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

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

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

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

Back to BINOMIAL THEOREM

Your Progress

0 / 951 Answered

Correct

Incorrect

Accuracy

JEE • MATHEMATICS • BINOMIAL THEOREM

Binomial theorem practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

Practice format and chapter quiz

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.

Take a chapter quiz →

Showing 661-690 of 951 questions

Preparing practice and saved progress…

661 Medium

661 of 951

Application +4 | -1

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

is equal to

View answer and explanation

Correct option: C

2^{n-1} \sin n x

Option C (

2^{n-1} \sin n x

) is correct because the summation simplifies to this expression using trigonometric identities and the binomial theorem.

662 Easy

662 of 951

Analyzing +4 | -1

If the coefficient of the middle term in the expansion of

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

p

and the coefficients of middle term in the expansion of

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

are

q

and

r

, then

View answer and explanation

Correct option: C

p = q + r

Option C (

p = q + r

) is correct because the middle term's coefficient in the expansion of

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

is the sum of the two middle terms' coefficients in the expansion of

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

663 Easy

663 of 951

Application +4 | -1

\left| x \right| < 1

, then the coefficient of

{x^6}

in the expansion of

{(1 + x + {x^2})^{ - 3}}

View answer and explanation

Correct option: A

Option A (3) is correct because the coefficient of x⁶ in the expansion of (1 + x + x²)^(-3) is calculated to be 3.

664 Medium

664 of 951

Application +4 | -1

If the sum of the 5th and 6th terms in the expansion of

(a-b)^n, n \geq 5

is zero, then

\frac{a}{b}=

View answer and explanation

Correct option: D

\frac{n-4}{5}

Option D is correct because setting the sum of the 5th and 6th terms to zero and simplifying the binomial coefficients leads to the given ratio.

665 Easy

665 of 951

Application +4 | -1

The value of

{\left( {0.99} \right)^{15}}

View answer and explanation

Correct option: B

0.8601

Option B (0.8601) is correct because using the binomial approximation, we can calculate the value of (0.99)¹⁵ to be approximately 0.8601.

666 Medium

666 of 951

Application +4 | -1

The value of the term independent of

x

in the expansion of

\left(1+\frac{x}{2}-\frac{2}{x}\right)^4, x \neq 0

is equal to

View answer and explanation

Correct option: C

-5

Option C (-5) is correct because it is the constant term obtained after expanding the given expression and simplifying.

667 Medium

667 of 951

Analyzing +4 | -1

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

, then

View answer and explanation

Correct option: C

a_{n-3}=a_{n+3}

Option D (

a_{{n-3}}=a_{{n+3}}

) is correct because binomial coefficients are symmetric, meaning

{ }^{2n}C_{{n-3}} = { }^{2n}C_{{n+3}}

668 Medium

668 of 951

Analyzing +4 | -1

The value of

\sum\limits_{0 \le i} {\sum\limits_{ < j \le n} {j{.^n}{C_j}} }

is equal to

View answer and explanation

Correct option: B

n(n+3) \cdot 2^{n-3}

Option B (

n(n+3) \cdot 2^{n-3}

) is correct because it is derived by simplifying the double summation involving binomial coefficients.

669 Easy

669 of 951

Application +4 | -1

The Binomial coefficient of the 4 th term in the expansion of

(x-q)^5

View answer and explanation

Correct option: C

Option C (10) is correct as it represents the binomial coefficient of the 4th term in the expansion of (x-q)⁵.

670 Medium

670 of 951

Application +4 | -1

The term independent of

x

in the expansion of

\left[\left(t^{-1}-1\right) x+\left(t^{-1}+1\right)^{-1} x^{-1}\right]^8

View answer and explanation

Correct option: C

70\left(\frac{1-t}{1+t}\right)^4

Option C (

70\left(\frac{{1-t}}{{1+t}}\right)^4

) is correct because the term independent of x is found by setting the power of x to zero and simplifying the resulting expression.

671 Medium

671 of 951

Application +4 | -1

In how many ways 12 identical pens can be distributed among 5 students?

View answer and explanation

Correct option: B

\left(\begin{array}{c}16 \\ 4\end{array}\right)

Option B

\left(\begin{array}{c}16 \\ 4\end{array}\right)

is correct because it represents the number of ways to distribute 12 identical pens among 5 students.

672 Easy

672 of 951

Application +4 | -1

The total number of terms in the expansion of

(a+b+c+d)^n, n \in N

View answer and explanation

Correct option: C

\frac{(n+1)(n+2)(n+3)}{6}

Option C (

\frac{{(n+1)(n+2)(n+3)}}{{6}}

) is correct because it represents the number of terms in the expansion of (a+b+c+d)^n, which is a combination with repetition problem.

673 Easy

673 of 951

Application +4 | -1

The smallest natural number

n

, such that the coefficient of x in the expansion of

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

^n{C_{23}}

, is

View answer and explanation

Correct option: D

Option D (38) is correct because it is the smallest natural number n that satisfies the condition for the coefficient of x in the expansion to be

^n{C_{23}}

674 Medium

674 of 951

Application +4 | -1

n > 1

then value of the expression is

C_0-2 C_1+3 C_2-4 C_3+\ldots+(-1)^n(n+1) C_n

View answer and explanation

Correct option: B

Option B (0) is correct because differentiating the binomial expansion and substituting x=-1 results in the given expression equaling zero.

675 Medium

675 of 951

Application +4 | -1

The integral part of

(\sqrt{2}+1)^6

View answer and explanation

Correct option: C

197

Option C (197) is correct because it is the integral part of the expression

(\sqrt{{2}}+1)^6

calculated using binomial expansion and conjugate properties.

Explore Other Chapters in MATHEMATICS

View All →

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

676 Hard

676 of 951

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.

677 Easy

677 of 951

Analyzing +4 | -1

\sum\limits_{\mathop {i,j = 0}\limits_{i \ne j} }^n {{\,^n}{C_i}^n{C_j}}

is equal to

View answer and explanation

Correct option: A

2^{2 n}-{ }^{2 n} C_n

Option A (

2^{2 n}-{ }^{2 n} C_n

) is correct because it represents the total number of ways to choose i and j from 0 to n, excluding the cases where i equals j.

678 Medium

678 of 951

Application +4 | -1

In the binomial

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

, if the ratio of the seventh term from the beginning of the expansion to the seventh term from its end is

1 / 6

then

\mathrm{n}=

View answer and explanation

Correct option: B

Option B (9) is correct because equating the ratio of the seventh terms from the beginning and end leads to n=9.

679 Easy

679 of 951

Analyzing +4 | -1

If the coefficients of the three successive terms in the binomial expansion of

(1+x)^{\mathrm{n}}

are in the ratio

1: 7: 42

, then the first of these terms in the expansion is:

View answer and explanation

Correct option: C

7^{\text {th }}

The correct answer is C (7th) because by setting up ratios of consecutive binomial coefficients and solving for r, we find that the first term is the 7th term.

680 Medium

680 of 951

Application +4 | -1

The coefficient of

y

in the expansion of

\left(y^2+c / y\right)^5

View answer and explanation

Correct option: A

10 c^3

Option A (

10 c^3

) is correct because it is the coefficient of

y

in the expansion of the given binomial.

681 Easy

681 of 951

Application +4 | -1

If sum of the coefficients in the expansion of

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

is 128, then coefficient of

x

in the expansion of

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

View answer and explanation

Correct option: C

Option D (35) is correct because the sum of coefficients being 128 implies n=7, and the coefficient of x in the expansion is 35.

682 Easy

682 of 951

Application +4 | -1

The value of

2\left(n c_1\right)+2^3\left(n c_3\right)+2^5\left(n c_5\right) \ldots . . .2 n

' being an integers is

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because it represents the closed-form expression for the given series, derived from the binomial theorem.

683 Easy

683 of 951

Application +4 | -1

The sum of coefficients in the expansion of

(7 x-8 y)^{149}

View answer and explanation

Correct option: B

-1

Option B (-1) is correct because the sum of the coefficients in the expansion is found by substituting x=1 and y=1 into the expression, which results in (7-8)¹⁴⁹ = (-1)¹⁴⁹ = -1.

684 Easy

684 of 951

Application +4 | -1

If in the expansion of

{\left( {a - 2b} \right)^n}.

. The sum of the 5th and 6th term is zero, then the value of

\frac{a}{b}

View answer and explanation

Correct option: B

\frac{{2\left( {n - 4} \right)}}{5}

Option B (

\frac{{2\left( {n - 4} \right)}}{5}

) is correct because it is the result obtained after simplifying the equation derived from the given condition that the sum of the 5th and 6th terms is zero.

685 Easy

685 of 951

Application +4 | -1

The expression

\frac{1}{{\sqrt[3]{{6 - 3x}}}}

is equal to

View answer and explanation

Correct option: B

{6^{ - 1/3}}\left[ {1 + \frac{x}{6} + \frac{{2{x^2}}}{{{6^2}}} + ...} \right]

Option B (

{6^{ - 1/3}}\left[ {1 + \frac{x}{6} + \frac{{2{x^2}}}{{{6^2}}} + ...} \right]

) is correct because by rewriting the expression and applying the binomial theorem, the expansion matches option B.

686 Medium

686 of 951

Application +4 | -1

The approximate value of

{\left( {7.995} \right)^{1/3}}

correct to four decimal places is

View answer and explanation

Correct option: B

1.9996

Option B (1.9996) is correct. Using binomial approximation, the value of (7.995)^(1/3) is approximately 1.9996.

687 Easy

687 of 951

Application +4 | -1

r

and

n

are positive integers such that

r > 1,n > 2

and coefficients of

{\left( {r + 2} \right)^{\mathrm{th}}}

term and

{\left( {3r} \right)^{\mathrm{th}}}

term in the expansion of

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

are equal, then

n

equals

View answer and explanation

Correct option: C

2r

Option C (2r) is correct because equating the coefficients and solving for n yields n = 2r.

688 Easy

688 of 951

Analyzing +4 | -1

The total number of irrational terms in the binomial expansion of

{\left( {{7^{1/5}} - {3^{1/10}}} \right)^{60}}

View answer and explanation

Correct option: C

Option C (54) is correct because by finding the number of rational terms in the binomial expansion, we can subtract that from the total number of terms to find the number of irrational terms.

689 Medium

689 of 951

Application +4 | -1

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

, then the value of

{C_0} + 2{C_1} + 3{C_2} + ... + \left( {n + 1} \right){C_n}

will be

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because differentiating the binomial expansion and substituting x=1 yields the given sum, which simplifies to

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

690 Hard

690 of 951

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.

Revision notes and key takeaways

Review BINOMIAL THEOREM in your textbook or class notes, then use this set to check your understanding.

Showing 661-690 of 951 questions

Your Progress

Binomial theorem practice · Mixed difficulty

Explore Other Chapters in MATHEMATICS

Continue Your Practice

Your practice