The range of the values of the term independent of $x$ in the expansion of ${\left( {x{{\sin }^{ - 1}}{\rm{\alpha }} + \frac{{{{\cos }^{ - 1}}{\rm{\alpha }}}}{x}} \right)^{10}},{\rm{\alpha }} \in \left[ { - 1,1} \right]$ is

$\left[ { - \frac{{{{\rm{\;}}^{10}}{{\rm{C}}_5}{\rm{.}}{{\rm{\pi }}^{10}}}}{{{2^5}}},{\rm{\;}}\frac{{{{\rm{\;}}^{10}}{{\rm{C}}_5}{\rm{.}}{{\rm{\pi }}^{10}}}}{{{2^{20}}}}} \right]$

$\left[ {\frac{{{{\rm{\;}}^{10}}{{\rm{C}}_5}{\rm{.}}{{\rm{\pi }}^{10}}}}{{{2^5}}},{\rm{\;}} - \frac{{{{\rm{\;}}^{10}}{{\rm{C}}_5}{{\rm{\pi }}^{10}}}}{{{2^{20}}}}} \right]$

$\left[ {\frac{{{{\rm{\;}}^{10}}{{\rm{C}}_5}{\rm{.}}{{\rm{\pi }}^5}}}{{{2^5}}},{\rm{\;}}\frac{{{{\rm{\;}}^{10}}{{\rm{C}}_5}{\rm{.}}{{\rm{\pi }}^5}}}{{{2^{20}}}}} \right]$

$\left[ { - \frac{{{{\rm{\;}}^{10}}{{\rm{C}}_5}{\rm{.}}{{\rm{\pi }}^5}}}{{{2^5}}},{\rm{\;}}\frac{{{{\rm{\;}}^{10}}{{\rm{C}}_5}{\rm{.}}{{\rm{\pi }}^5}}}{{{2^{20}}}}} \right]$

If $n$ is a positive integer and ${C_k} = {\;^n}{C_k},$ then $\mathop \sum \limits_{k = 1}^n {k^3}{\left( {\frac{{{C_k}}}{{{C_{k - 1}}}}} \right)^2}$ equals

$\frac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{{12}}$

$\frac{{n\left( {n + 1} \right){{\left( {n + 2} \right)}^2}}}{{12}}$

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JEE • MATHEMATICS • BINOMIAL THEOREM

Binomial theorem practice · Hard

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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 1-30 of 49 questions

1 Hard

1 of 49

Application +4 | -1

For natural numbers

m,n

{\left( {1 - y} \right)^m}{\left( {1 + y} \right)^n} = 1 + {a_1}{y^2} +

….and

{a_1} = {a_2} = 10,\;

then

\left( {m,n} \right)

View answer and explanation

Correct option: C

\left( {35,45} \right)

Option C ((35,45)) is correct because these values satisfy the equations derived from the given conditions on the binomial expansion.

2 Hard

2 of 49

Understanding +4 | -1

{ }^{\mathrm{n}} \mathrm{C}_0-\frac{1}{2}{ }^{\mathrm{n}} \mathrm{C}_1+\frac{1}{3}{ }^{\mathrm{n}} \mathrm{C}_2-\ldots \ldots .+(-1)^{\mathrm{n}} \frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}}}{\mathrm{n}+1}=

View answer and explanation

Correct option: C

\frac{1}{n+1}

Option C (

\frac{{1}}{{n+1}}

) is correct because the given expression is the result of integrating the binomial expansion of

(1+x)^n

and then substituting

x = -1

3 Hard

3 of 49

Application +4 | -1

Coefficient of

x^{10}

in the expansion of

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

View answer and explanation

Correct option: A

476

Option A (476) is correct because the coefficient of

x^{10}

is found by expanding the given expression using the binomial theorem and identifying the relevant terms.

4 Hard

4 of 49

Application +4 | -1

The number of solutions of the equation

x+y+z=20, x \geq 1, y \geq 2, z \geq 3, x, y, z \in \mathrm{W}

View answer and explanation

Correct option: C

120

Option B, 5456, is correct because it represents the number of non-negative integral solutions to the inequality.

5 Hard

5 of 49

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.

6 Hard

6 of 49

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.

7 Hard

7 of 49

Application +4 | -1

Let

a,b

be integers and

f\left( x \right)

be a polynomial with integer coefficients such that

f\left( b \right) - f\left( a \right) = 1.

Then, the value of

b - a,

View answer and explanation

Correct option: C

1, - 1

Option C (

1, -1

) is correct because the difference

b-a

must be a factor of 1, which can only be 1 or -1.

8 Hard

8 of 49

Analyzing +4 | -1

If the coefficients of three consecutive terms in the expansion of

(1+x)^{n}

are in the ratio

1: 5: 20

, then the coefficient of the fourth term is

View answer and explanation

Correct option: D

3654

Option C (0) is correct because after simplifying the expression for y, it becomes a constant, and the second derivative of a constant is zero.

9 Hard

9 of 49

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=1

for all

k \geq n

, then

b_n

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 it represents the coefficient of

(x-3)^n

after transforming the given polynomial expression.

10 Hard

10 of 49

Analyzing +4 | -1

The coefficient of

{x^{50}}

in the expression

{\left( {1 + x} \right)^{1000}} + 2\;x{\left( {1 + x} \right)^{999}} + 3\;{x^2}{\left( {1 + x} \right)^{998}} + \ldots + 1001\;{x^{1000}}

View answer and explanation

Correct option: C

{\;^{1002}}{C_{50}}

The correct answer is C (

{\;^{1002}}{C_{50}}

) because the given series can be simplified to a single binomial term using summation techniques.

11 Hard

11 of 49

Analyzing +4 | -1

{C_\mathrm{r}}

stands for

{\;^n}{C_r}

, the sum of the given series

\frac{{2\left( {\frac{n}{2}} \right)\;!\;\left( {\frac{n}{2}} \right)\;!}}{{n!}}\;.\left[ {C_0^2 - 2C_1^2 + 3C_2^2 - ... + {{\left( { - 1} \right)}^n}\;\left( {n + 1} \right)C_n^2} \right],

where

n

is an even positive integer, is

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because simplifying the given series using properties of binomial coefficients leads to the result

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

12 Hard

12 of 49

Application +4 | -1

The term independent of

x

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

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the term independent of

x

requires the power of

x

to be zero, but the resulting value of

r

is not an integer, meaning there is no constant term.

13 Hard

13 of 49

Analyzing +4 | -1

The range of the values of the term independent of

x

in the expansion of

{\left( {x{{\sin }^{ - 1}}\mathrm{\alpha } + \frac{{{{\cos }^{ - 1}}\mathrm{\alpha }}}{x}} \right)^{10}},\mathrm{\alpha } \in \left[ { - 1,1} \right]

View answer and explanation

Correct option: B

\left[ { - \frac{{{\mathrm{\;}^{10}}{\mathrm{C}_5}\mathrm{.}{\mathrm{\pi }^{10}}}}{{{2^5}}},\mathrm{\;}\frac{{{\mathrm{\;}^{10}}{\mathrm{C}_5}\mathrm{.}{\mathrm{\pi }^{10}}}}{{{2^{20}}}}} \right]

Option B is correct because the range of the term independent of x is found by analyzing the term and using properties of inverse trigonometric functions.

14 Hard

14 of 49

Analyzing +4 | -1

a,\;b,\;c,\;d

be four consecutive coefficients in the binomial expansion of

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

, then the value of the expression

\left\{ {{{\left( {\frac{b}{{b + c}}} \right)}^2} - \frac{{ac}}{{\left( {a + b} \right)\left( {c + d} \right)}}} \right\}(\mathrm{where}\;x > 0)

View answer and explanation

Correct option: B

> 0

Option B (

> 0

) is correct because, based on the relationship between arithmetic and geometric means, the expression is always greater than zero.

15 Hard

15 of 49

Analyzing +4 | -1

Let

a_n=(\log 3)^n \sum\limits_{{r=1}}^n \frac{r^2}{r !(n-r) !}, n \in N

. Then, the sum of the series

a_1+a_2+a_3+a_4+\ldots

View answer and explanation

Correct option: A

(1+\log 3)(9 \log 3)

Option A ((1+log 3)(9 log 3)) is correct because it is the sum of the series after applying binomial theorem and simplifying 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

16 Hard

16 of 49

Application +4 | -1

The digit at unit’s place in the number

{17^{1995}} + {11^{1995}} - {7^{1995}},

View answer and explanation

Correct option: B

Option B is correct. The unit digit is determined by considering the unit digits of each term and their cyclical patterns.

17 Hard

17 of 49

Application +4 | -1

Find the coefficient of

x^5

in the expansion of

\left(1+x+x^3\right)^9

View answer and explanation

Correct option: A

378

Option A (378) is the correct answer because it is the calculated coefficient of

x^5

in the expansion of the given expression.

18 Hard

18 of 49

Application +4 | -1

The coefficient of

{x^5}

in the expansion of

{\left( {1 + {x^2}} \right)^5}{\left( {1 + x} \right)^4}

View answer and explanation

Correct option: B

Option B (60) is correct because by expanding the given expression and finding the coefficient of

x^5

, we get 60.

19 Hard

19 of 49

Application +4 | -1

k

and

n

are positive integer and

s_k=1^k+2^k+\ldots .+n^k

, then

\sum\limits_{{r=1}}^m{ }^{m+1} C_r s_r

is equal to

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because the given summation simplifies to this expression using binomial theorem and telescoping series.

20 Hard

20 of 49

Analyzing +4 | -1

The coefficient of

x^{10}

in the expansion of

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

\left(1+x^{3}\right)^{4}

is equal to

View answer and explanation

Correct option: A

The correct answer is A (52) because by expanding the given expression and identifying the combinations that result in x¹⁰, the sum of their coefficients is 52.

21 Hard

21 of 49

Application +4 | -1

C_0, C_1, C_2, \cdots, C_n

are binomial coefficients, then

\lim \limits_{{n \rightarrow \infty}}\left[C_n-\left(\frac{2}{3}\right) C_{n-1}+\left(\frac{2}{3}\right)^2 C_{n-2}+\cdots+(-1)^n\left(\frac{2}{3}\right)^n C_0\right]

View answer and explanation

Correct option: A

Option A (0) is the correct answer because the limit of the given expression as n approaches infinity is 0.

22 Hard

22 of 49

Analyzing +4 | -1

The coefficient of

{x^n}\;{y^n}

in the expansion of

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

View answer and explanation

Correct option: D

\sum \limits_{r = 0}^n C_r^3

Option D (

\sum \limits_{r = 0}^n C_r^3

) is correct because the coefficient of x^n y^n in the given expansion is the sum of the cubes of the binomial coefficients.

23 Hard

23 of 49

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.

24 Hard

24 of 49

Analyzing +4 | -1

n

is a positive integer and

{C_k} = {\;^n}{C_k},

then

\sum \limits_{k = 1}^n {k^3}{\left( {\frac{{{C_k}}}{{{C_{k - 1}}}}} \right)^2}

equals

View answer and explanation

Correct option: B

\frac{{n{{\left( {n + 1} \right)}^2}\left( {n + 2} \right)}}{{12}}

Option B (

\frac{{n{{\left( {n + 1} \right)}^2}\left( {n + 2} \right)}}}{{12}}

) is correct because, after simplifying the expression and using the formulas for the sum of first n integers, squares, and cubes, we arrive at the given result.

25 Hard

25 of 49

Application +4 | -1

The coefficient of

x^3

in the expansion of

\left(1-x+x^2\right)^5

View answer and explanation

Correct option: D

-30

Option D (-30) is correct because it is the coefficient of x³ in the expansion of (1-x+x²)⁵, obtained by considering relevant terms in the expansion.

26 Hard

26 of 49

Analyzing +4 | -1

\left( {\begin{array}{cccccccccccccccccccc} {30}\\ 0 \end{array}} \right)\left( {\begin{array}{cccccccccccccccccccc} {30}\\ {10} \end{array}} \right) - \left( {\begin{array}{cccccccccccccccccccc} {30}\\ 1 \end{array}} \right)\left( {\begin{array}{cccccccccccccccccccc} {30}\\ {11} \end{array}} \right) + ... + \left( {\begin{array}{cccccccccccccccccccc} {30}\\ {20} \end{array}} \right)\left( {\begin{array}{cccccccccccccccccccc} {30}\\ {30} \end{array}} \right)

is equal to

View answer and explanation

Correct option: C

{\;^{30}}{C_{10}}

Option C is correct because the given expression is the coefficient of x²⁰ in (1-x²)³⁰, which simplifies to ^{30}C_{10}.

27 Hard

27 of 49

Application +4 | -1

The value of

1 \times 2 \times 3 \times 4 + 2 \times 3 \times 4 \times 5 + 3 \times 4 \times 5 \times 6 + \ldots + n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right),

View answer and explanation

Correct option: B

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

Option C (

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

) is correct because it represents the sum of the given series, which can be derived using summation properties and binomial coefficients.

28 Hard

28 of 49

Application +4 | -1

In the expansion of the following expression

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

, the coefficient of

{x^4}\left( {0 \le k \le n} \right)

View answer and explanation

Correct option: A

{\;^{n + 1}}{C_{k + 1}}

Option A (

{\;^{n + 1}}{C_{k + 1}}

) is correct because the sum of the geometric progression can be simplified, and the coefficient of x^k is then found using the binomial theorem.

29 Hard

29 of 49

Analyzing +4 | -1

The coefficient of

{x^{n - 2}}

in the polynomial

\left( {x - 1} \right)\left( {x - 2} \right) \ldots \left( {x - n} \right)

View answer and explanation

Correct option: B

\frac{1}{{24}}n\left( {{n^2} - 1} \right)\left( {3\;n + 2} \right)

Option B (

\frac{1}{{24}}n\left( {{n^2} - 1} \right)\left( {3\;n + 2} \right)

) is correct because it is the correct coefficient of

{x^{n - 2}}

when expanding the given polynomial.

30 Hard

30 of 49

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.

Revision notes and key takeaways

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Key Takeaways

✓ This set contains 49 single-correct-answer MCQs on BINOMIAL THEOREM.
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