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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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511 Medium

511 of 951

Application +4 | -1

Let

{\left( {1 + x} \right)^n} = \mathop {\sum \,}\limits_{r = 0}^n {a_r}\;{x^r}.

Then,

\left( {1 + \frac{{{a_1}}}{{{a_0}}}} \right)\left( {1 + \frac{{{a_2}}}{{{a_1}}}} \right) \ldots \left( {1 + \frac{{{a_n}}}{{{a_{n - 1}}}}} \right)\;

is equal to

View answer and explanation

Correct option: B

\frac{{{{\left( {n + 1} \right)}^n}}}{{n\;!}}

Option B is correct because the given expression simplifies to

\frac{{{{\left( {n + 1} \right)}^n}}}{{n\;!}}

after applying the binomial coefficient properties and simplifying the product.

512 Easy

512 of 951

Application +4 | -1

The term independent of

x

in the expansion of

{\left( {\sqrt {\frac{x}{3}} + \frac{3}{{2{x^2}}}} \right)^{10}}

will be

View answer and explanation

Correct option: B

\frac{5}{4}

Option B (

\frac{5}{4}

) is correct because it is the constant term obtained after expanding the given binomial and setting the power of x to zero.

513 Medium

513 of 951

Analyzing +4 | -1

The value of

1+\left(\frac{1}{2}+\frac{1}{3}\right) \frac{1}{4}+\left(\frac{1}{4}+\frac{1}{5}\right) \frac{1}{4^2}+\left(\frac{1}{6}+\frac{1}{7}\right) \frac{1}{4^3}+\ldots

View answer and explanation

Correct option: A

\log _e \sqrt{12}

Option A (

\log _e \sqrt{{12}}

) is correct because the given infinite series can be simplified and expressed in terms of logarithms, ultimately resulting in

\log _e \sqrt{{12}}

514 Easy

514 of 951

Application +4 | -1

If sum of the coefficients of the first, second and third terms of the expansion of

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

is 46, then the coefficient of the term that does not contain

x

View answer and explanation

Correct option: A

Option A (84) is correct because after finding m=9, the coefficient of the term independent of x in the expansion is calculated as

^{9}C_6 = 84

515 Medium

515 of 951

Application +4 | -1

The greatest term in the expansion of

\sqrt 3 {\left( {1 + \frac{1}{{\sqrt 3 }}} \right)^{20}}

View answer and explanation

Correct option: C

\frac{{25840}}{9}

Option C (

\frac{{25840}}{9}

) is correct because it is the greatest term calculated by finding the value of r for which T_{r+1} >= T_r in the binomial expansion.

516 Medium

516 of 951

Application +4 | -1

Coeffiecient of

t^{24}

in the expansion of

\left(1+t^2\right)^{12}\left(1+t^{12}\right)\left(1+t^{24}\right)

View answer and explanation

Correct option: A

{ }^{12} C_6+2

Option A (

{ }^{12} C_6+2

) is correct because the coefficient of t²⁴ is found by considering the different ways to obtain t²⁴ from the product of the three terms.

517 Easy

517 of 951

Application +4 | -1

The sum of the real values of x for which the middle term in the binomial expansion of

{\left( {\frac{{{x^3}}}{3} + \frac{3}{x}} \right)^8}

equals 5670 is

View answer and explanation

Correct option: B

Option B (0) is correct because the real values of x are

\pm \sqrt{3}

, and their sum is 0.

518 Medium

518 of 951

Application +4 | -1

The sum of coefficients of integral powers of

x

in the binomial expansion

{(1 - 2\sqrt x )^{50}}

View answer and explanation

Correct option: A

\frac{1}{2}({3^{50}} + 1)

Option A (

\frac{1}{2}({3^{50}} + 1)

) is correct because, by using the binomial theorem and considering only the integral powers of x, the sum of the coefficients simplifies to the given expression.

519 Easy

519 of 951

Application +4 | -1

Coefficient of the term independent of

x

in the expansion

{\left( {x + \frac{1}{{{x^2}}}} \right)^6}

is equal to

View answer and explanation

Correct option: B

Option B (15) is correct because it is the coefficient of the term where the power of x is zero in the given binomial expansion.

520 Medium

520 of 951

Application +4 | -1

The first integral term in the expansion of

(\sqrt{3}+\sqrt[3]{2})^9

, is the

View answer and explanation

Correct option: C

4^{\text {th}}

term

Option C,

4^{\text {th}}

term, is correct because the first integral term occurs when r=3 in the binomial expansion.

521 Medium

521 of 951

Analyzing +4 | -1

In the expansion of

(x+y+z)^{25}

View answer and explanation

Correct option: A

every term is of the form

{ }^{25} C_r \cdot{ }^r C_k \cdot x^{25-r} \cdot y^{r k} \cdot z^k

Option A is correct because it accurately represents the general form of terms in the multinomial expansion of (x+y+z)²⁵.

522 Easy

522 of 951

Application +4 | -1

The number of real negative terms in the binomial expansion of

(1+i x)^{4 n-2}, n \in N, n>0, i=\sqrt{-1}

, is

View answer and explanation

Correct option: A

n

Option A (

n

) is correct because the number of real negative terms in the binomial expansion of

(1+ix)^{4n-2}

can be determined by analyzing the powers of

i

and identifying when they result in real negative values.

523 Medium

523 of 951

Analyzing +4 | -1

The coefficient of

{x^4}

in the expansion of

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

View answer and explanation

Correct option: D

{\;^n}{C_4} + {\;^n}{C_2} + {\;^n}{C_1}.{\;^n}{C_2}

The correct answer is D (

{\;^n}{C_4} + {\;^n}{C_2} + {\;^n}{C_1}.{\;^n}{C_2}

) because it is found by expanding the given expression and identifying the coefficient of

{x^4}

524 Easy

524 of 951

Application +4 | -1

In the expansion of

{\left( {x - \frac{1}{x}} \right)^6}

, the coefficient of

{x^0}

View answer and explanation

Correct option: B

- 20

Option B (-20) is correct. The constant term in the expansion of (x - 1/x)⁶ is -20, found by determining the term where the power of x is 0.

525 Easy

525 of 951

Analyzing +4 | -1

\frac{1}{{1!\left( {n - 1} \right)!}} + \frac{1}{{3!\left( {n - 3} \right)!}} + \frac{1}{{5!\left( {n - 5} \right)!}} + ...

is equal to

View answer and explanation

Correct option: B

\frac{{{2^{n - 1}}}}{{n!}}

The correct answer is B (

\frac{{{2^{n - 1}}}}{{n!}}

) because it is derived from recognizing the sum as part of the binomial expansion of

(1+1)^n

and

(1-1)^n

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526 Hard

526 of 951

Analyzing +4 | -1

The coefficient of

x^{50}

in the expansion

(1+x)^{1000}+2 x(1+x)^{999}+3x^2(1+x)^{998} +\ldots+1001 x^{1000} \text { is }

View answer and explanation

Correct option: A

{ }^{1002} C_{50}

Option A (

{ }^{{1002}} C_{{50}}

) is correct because the coefficient of

x^{{50}}

is found by simplifying the given series into a binomial expansion and then extracting the required coefficient.

527 Medium

527 of 951

Analyzing +4 | -1

(1+x)^n=C_0+C_1 x+C_2 x^2+\ldots+C_n x^n

, then

\left(C_0+C_1\right)\left(C_1+C_2\right) \ldots\left(C_{n-1}+C_n\right)

=k \cdot C_1 C_2 C_3 \ldots C_n

, where

k

is equal to

View answer and explanation

Correct option: A

\frac{(n+1)^n}{n !}

Option A (

\frac{{(n+1)^n}}{{n !}}

) is correct because the given expression can be simplified using binomial coefficient identities.

528 Easy

528 of 951

Application +4 | -1

If the constant term in the expansion of

\left(\frac{\sqrt[5]{3}}{x}+\frac{2 x}{\sqrt[3]{5}}\right)^{12}, x \neq 0,

\alpha \times 2^8 \times \sqrt[5]{3},

then

25 \alpha

is equal to:

View answer and explanation

Correct option: C

693

The correct answer is C (693). The constant term in the binomial expansion is found by setting the power of x to zero and solving for the term.

529 Medium

529 of 951

Application +4 | -1

If the coefficients of

x^{7}

\left(a x^{2}+\frac{1}{2 b x}\right)^{11}

and

x^{-7}

\left(a x-\frac{1}{3 b x^{2}}\right)^{11}

are equal, then

View answer and explanation

Correct option: A

729 a b=32

Option A (

729 a b=32

) is correct because equating the coefficients of

x^7

and

x^{-7}

and simplifying leads to the relationship

729ab = 32

530 Medium

530 of 951

Analyzing +4 | -1

The expansion

\left[x+\left(x^3-1\right)^{1 / 2}\right]^5+\left[x-\left(x^3-1\right)^{1 / 2}\right]^5

is a polynomial of degree

View answer and explanation

Correct option: C

Option C (7) is correct because after expanding and simplifying the expression, the highest power of

x

is 7.

531 Easy

531 of 951

Application +4 | -1

The coefficient of

x^{53}

\sum\limits_{{r=0}}^{100}{ }^{100} C_r(x-3)^{100-r} 2^r

View answer and explanation

Correct option: C

{ }^{-100} C_{53}

Option C (

{}^{-100} C_{53}

) is correct because the given expression simplifies to (x-1)¹⁰⁰, and the coefficient of x⁵³ is found using the binomial theorem.

532 Medium

532 of 951

Understanding +4 | -1

The number of terms in

\left(x^3+1+\frac{1}{x^3}\right)^{100}

View answer and explanation

Correct option: D

201

Option D is correct because the expansion results in terms from x^-300 to x³⁰⁰, with a difference of 3 between each power, leading to 201 terms.

533 Medium

533 of 951

Analyzing +4 | -1

n \in N

and

n

is not a multiple of three and

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

, then the value of

\sum\limits_{{r=0}}^n(-1)^r \cdot a_r \cdot{ }^n C_r

is equal to

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the sum evaluates to 0, which is not listed.

534 Medium

534 of 951

Application +4 | -1

The coefficient of

x^4

in the expansion of

\left(1+x+x^2+x^3\right)^n

View answer and explanation

Correct option: D

{ }^n C_4+{ }^n C_2+{ }^n C_1 \cdot{ }^n C_2

Option D (

{ }^n C_4+{ }^n C_2+{ }^n C_1 \cdot{ }^n C_2

) is correct because expanding (1+x+x²+x³)^n as (1+x)^n(1+x²)^n and finding the x⁴ term gives the correct coefficient.

535 Medium

535 of 951

Application +4 | -1

The coefficient of

x^{-2}

in the expansion of

(1+x)^{40}\left(1+\frac{1}{x}\right)^{40}

View answer and explanation

Correct option: D

\left(\begin{array}{l}80 \\ 38\end{array}\right)

Option D

\left(\begin{array}{l}80 \\ 38\end{array}\right)

is correct because the coefficient of

x^{-2}

is equivalent to the coefficient of

x^{38}

in the expansion of

(1+x)^{80}

536 Medium

536 of 951

Application +4 | -1

Let

m

and

n

be the coefficient of

7^{\text {th}}

and

13^{\text {th}}

term in expansion of

\left(\frac{1}{3} x^{1 / 3}+\frac{1}{2 x^{2 / 5}}\right)^{18}

, then

\left(\frac{m}{n}\right)^{1 / 3}

View answer and explanation

Correct option: D

\frac{4}{9}

Option D (4/9) is correct because after calculating the coefficients m and n of the 7th and 13th terms in the binomial expansion, the expression (m/n)^(1/3) simplifies to 4/9.

537 Easy

537 of 951

Application +4 | -1

In the expansion of

{(1 + x)^{30}}

, the sum of the coefficients of odd powers of

x

View answer and explanation

Correct option: D

{2^{29}}

Option D (

{2^{29}}

) is correct because the sum of coefficients of odd powers of x in the expansion of

{(1 + x)^{30}}

is equal to

{2^{n-1}}

, where n is the power.

538 Medium

538 of 951

Application +4 | -1

If for positive integers

r > 1,\;n > 2

, the coefficient of the

\left( {3r} \right)

th and

\left( {r + 2} \right)

th powers of

x

in the expansion of

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

are equal, then

View answer and explanation

Correct option: C

n = 2r + 1

Option C (

n = 2r + 1

) is correct because it satisfies the condition for equality of binomial coefficients in the expansion of

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

when

r > 1

and

n > 2

539 Medium

539 of 951

Application +4 | -1

In how many ways can 15 objects be placed in 3 boxes? (Each box can hold atmost 15 objects).

View answer and explanation

Correct option: D

136

Option D (136) is correct because it represents the number of ways to distribute 15 identical objects into 3 distinct boxes, which can be calculated using stars and bars method.

540 Easy

540 of 951

Application +4 | -1

The coefficient of the middle term in the expansion of

{\left( {x + 2y} \right)^6}

View answer and explanation

Correct option: B

8\left( {{\;^6}{C_3}} \right)

Option B (8(^6C_3)) is correct because the coefficient of the middle term in the expansion of (x + 2y)⁶ is 8 times 6C3.

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