The value of ${ }^n C_3+{ }^n C_7+{ }^n C_{11}+\ldots$ is

$\frac{1}{4}\left\{2^{n+1}-2^{n / 2} \sin \frac{n \pi}{4}\right\}$

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 · 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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Showing 721-750 of 951 questions

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721 Easy

721 of 951

Application +4 | -1

Given positive integers

r>1, n>2

and the coefficients of (3r)th term and

(r+2)

th term in the binomial expansion of

(1+x)^{2 n}

are equal, then

r=

View answer and explanation

Correct option: A

\frac{n}{2}, n

even

Option A (3) is correct because the minimum value of the sum of absolute differences occurs at the median, which is 3 in this case, and the result is |3-2| + |3-5| = 1 + 2 = 3.

722 Medium

722 of 951

Application +4 | -1

The value of

\sum\limits_{{r=0}}^7 \frac{{ }^7 C_r}{{ }^{r+2} C_2}

is equal to

View answer and explanation

Correct option: B

\frac{251}{18}

Option B,

\frac{251}{18}

, is correct because it is the result of simplifying the summation using binomial coefficient identities.

723 Medium

723 of 951

Application +4 | -1

The coefficient of

{a^5}{b^6}{c^7}

in the expansion of

{\left( {bc + ca + ab} \right)^9}

View answer and explanation

Correct option: D

1260

Option D (1260) is correct because it is the coefficient of

{a^5}{b^6}{c^7}

obtained by expanding the given expression using binomial theorem properties.

724 Medium

724 of 951

Analyzing +4 | -1

For

r=0,1, \ldots, 10

, let

A_r, B_r

and

C_r

denote, respectively the coefficient of

x^r

in the expansions of

(1+x)^{10},(1+x)^{20}

and

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

Then,

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

is equal to

View answer and explanation

Correct option: D

C_{10}=B_{10}

Option D (

C_{10}=B_{10}

) is correct because simplifying the summation using properties of binomial coefficients leads to the conclusion that

C_{10} - B_{10} = 0

, thus

C_{10} = B_{10}

725 Easy

725 of 951

Application +4 | -1

n \in N

such that

{\left( {7 + 4\sqrt 3 } \right)^n} = I + F,

where

I \in N

and

0 < F < 1.

Then, the value of

\left( {I + F} \right)\left( {1 - F} \right)

View answer and explanation

Correct option: B

Option B (1) is correct because by defining G as (7-4√3)^n, we can show that F+G=1, leading to (I+F)(1-F) = (I+F)G = 1.

726 Easy

726 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,

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

is approximately 0.8601 when calculated to four decimal places.

727 Medium

727 of 951

Analyzing +4 | -1

Statement I:The integral part of

(8+3 \sqrt{7})^{20}

is an even integer.

Statement II:

(8+3 \sqrt{7})^{20}+(8-3 \sqrt{7})^{20}

is an even integer.

View answer and explanation

Correct option: D

Statement-I is false, statement-II is true

Option D (Statement-I is false, statement-II is true) is correct because the integral part of the expression is odd, and the given sum is an even integer.

728 Easy

728 of 951

Application +4 | -1

The sum of all rational terms in the expansion of

\left(2^{\frac{1}{5}}+5^{\frac{1}{3}}\right)^{15}

is equal to:

View answer and explanation

Correct option: A

3133

Option A (3133) is correct because it sums the rational terms in the binomial expansion, which occur when the powers of both terms are integers.

729 Medium

729 of 951

Analyzing +4 | -1

The value of

\sum\limits_{{r=0}}^{2 n}(-1)^r \cdot\left({ }^{2 n} C_r\right)^2

is equal to

View answer and explanation

Correct option: C

(-1)^n \cdot{ }^{2 n} C_n

Option C (

(-1)^n \cdot{ }^{2 n} C_n

) is correct because it is obtained by recognizing the sum as the coefficient of a term in the product of two binomial expansions.

730 Medium

730 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_{{r=0}}^n \sum\limits_{{s=0}}^n(r+s) C_r C_s

is equal to

View answer and explanation

Correct option: B

n \cdot 2^{2 n}

Option B (

n \cdot 2^{2 n}

) is correct because the summation can be separated and simplified using properties of binomial coefficients.

731 Easy

731 of 951

Application +4 | -1

The remainder when

7^{128}

is divided by 10 is

View answer and explanation

Correct option: A

Option A (1) is correct because the remainder when

7^{128}

is divided by 10 is 1, which can be found by using modular arithmetic and simplifying the expression.

732 Medium

732 of 951

Analyzing +4 | -1

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

then the value of

{C_0}{C_1} - {C_1}{C_2} + {C_2}{C_3} - \ldots ,

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the value of the expression is 0, which is not equal to any of the given options.

733 Medium

733 of 951

Application +4 | -1

If in the expansion of

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

, the coefficient of

x

and

{x^2}

are 3 and

- 6

respectively, then

m

View answer and explanation

Correct option: C

Option C (12) is correct because by expanding the product of the two binomials and equating the coefficients of

x

and

x^2

to the given values, we can solve for

m

734 Medium

734 of 951

Application +4 | -1

A ratio of the 5 th term from the beginning to the

5^{\text {th }}

term from the end in the binomial expansion of

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

View answer and explanation

Correct option: D

4(36)^{\frac{1}{3}}: 1

Option D (

4(36)^{\frac{{1}}{{3}}}: 1

) is correct because the ratio of the 5th term from the beginning to the 5th term from the end is calculated by using the binomial theorem and simplifying the expression.

735 Easy

735 of 951

Application +4 | -1

The largest term in the expansion of

{\left( {3 + 2x} \right)^{50}}

, where

x = \frac{1}{5}

View answer and explanation

Correct option: C

7th

Option C (7th) is correct because by finding the ratio of consecutive terms and setting it greater than or equal to 1, we find that the largest term occurs at r <= 6, meaning the 7th term is the largest.

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736 Easy

736 of 951

Application +4 | -1

In the expansion of

\left(1+3 x+2 x^2\right)^6

the coefficient of

x^{11}

{k} \times 2^6

, then

{k}

View answer and explanation

Correct option: A

Option A (9) is correct because by expanding the given expression and finding the coefficient of x¹¹, k is found to be 9.

737 Medium

737 of 951

Analyzing +4 | -1

The value of

{ }^n C_3+{ }^n C_7+{ }^n C_{11}+\ldots

View answer and explanation

Correct option: A

\frac{1}{2}\left\{2^{n-1}-2^{n / 2} \sin \frac{n \pi}{4}\right\}

Option A (

\frac{{1}}{{2}}\left{2^{{n-1}}-2^{{n / 2}} \sin \frac{{n \pi}}{{4}}\right}

) is correct because by using the binomial theorem and complex numbers, the given series can be expressed and simplified to

\frac{{1}}{{2}}\left{2^{{n-1}}-2^{{n / 2}} \sin \frac{{n \pi}}{{4}}\right}

738 Easy

738 of 951

Application +4 | -1

\left(1+x-2 x^2\right)^6=1+a_1 x+a_2 x^2+a_3 x^3+\ldots,

then the value of

a_2+a_4+a_6+\ldots+a_{12}

will be

View answer and explanation

Correct option: B

Option B (31) is correct because substituting x=1 and x=-1 into the given polynomial and solving the resulting equations yields the value of the desired sum.

739 Easy

739 of 951

Analyzing +4 | -1

The sum of rational terms in

(\sqrt{2}+\sqrt{3}+\sqrt[6]{5})^{10}

, is

View answer and explanation

Correct option: A

12632

Option A (12632) is correct because the sum of rational terms in the expansion is 12632, obtained by considering the possible combinations of powers that result in rational terms.

740 Medium

740 of 951

Application +4 | -1

The value of

\sum_{r=0}^{20}{ }^{50-r} C_{6}

is equal to :

View answer and explanation

Correct option: A

{ }^{51} C_{7}-{ }^{30} C_{7}

Option A (

{}^{51} C_{7}-{ }^{30} C_{7}

) is correct because it uses the Hockey Stick Identity to simplify the summation of binomial coefficients.

741 Medium

741 of 951

Application +4 | -1

If 21 st and 22 nd terms in the expansion of

(1+x)^{44}

are equal, then

x

is equal to

View answer and explanation

Correct option: C

7 / 8

Option C (7 / 8) is correct because the equality of consecutive terms in a binomial expansion leads to the equation x = 7/8.

742 Medium

742 of 951

Application +4 | -1

If for

1 \leq m \leq n, f(m, n)=C_0-C_1+C_2-

\cdots+(-1)^{m-1} C_{m-1}

, then

\frac{f(9,5)}{10}

View answer and explanation

Correct option: C

Option C (7) is correct because the alternating sum of binomial coefficients is simplified by relating it to the coefficient of a term in a binomial expansion.

743 Hard

743 of 951

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.

744 Easy

744 of 951

Analyzing +4 | -1

If the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of

\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^{n}

\sqrt{6}: 1

, then the third term from the beginning is:

View answer and explanation

Correct option: D

60 \sqrt{3}

Option D is correct. By setting up the ratio of the fifth terms from the beginning and end, we find

n=10

, and then we can calculate the third term from the beginning.

745 Easy

745 of 951

Understanding +4 | -1

n

is an odd natural number and

^n{C_0}{ < ^n}{C_1}{ < ^n}{C_2} < {....^n}{C_r}{ > ^n}{C_{r + 1}}{ > ^n}{C_{r + 2}} > {....^n}{C_n},

then

r =

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because when n is odd, the greatest binomial coefficient occurs at the two middle terms, which are

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

and

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

746 Medium

746 of 951

Application +4 | -1

The coefficient of

x^4

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

View answer and explanation

Correct option: B

\frac{405}{256}

Option B is correct because by finding the general term in the binomial expansion and setting the power of x to 4, we can find the coefficient of x⁴.

747 Medium

747 of 951

Application +4 | -1

The coefficient of

x^5

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

View answer and explanation

Correct option: D

None of the above

Option D (None of the above) is correct because the coefficient of

x^5

in the simplified expression is 0.

748 Medium

748 of 951

Application +4 | -1

There are numerous balls of red, yellow and blue colours. In how many ways 12 balls can be chosen from them?

View answer and explanation

Correct option: C

Option C, 91, is correct because it represents the number of ways to choose 12 balls from the three colors.

749 Hard

749 of 951

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.

750 Medium

750 of 951

Application +4 | -1

(a+b x)^{-3}=\frac{1}{27}+\frac{1}{3} x+\ldots

, then the ordered pair

(a, b)

equals

View answer and explanation

Correct option: D

(3,-9)

Option D ((3,-9)) is correct because solving the equations derived from the binomial expansion yields a=3 and b=-9.

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