Let $n$ be an odd integer. If $\sin n{\rm{\theta }} = \mathop \sum \limits_{r = 0}^n {b_r}{\sin ^r}{\rm{\theta }}$ for every value of $\theta $, then

${b_0} = 0,\;{b_1} = {n^2} - 3n + 3$

If the expansion in power of $x$ of the function $\frac{1}{{\left( {1 - ax} \right)\left( {1 - bx} \right)}}\;{\rm{is}}\,\;{a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3} + ...,\;{\rm{then\;}}{a_n}\;{\rm{is}}$

$\frac{{{a_n} - {b^n}}}{{b - a}}$

\[\frac{{{a^{n + 1}} - {b^{n + 1}}}}{{b - a}}\]

$\frac{{{b^n} - {a^n}}}{{b - a}}$

If ${\;^{n - 1}}{C_r} = \left( {{k^2} - 3} \right).{\;^n}{C_{r + 1}}$, then $k$ is belongs to

$\left( { - \infty , - 2} \right]$

$\left[ { - \sqrt 3 ,\sqrt 3 } \right]$

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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 31-49 of 49 questions

31 Hard

31 of 49

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}}

32 Hard

32 of 49

Analyzing +4 | -1

Let

n

be an odd integer. If

\sin n\mathrm{\theta } = \sum \limits_{r = 0}^n {b_r}{\sin ^r}\mathrm{\theta }

for every value of

\theta

, then

View answer and explanation

Correct option: B

{b_0} = 0,\;{b_1} = n

The correct answer is B (

{b_0} = 0,\;{b_1} = n

) because it is derived from the expansion of

\sin n\theta

using trigonometric identities and comparing coefficients.

33 Hard

33 of 49

Application +4 | -1

If the expansion in power of

x

of the function

\frac{1}{{\left( {1 - ax} \right)\left( {1 - bx} \right)}}\;\mathrm{is}\,\;{a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3} + ...,\;\mathrm{then\;}{a_n}\;\mathrm{is}

View answer and explanation

Correct option: C

\frac{{{b^{n + 1}} - {a^{n + 1}}}}{{b - a}}

Option C (

\frac{{{b^{n + 1}} - {a^{n + 1}}}}{{b - a}}

) is correct because by expressing the given function as a product of two series and finding the coefficient of

x^n

, we arrive at this expression.

34 Hard

34 of 49

Application +4 | -1

{\;^{n - 1}}{C_r} = \left( {{k^2} - 3} \right).{\;^n}{C_{r + 1}}

, then

k

is belongs to

View answer and explanation

Correct option: D

\left( {\sqrt 3 ,2} \right]

Option D (

\left( {\sqrt 3 ,2} \right]

) is correct because it represents the valid range for k based on the given condition and binomial coefficient properties.

35 Hard

35 of 49

Application +4 | -1

n

is a multiple of 3, then

\sum\limits_{{r=0}}^{\frac{n}{3}}\left(\begin{array}{c}n \\ 3 r\end{array}\right)=

View answer and explanation

Correct option: A

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

Option A is correct because the sum of the binomial coefficients with indices that are multiples of 3 can be found using complex numbers and the binomial theorem.

36 Hard

36 of 49

Application +4 | -1

{(1 + 2x + 3{x^2})^{10}} = {a_0} + {a_1}x + {a_2}{x^2} + ... + {a_{20}}{x^{20}},

then

{a_1}

equals

View answer and explanation

Correct option: B

Option B (20) is correct because the coefficient of x in the expansion can be found by considering the term with x or by differentiating the given equation and substituting x=0.

37 Hard

37 of 49

Understanding +4 | -1

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

can be expanded as a power series of

x

View answer and explanation

Correct option: B

|x| < 2 / \sqrt{5}

Option B (

|x| < 2 / \sqrt{5}

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

38 Hard

38 of 49

Analyzing +4 | -1

The coefficient of

{x^n}

in the polynomial

\left( {x + {\;^n}{C_0}} \right)\left( {x + \;3{\;^n}{C_1}} \right)\left( {x + \;5{\;^n}{C_2}} \right) \ldots \left[ {x + \left( {2n + 1} \right){\;^n}{C_n}} \right]

View answer and explanation

Correct option: C

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

The correct answer is C ((

n + 1

)

2^n

) because the coefficient of x^n is the sum of the roots of the polynomial, which can be simplified using binomial properties.

39 Hard

39 of 49

Application +4 | -1

If the ninth term in the expansion of

{\left\{ {{3^{{{\log }_3}\sqrt {{{25}^{x - 1}} + 7} }} + {3^{ - 1/8{{\log }_3}\left( {{5^{x - 1}} + 1} \right)}}} \right\}^{10}}

is equal to 180 and

x > 1,

then

x

equals

View answer and explanation

Correct option: B

{\log _5}15

Option B (

{\log _5}15

) is correct because it is the solution to the equation derived from the given binomial expansion and condition.

40 Hard

40 of 49

Application +4 | -1

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

, then the value of

a_0+a_3+a_6+\cdots

View answer and explanation

Correct option: A

a_1+a_4+a_7+\ldots

Option A (

a_1+a_4+a_7+\ldots

) is correct because by substituting complex roots of unity into the given expression, we can show the equality of the sums.

41 Hard

41 of 49

Application +4 | -1

\mathrm{C}_0, \mathrm{C}_1, \mathrm{C}_2 \ldots \ldots \mathrm{C}_{\mathrm{n}}

are binomial coefficients in the expansion of

(1+\mathrm{x})^{\mathrm{n}}, \mathrm{n} \in \mathrm{N}

, then

\frac{C_1}{2}+\frac{C_3}{4}+\frac{C_5}{6}+\ldots \ldots+\frac{C_b}{n+1}

is equal to -

View answer and explanation

Correct option: C

\frac{2^{\mathrm{n}}-1}{\mathrm{n}+1}

Option C (

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

) is correct because integrating the binomial expansion and then using specific values for x allows us to isolate the desired sum.

42 Hard

42 of 49

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.

43 Hard

43 of 49

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}}

44 Hard

44 of 49

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.

45 Hard

45 of 49

Application +4 | -1

\sum_{r=0}^m{ }^{n+r} \mathrm{C}_n

is equal to :

View answer and explanation

Correct option: A

{ }^{n+m+1} \mathrm{C}_{n+1}

Option A (

{ }^{n+m+1} \mathrm{C}_{n+1}

) is correct because by repeatedly applying the identity

{ }^n \mathrm{C}_r+{ }^n \mathrm{C}_{r+1}={ }^{n+1} \mathrm{C}_{r+1}

, the summation simplifies to the given expression.

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46 Hard

46 of 49

Analyzing +4 | -1

Let

S\left( k \right) = 1 + 3 + 5 + ... + \left( {2k - 1} \right) = 3 + {k^2}

. Then, which of the following is true?

View answer and explanation

Correct option: B

S\left( k \right) \Rightarrow S\left( {k + 1} \right)

Option B (

S(k) \Rightarrow S(k + 1)

) is correct because if we assume

S(k)

is true, then

S(k+1)

is also true, showing the implication holds.

47 Hard

47 of 49

Application +4 | -1

n \in \mathrm{N}-\{1\}

, then

\sum\limits_{{r=0}}^n\left\{(-1)^r(100-r)\left(\begin{array}{l}n \\ r\end{array}\right)\right\}=

View answer and explanation

Correct option: D

Option D (0) is correct because the given summation simplifies to zero when

n > 1

due to the properties of binomial coefficients.

48 Hard

48 of 49

Analyzing +4 | -1

(1+x)^n=\sum\limits_{{r=0}}^n{ }^n C_r x^n

, then

\frac{C_0}{1 \cdot 2} 2^2+\frac{C_1}{2 \cdot 3} 2^3

+\frac{C_2}{3 \cdot 4} 2^4+\ldots+\frac{C_n}{(n+1)(n+2)} 2^{n+2}

is equal to

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because it is the sum of the series after applying binomial theorem and simplifying the expression.

49 Hard

49 of 49

Application +4 | -1

If the sum of the coefficients in the expansion of

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

is equal to the sum of the coefficient in the expansion of

{\left( {x - \alpha \;y} \right)^{35}},

then

\alpha =

View answer and explanation

Correct option: B

Option B (1) is correct because setting the sum of coefficients equal in both expressions leads to the equation (

\alpha

- 1)³⁵ = -(

\alpha

- 1)³⁵, which is only true when

\alpha

= 1.

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