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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 361-390 of 951 questions

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361 Medium

361 of 951

Analyzing +4 | -1

Let

P=(8 \sqrt{3}+13)^{13}, Q=(6 \sqrt{2}+9)^9

then (where [] represents greatest integer function)

View answer and explanation

Correct option: D

[P]+[Q]=

Even

Option D (

[P]+[Q]=

Even) is correct because by analyzing the binomial expansion and conjugate terms, we can determine that both [P] and [Q] are even.

362 Easy

362 of 951

Understanding +4 | -1

The coefficient of

x^k(0 \leq k \leq n)

in the expansion of

E=1+(1+x)+(1+x)^2+\cdots+(1+x)^n

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because it represents the coefficient of

x^k

after summing the geometric series and simplifying.

363 Medium

363 of 951

Application +4 | -1

Number of ways of selection of 8 letters from 24 letters of which 8 are

a, 8

are

b

and the rest unlike, is given by

View answer and explanation

Correct option: C

10.2^7

Option C (

10.2^7

) is correct because it represents the number of ways to select 8 letters from the given set, calculated using generating functions and differentiation.

364 Medium

364 of 951

Application +4 | -1

If sum of coefficients in expansion of

(1+2 \mathrm{x})^{\mathrm{m}} \&(2+\mathrm{x})^{\mathrm{n}}

are respectively

6561 \& 243

, then position of point

(\mathrm{m}, \mathrm{n})

w.r.t. circle

\mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}-6 \mathrm{y}-32=0

View answer and explanation

Correct option: B

inside the circle

Option B (inside the circle) is correct because substituting m=8 and n=5 into the circle's equation results in a negative value, indicating the point lies inside the circle.

365 Medium

365 of 951

Application +4 | -1

The sum of the coefficient of

x^{2 / 3}

and

x^{-2 / 5}

in the binomial expansion of

\left(x^{2 / 3}+\frac{1}{2} x^{-2 / 5}\right)^9

is:

View answer and explanation

Correct option: A

21-Apr

Option A (21/4) is correct because it is the sum of the coefficients of x^(2/3) and x^(-2/5) in the binomial expansion.

366 Medium

366 of 951

Application +4 | -1

The coefficient of

x^2 y^3

in the expansion of

(1-x+y)^{20}

View answer and explanation

Correct option: D

None of these

Option D is correct because the coefficient of x²y³ is found using the multinomial theorem, and the correct coefficient is not among the given options.

367 Medium

367 of 951

Analyzing +4 | -1

Coefficient of

x^{11}

in the expansion of

\left(1+x^2\right)^4\left(1+x^3\right)^7\left(1+x^4\right)^{12}

View answer and explanation

Correct option: C

1113

Option C (1113) is correct because by considering all possible combinations of powers of x that sum to 11, the coefficient is calculated to be 1113.

368 Easy

368 of 951

Application +4 | -1

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

then

\frac{{{\;^n}{C_1}}}{{{\;^n}{C_0}}} + \frac{{2{\;^n}{C_2}}}{{{\;^n}{C_1}}} + \frac{{\;{3^n}{C_3}}}{{{\;^n}{C_2}}} + ... + \frac{{n{\;^n}{C_n}}}{{{\;^n}{C_{n - 1}}}}

is equal to

View answer and explanation

Correct option: C

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

The correct answer is

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

because substituting values for n and checking the options reveals that only option C satisfies the condition.

369 Medium

369 of 951

Analyzing +4 | -1

\frac{e^x}{1-x}=B_0+B_1 x+B_2 x^2+\ldots+B_n x^n+\ldots

, then

B_n-B_{n-1}

equals

View answer and explanation

Correct option: A

\frac{1}{n !}

Option A (

\frac{{1}}{{n !}}

) is correct because the coefficient difference can be found by expanding the given expression and comparing coefficients.

370 Medium

370 of 951

Application +4 | -1

If in the expansion of

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

, the coefficient of

5^{\text {th }}, 6^{\text {th }}

and

7^{\text {th }}

terms are in A.P., then

\mathrm{n}

is equal to-

View answer and explanation

Correct option: B

7,14

Option B (7,14) is correct because setting up the arithmetic progression equation for the coefficients and solving the quadratic equation yields these values for n.

371 Easy

371 of 951

Application +4 | -1

(1+\mathrm{x})^n=1+\mathrm{C}_1 \mathrm{x}+\mathrm{C}_2 \mathrm{x}^2+\ldots .+\mathrm{C}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}}

, then

\mathrm{C}_1+\mathrm{C}_3+\mathrm{C}_5+\ldots .

. is equal to-

View answer and explanation

Correct option: D

2^{n-1}

Option D (

2^{n-1}

) is correct because adding and subtracting the binomial expansion with x=1 and x=-1 isolates the sum of odd-indexed coefficients.

372 Easy

372 of 951

Analyzing +4 | -1

{(1 + x - 2{x^2})^6} = 1 + {a_1}x + {a_2}{x^2} + ... + {a_{12}}{x^{12}},

then the value of

{a_2} + {a_4} + ... + {a_{12}}\;

View answer and explanation

Correct option: A

The correct answer is A (31) because it is obtained by substituting x = 1 and x = -1 into the given equation and solving for the required sum.

373 Medium

373 of 951

Application +4 | -1

\sin n x=\sum\limits_{{r=0}}^n a_r \sin ^r x

, where

n

is an odd natural number. Then

View answer and explanation

Correct option: C

a_0=0, a_1=n

Option C (

a_0=0, a_1=n

) is correct because by expanding

\sin nx

using the imaginary part of Euler's formula, the constant term is zero and the coefficient of

\sin x

n

374 Easy

374 of 951

Analyzing +4 | -1

The value of

\frac{19^3+6^3+(3)(19)(6)(25)}{3^6+6(243)(2)+(15)(81)(4)+(20)(27)(8)+(15)(9)(16)+(6)(3)(32)+2^6}

View answer and explanation

Correct option: D

Option D (1) is correct because the numerator can be written as

(19+6)^3 = 25^3

and the denominator as

(3+2)^6 = 5^6

, so the fraction simplifies to 1.

375 Medium

375 of 951

Application +4 | -1

The unit digit of

17^{2009}+11^{2009}-7^{2009}

View answer and explanation

Correct option: A

Option A (97) is correct because it calculates the total number of terms and subtracts the number of rational terms to find the number of irrational terms.

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376 Medium

376 of 951

Application +4 | -1

One value of

\alpha

for which the coefficients of the middle terms in the expansion of

(1+\alpha x)^4

and

(1-\alpha x)^6

are equal,

View answer and explanation

Correct option: A

Option A (0) is correct because by equating the coefficients of the middle terms in both expansions, we find that one possible value for alpha is 0.

377 Medium

377 of 951

Analyzing +4 | -1

n

is an integer between 0 and 21 , then the minimum value of

n !(21-n)

! is attained for

n

, is equal to

View answer and explanation

Correct option: B

Option B (10) is correct because

n!(21-n)!

is minimized when

{ }^{21} C_n

is maximized, which occurs when

n

is closest to

\frac{21}{2}

, i.e.,

n=10

n=11

378 Medium

378 of 951

Application +4 | -1

The coefficient of

x^m

(1+x)^r+(1+x)^{r+1}+(1+x)^{r+2}+\cdots+(1+x)^n, r \leq m \leq n

View answer and explanation

Correct option: A

{ }^{n+1} C_{m+1}

Option A (

{ }^{n+1} C_{m+1}

) is correct because the sum of the binomial series can be simplified to a single binomial coefficient using the hockey-stick identity.

379 Easy

379 of 951

Application +4 | -1

The values of

a

and

b

so that the coefficient of

x^n

in the expansion of

\frac{a+b x}{(1-x)^2}

may be

2 n+1

, are

View answer and explanation

Correct option: A

1,1

Option A (1,1) is correct because by expanding the given expression and equating the coefficient of x^n to 2n+1, we get a=1 and b=1.

380 Easy

380 of 951

Analyzing +4 | -1

The term independent of

x

in the expansion of

{\left( {\frac{{2\sqrt x }}{5} - \frac{1}{{2x\sqrt x }}} \right)^{11}}\mathrm{is}

View answer and explanation

Correct option: D

No term

Option D (No term) is correct because the power of x in the general term is never zero for any integer value of r, indicating no term is independent of x.

381 Medium

381 of 951

Application +4 | -1

{ }^n C_0,{ }^n C_1,{ }^n C_2, \ldots,{ }^n C_n

denote the binomial coefficients in the expansion of

(1+x)^n

and

p+q=1

, then

\sum\limits_{{r=0}}^n r^2{ }^n C_r p^r q^{n-r}

View answer and explanation

Correct option: C

n^2 p^2+n p q

Option C (n² p²⁺n p q) is correct because it is derived by manipulating the summation and using the binomial theorem with the given condition p+q=1.

382 Easy

382 of 951

Application +4 | -1

Coefficient of

x^5

in the expansion of

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

View answer and explanation

Correct option: D

Option D (60) is correct because the coefficient of

x^5

in the expansion of

(1+x^2)^5(1+x)^4

is obtained by multiplying the relevant terms and summing them, resulting in 60.

383 Easy

383 of 951

Analyzing +4 | -1

The

\operatorname{sum} S={ }^{20} C_2+2 \cdot{ }^{20} C_3+3 \cdot{ }^{20} C_4+\ldots+19 \cdot{ }^{20} C_{20}

is equal to

View answer and explanation

Correct option: C

1+9 \cdot 2^{20}

The correct answer is C (

1+9 \cdot 2^{20}

) because it is derived by differentiating the binomial expansion and substituting x=1.

384 Easy

384 of 951

Application +4 | -1

Coefficient of

x^n

in the expansion of

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

is-

View answer and explanation

Correct option: C

\frac{2^n}{n !}

Option C (

\frac{2^n}{n !}

) is correct because the coefficient of

x^n

in the expansion is the sum of products of coefficients, which simplifies to the given expression.

385 Easy

385 of 951

Analyzing +4 | -1

The value of

\sum\limits_{{k=0}}^n\left({ }^{4 n+1} C_k+{ }^{4 n+1} C_{2 n-k}\right)

View answer and explanation

Correct option: A

2^{4 n}+{ }^{4 n+1} C_{2 n}

Option A (

2^{4 n}+{ }^{4 n+1} C_{2 n}

) is correct because the summation can be simplified to

2^{4n} + { }^{4n+1}C_{2n}

using properties of binomial coefficients.

386 Easy

386 of 951

Analyzing +4 | -1

If coefficient of

x^n

in the expansion of

(1+x)^{101}

\left(1-x+x^2\right)^{100}

is non-zero, then

n

cannot be of the form

View answer and explanation

Correct option: C

3 \lambda+2

Option C (

3 \lambda+2

) is correct because the expansion of the given expression only contains terms of the form

x^{3 \lambda}

x^{3 \lambda+1}

387 Easy

387 of 951

Application +4 | -1

If in the expansion of

\left(2^x+\frac{1}{4^x}\right)^n, T_3 / T_2=7

and the sum of the coefficient of

2^{\text {nd }}

and

3^{\text {rd }}

terms is 36 , then the value of

x

View answer and explanation

Correct option: A

-1 / 3

Option A (-1/3) is correct because solving the given conditions

{ }^n C_1+{ }^n C_2=36

and

\frac{{ }^n C_2(2^x)^{n-2} (\frac{{1}}{{4^x}})^2}{{ }^n C_1(2^x)^{n-1} (\frac{{1}}{{4^x}})^1}=7

leads to

x = -1/3

388 Medium

388 of 951

Analyzing +4 | -1

The value of

\sum\limits_{{k=8}}^{23}(-1)^{k+1} \, \ { }^{31} C_k

is equal to

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the given summation simplifies to 0 using properties of binomial coefficients and their alternating sums.

389 Medium

389 of 951

Application +4 | -1

\log x=\log _{10} x

, the values of

x

for which 4 th term in the expansion of

\left[\sqrt{x^{2 /(1+\log x)}}+x^{1 / 12}\right]^6

is 200 are

View answer and explanation

Correct option: B

0.0001,10

Option B (0.0001, 10) is correct because solving the equation resulting from the binomial expansion leads to these values of x.

390 Medium

390 of 951

Analyzing +4 | -1

{\left( {1 + x - 2{x^2}} \right)^6} = 1 + {a_1}x + {a_2}{x^2} + ... + {a_{12}}{x^{12}}

, then the expression

{a_2} + {a_4} + {a_6} + ... + {a_{12}}

has the value

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the value of the expression is 31, which is not among the provided options.

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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