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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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841 Medium

841 of 951

Application +4 | -1

The value of

\sum\limits_{{r=0}}^n{ }^n C_r \sin r x \cdot \cos (n-r) x

is equal to

View answer and explanation

Correct option: A

2^{n-1} \cdot \sin (n x)

Option A (

2^{n-1} \cdot \sin (n x)

) is correct because the summation can be simplified using trigonometric identities and binomial theorem.

842 Hard

842 of 951

Analyzing +4 | -1

Let

\left(1-x-2 x^2\right)^6=1+a_1 x+a_2 x^2+

\ldots .+a_{12} \mathrm{x}^{12}

. Then

\frac{a_2}{2^2}+\frac{a_4}{2^4}+\frac{a_6}{2^6}+\ldots \ldots+\frac{a_{12}}{2^{12}}

is equal to

View answer and explanation

Correct option: B

-1 / 2

Option B (-1 / 2) is correct because by substituting x = 1/2 and x = -1/2 into the given equation and adding the resulting equations, we can isolate the desired sum and find its value to be -1/2.

843 Medium

843 of 951

Analyzing +4 | -1

Let

f(x)=\left(\sqrt{x^2+1}+\sqrt{x^2-1}\right)^6

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

. Then,

View answer and explanation

Correct option: B

f(x)

has exactly two terms

Option B (

f(x)

has exactly two terms) is correct because after expanding and simplifying the expression,

f(x)

reduces to

64x^6 - 48x^2

, which has two terms.

844 Medium

844 of 951

Application +4 | -1

The largest term in the expansion of

(2+3 x)^{25}

where

x=2

is its.

View answer and explanation

Correct option: C

20^{\text {th }}

term

Option C (

20^{\text {th }}

term) is correct because the largest term is found by determining when the ratio of consecutive terms is greater than or equal to 1.

845 Easy

845 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}{C_1} + {C_1}{C_2} + ... + {C_{n - 1}}{C_n}\;

is equal to

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because multiplying the two binomial expansions and comparing coefficients of x^(n-1) leads to this expression.

846 Medium

846 of 951

Application +4 | -1

The approximate value of

\frac{1}{\sqrt{23}}

correct to four decimal places is

View answer and explanation

Correct option: B

0.2085

Option B (0.2085) is the correct approximate value of

\frac{1}{\sqrt{23}}

correct to four decimal places, obtained through binomial approximation.

847 Easy

847 of 951

Application +4 | -1

The coefficient of the middle term in the binomial expansion in powers of

x

{(1 + \mathrm{\alpha }x)^4}

and of

{(1 - \mathrm{\alpha }x)^6}

is the same, if

\alpha

equals

View answer and explanation

Correct option: C

- \frac{3}{{10}}

Option C (

- \frac{3}{{10}}

) is correct because it is the value of

\alpha

that makes the coefficients of the middle terms in the two binomial expansions equal.

848 Medium

848 of 951

Application +4 | -1

{\left( {1 + ax} \right)^n} = 1 + 8x + 24{x^2} + ...,

then the values of

a

and

n

are

View answer and explanation

Correct option: A

\left( {2,4} \right)

Option A ((2,4)) is correct because by comparing the coefficients of x and x² in the given expansion with the binomial expansion of (1 + ax)^n, we can solve for a and n.

849 Medium

849 of 951

Application +4 | -1

Value of

S=1 \times 2 \times 3 \times 4+2 \times 3 \times 4 \times 5 +\ldots+n(n+1)(n+2)(n+3)

View answer and explanation

Correct option: A

\frac{1}{5} n(n+1)(n+2)(n+3)(n+4)

Option A (

\frac{1}{5} n(n+1)(n+2)(n+3)(n+4)

) is correct because the summation can be expressed as a telescoping sum of combinations, leading to the given formula.

850 Easy

850 of 951

Application +4 | -1

The coefficient of

{x^{ - 17}}

in the expansion of

{\left( {{x^4} - \frac{1}{{{x^3}}}} \right)^{15}}\;\mathrm{is}

View answer and explanation

Correct option: C

- {\;^{15}}{C_{11}}\;

Option C (-^15C_11) is correct because the coefficient of x^(-17) in the expansion of (x⁴ - 1/x³)¹⁵ is -^15C_11.

851 Medium

851 of 951

Understanding +4 | -1

Value of the expression

C_0^2+C_1^2+C_2^2+\cdots+C_n^2 \text { is }

View answer and explanation

Correct option: C

{ }^{2 n} \mathrm{C}_n

Option C (

{ }^{2 n} \mathrm{C}_n

) is correct because the given expression represents the coefficient of

x^n

in the expansion of

(1+x)^n(1+x)^n

852 Medium

852 of 951

Analyzing +4 | -1

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

, then

{\sum\limits_{0 \le i} {\sum\limits_{ < j \le n} {({C_i} - {C_j})} } ^2}

View answer and explanation

Correct option: A

(n+1){ }^{2 n} C_n-2^{2 n}

Option A is correct because the given summation simplifies to

(n+1){ }^{2 n} C_n-2^{2 n}

using properties of binomial coefficients.

853 Easy

853 of 951

Analyzing +4 | -1

The coefficient of

{x^{15}}

in the product

\left( {1 - x} \right)\left( {1 - 2x} \right)\left( {1 - {2^2}x} \right)\left( {1 - {2^3}x} \right) \ldots \left( {1 - {2^{15}}x} \right)

is equal to

View answer and explanation

Correct option: A

{2^{105}} - {2^{121}}

Option A (

{2^{105}} - {2^{121}}

) is correct because after expanding the product and identifying the coefficient of

{x^{15}}

, it simplifies to

{2^{105}} - {2^{121}}

854 Easy

854 of 951

Application +4 | -1

The value of the sum of the series

{ }^{14} {C}_0 \cdot{ }^{15} {C}_1+{ }^{14} {C}_1 \cdot{ }^{15} {C}_2+{ }^{14} {C}_2 \cdot{ }^{15} {C}_3+\ldots+{ }^{14} {C}_{14} \cdot{ }^{15} {C}_{15}

View answer and explanation

Correct option: C

{ }^{29} {C}_{14}

Option C (

{ }^{29} {C}_{14}

) is correct because the given sum represents the coefficient of x¹⁴ in the product of two binomial expansions, (1+x)¹⁴ and (x+1)¹⁵, which simplifies to (1+x)²⁹, and the coefficient of x¹⁴ is

{ }^{29} {C}_{14}

855 Hard

855 of 951

Analyzing +4 | -1

The coefficient of

{x^{20}}

in the expansion of

{\left( {1 + {x^2}} \right)^{40}}{\left( {{x^2} + 2 + \frac{1}{{{x^2}}}} \right)^{ - 5}},

View answer and explanation

Correct option: B

{\;^{30}}{C_{25}}

Option D (

{{\;^{30}}C_{25}}

) is correct because simplifying the expression and finding the coefficient of

x^{10}

(1+x^2)^{30}

leads to

{{\;^{30}}C_5}

which is equal to

{{\;^{30}}C_{25}}

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856 Easy

856 of 951

Application +4 | -1

In the expansion of

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

, the 5 th term from the end is

View answer and explanation

Correct option: B

\frac{17010}{x^8}

Option B (

\frac{{17010}}{{x^8}}

) is correct because it is the 5th term from the end in the expansion.

857 Easy

857 of 951

Analyzing +4 | -1

x^{10}+10 x^8 a+40 x^6 a^2+80 x^4 a^3+80 x^2 a^4+32 a^5

is equal to

View answer and explanation

Correct option: C

\left(x^2+2 a\right)^5

Option C (

(x^2+2 a)^5

) is correct because it matches the binomial expansion of the given expression.

858 Easy

858 of 951

Analyzing +4 | -1

f(x)=\sum\limits_{r=1}^n\left\{r^2\left({ }^n C_r-{ }^n C_{r-1}\right)+(2 r+1)^n C_r\right\}

and

f(30)=30(2)^\lambda

, then the value of

\lambda

View answer and explanation

Correct option: C

Option C (5) is correct, but the provided solution steps are incomplete. A full solution is not possible without further information or context.

859 Medium

859 of 951

Analyzing +4 | -1

(1+x)^{10}=a_0+a_1 x+a_2 x^2+\ldots .+a_{10} x^{10}

, then

\left(a_0-a_2+a_4-a_6+a_8-a_{10}\right)^2 +\left(a_1-a_3+a_5-a_7+a_9\right)^2 \text { is equal to }

View answer and explanation

Correct option: B

2^{10}

Option B (

2^{10}

) is correct because by substituting x = i and x = -i into the binomial expansion and multiplying the results, we can simplify the expression to

2^{10}

860 Easy

860 of 951

Understanding +4 | -1

1 + \frac{1}{3}x + \frac{{1.4}}{{3.6}}{x^2} + \frac{{1.4.7}}{{3.6.9}}{x^3} + ...

is equal to

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because the given series matches the binomial expansion of

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

861 Medium

861 of 951

Analyzing +4 | -1

\sum\limits_{{r=0}}^{2 n} a_r(x-2)^r=\sum\limits_{{r=0}}^{2 n} b_r(x-3)^r

and

a_k=\frac{1}{{ }^{n+k} C_n}

for

0 \leq k \leq n

then

b_n

equals

View answer and explanation

Correct option: D

n+1

Option D (

n+1

) is correct because by substituting

x-3=t

, the coefficient

b_n

is found to be the sum of

a_k { }^{n+k} C_k

from k=0 to n, which simplifies to

n+1

862 Medium

862 of 951

Application +4 | -1

\left(\sqrt[6]{3} \sqrt{2}+\frac{1}{\sqrt[3]{3}}\right)^n, \frac{t_7 \text { from the } 1 \mathrm{st}}{t_7 \text { from the last }}=\frac{1}{6}, n

is equal to

View answer and explanation

Correct option: B

Option B (10) is correct because by equating the ratio of the 7th term from the beginning and the 7th term from the end to 1/6, we can solve for n and find that n=10.

863 Hard

863 of 951

Analyzing +4 | -1

For

r = 0,\;

…10 let

{A_r},\;{B_r}\;\mathrm{and\;}{C_r}

denotes, respectively, the coefficient of

{x^r}

in the

{\left( {1 + x} \right)^{10}},\;{\left( {1 + x} \right)^{20}},\;

and

{\left( {1 + x} \right)^{30}}

. Then

\sum \limits_{r = 1}^{10} {A_r}\left( {{B_{10}}{B_r} - {C_{10}}{A_r}} \right)\mathrm{\;}

is equal to

View answer and explanation

Correct option: D

{C_{10}} - {B_{10}}

Option D (

{C_{10}} - {B_{10}}

) is correct because the summation simplifies to the difference between the coefficients of x¹⁰ in (1+x)³⁰ and (1+x)²⁰.

864 Easy

864 of 951

Application +4 | -1

For

\left| x \right| < 1

, the constant term in the expansion of

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

View answer and explanation

Correct option: D

- \frac{1}{2}

Option D (

- \frac{1}{2}

) is correct because it represents the constant term obtained after expanding the given expression using partial fractions and binomial expansions.

865 Easy

865 of 951

Application +4 | -1

({2^{3n}} - 1)

will be divisible by

\left( {\forall n \in N} \right)

View answer and explanation

Correct option: C

Option C (7) is correct because

{2^{3n}} - 1

can be rewritten as

{8^n} - 1

, which is always divisible by 7.

866 Medium

866 of 951

Analyzing +4 | -1

The value of

{ }^n C_0 \cdot{ }^{2 n} C_r-{ }^n C_1,{ }^{2 n-2} C_r

+{ }^n C_2 \cdot{ }^{2 n-4} C_r+\ldots+(-1)^n{ }^n C_n{ }^{2 n-2 n} C_n

View answer and explanation

Correct option: A

\begin{cases}2^{2 n-r}{ }^n C_{r-n}, & \text { if } r>n \\ 0, & \text { if } r < n\end{cases}

Option A is correct because the expression simplifies to the coefficient of

x^{r-n}

(2x+x^2)^n

, which is

2^{2n-r} { }^n C_{r-n}

r>n

and 0 if

r < n

867 Easy

867 of 951

Application +4 | -1

In the expansion of

{\left( {\frac{a}{x} + bx} \right)^{12}}

,the coefficient of

{x^{ - 10}}

will be

View answer and explanation

Correct option: C

12{a^{11}}b

Option C (

12{a^{11}}b

) is correct because it is the coefficient of

{x^{ - 10}}

obtained from the binomial expansion of the given expression.

868 Easy

868 of 951

Application +4 | -1

The 4th term from the end in the expansion of

\left(\frac{x^3}{2}-\frac{2}{x^2}\right)^9

View answer and explanation

Correct option: C

\frac{672}{x^3}

Option C (672/x³) is correct because it is the 7th term from the beginning, calculated using the binomial theorem.

869 Easy

869 of 951

Application +4 | -1

The coefficients of

{x^2}{y^2},\;yz{t^2}

and

\mathrm{\;}xyzt

in the expansion of

{\left( {x + y + z + t} \right)^4}

are in the ratio

View answer and explanation

Correct option: B

1:2:4

Option B (

1:2:4

) is correct because the coefficients can be calculated using the multinomial theorem, and their ratio is

6:12:24

which simplifies to

1:2:4

870 Medium

870 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 simplifying the summation leads to (x-1)¹⁰⁰, and the coefficient of x⁵³ is -

{ }^{100} C_{53}

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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Binomial theorem practice · Mixed difficulty

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