)

) is correct because it is derived by substituting x = 1 and x = -1 into the polynomial and then subtracting the resulting equations.

655 Easy

655 of 951

Application +4 | -1

Let

{\left( {1 + x} \right)^n} = 1 + {a_1}x + {a_2}{x^2} + .... + {a_n}{x^n}

. If

{a_1},\;{a_2}\,\mathrm{and\;}{a_3}

are in AP, then the value of

n

View answer and explanation

Correct option: D

Option D (7) is correct because by equating the given condition that a1, a2, and a3 are in AP, we can solve for n, which results in n=7.

656 Medium

656 of 951

Application +4 | -1

{\left( {5 + 2\sqrt 6 } \right)^n} = I + f;n,\;I \in N

and

0 \le f < 1,

then

I

equals

View answer and explanation

Correct option: D

\frac{1}{{1 - f}} - f

Option D (

\frac{1}{{1 - f}} - f

) is correct because it is derived from the properties of conjugate surds and integer/fractional parts.

657 Easy

657 of 951

Application +4 | -1

Let

{C_1},\;{C_2},\;{C_3}

are the usual binomial coefficients. Let

S = {C_1} + 2{C_2} + 3{C_3} + ... + n{C_n}

, then

S

equals

View answer and explanation

Correct option: C

n{2^{n - 1}}

Option C (

n{2^{n - 1}}

) is correct because it is the sum of the given series of binomial coefficients.

658 Medium

658 of 951

Analyzing +4 | -1

x=(2+\sqrt{3})^n

, then the value of

x-x^2+x[x]

, where [] denotes the greatest integer function, is equal to

View answer and explanation

Correct option: A

Option A (1) is correct because by considering the conjugate and using the properties of greatest integer function, we can simplify the expression to 1.

659 Easy

659 of 951

Application +4 | -1

{\left( {1 - x + {x^2}} \right)^n} = {a_0} + {a_1}x + ....{a_{2n}}{x^{2n}}

then the value of

{a_0} + {a_2} + {a_4} + .... + {a_{2n}}

View answer and explanation

Correct option: D

\frac{{{3^n} + 1}}{2}

Option D (

\frac{{{3^n} + 1}}{2}

) is correct because substituting x = 1 and x = -1 into the given expression and adding them allows us to isolate the desired sum.

660 Medium

660 of 951

Application +4 | -1

\left(1+x-3 x^2\right)^{2145}=a_0+a_1 x+a_2 x^2+\ldots

, then

a_0-a_1+a_2-a_3+\ldots

ends with___.

View answer and explanation

Correct option: A

Option A (3) is correct because substituting x = -1 into the given expression gives (-3)²¹⁴⁵, and the last digit of this number is 3.

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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