If ${\left( {1 + 2\;x + {x^2}} \right)^n} = \mathop \sum \limits_{r = 0}^{2\;n} a,\;{x^r},$ Then ${a_r} = $

${\left( {{\;^n}{C_r}} \right)^2}$

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JEE • MATHEMATICS • BINOMIAL THEOREM

Binomial theorem practice · Medium

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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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211 Medium

211 of 468

Application +4 | -1

The coefficient of

x^{10}

in the expansion of

\left(3 x^2-\frac{1}{x^2}\right)^{15}

View answer and explanation

Correct option: B

-\frac{15 !}{10 ! 5 !} 3^{10}

Option B is correct because the coefficient of x¹⁰ in the expansion of (3x² - 1/x²)¹⁵ is - (15! / (10! 5!)) * 3¹⁰, which is found using the binomial theorem.

212 Medium

212 of 468

Application +4 | -1

{\;^{15}}{C_0}{.^5}{C_5}{ + ^{15}}{C_1}{.^5}{C_4}{ + ^{15}}{C_2}{.^5}{C_3}{ + ^{15}}{C_3}{.^5}{C_2}{ + ^{15}}{C_4}{.^5}{C_1}

is equal to

View answer and explanation

Correct option: D

\frac{{20!}}{{5!15!}} - \frac{{15!}}{{5!10!}}

Option D (

\frac{{20!}}{{5!15!}} - \frac{{15!}}{{5!10!}}

) is correct because it represents the coefficient of x¹⁵ in the expansion of (1+x)²⁰ minus the term where both binomials contribute to x¹⁵.

213 Medium

213 of 468

Analyzing +4 | -1

9^7+7^9

is divisible by

View answer and explanation

Correct option: C

Option C (64) is correct because

9^7 + 7^9

is divisible by 64.

214 Medium

214 of 468

Application +4 | -1

The term that is independent of

x

, in the expansion of

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

View answer and explanation

Correct option: B

{ }^9 C_6\left(\frac{1}{6}\right)^3

Option B (

{ }^9 C_6\left(\frac{{1}}{{6}}\right)^3

) is correct because it represents the constant term in the expansion, obtained by setting the power of x to zero and simplifying.

215 Medium

215 of 468

Application +4 | -1

For any positive integers

m, n

with

n \geq m

, let

\left(\begin{array}{l}n \\ m\end{array}\right)={ }^n C_m

. Then,

\left(\begin{array}{l}n \\ m\end{array}\right)+\left(\begin{array}{c}n-1 \\ m\end{array}\right)+\left(\begin{array}{c}n-2 \\ m\end{array}\right)+\ldots+\left(\begin{array}{l}m \\ m\end{array}\right)

is equal to

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because the sum of binomial coefficients can be simplified using the identity

{ }^n C_r + { }^n C_{r-1} = { }^{n+1} C_r

repeatedly.

216 Medium

216 of 468

Application +4 | -1

The coefficient of

x^6

\left\{(1+x)^6+(1+x)^7+\ldots+(1+x)^{15}\right\}

View answer and explanation

Correct option: A

{ }^{16} C_9

Option A (

{ }^{16} C_9

) is correct because the given expression simplifies to a single binomial expansion, and the coefficient of x⁶ is

{ }^{16} C_7

which equals

{ }^{16} C_9

217 Medium

217 of 468

Analyzing +4 | -1

Value of

\left({ }^{40} C_0\right)\left({ }^{40} C_{15}\right)-\left({ }^{40} C_1\right)\left({ }^{40} C_{16}\right)+\left({ }^{40} C_2\right)\left({ }^{40} C_{12}\right) \cdots

-\left({ }^{40} C_{25}\right)\left({ }^{40} C_{40}\right)

equals

View answer and explanation

Correct option: A

Option A (0) is correct because the given expression represents the coefficient of x²⁵ in (1-x)⁴⁰(1+x)⁴⁰ which simplifies to (1-x²)⁴⁰, and thus the coefficient of x²⁵ is 0.

218 Medium

218 of 468

Application +4 | -1

The coefficient of

x^7

in the expansion of

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

View answer and explanation

Correct option: B

-144

Option B (-144) is correct because it is the result of calculating the coefficient of

x^7

after factoring and expanding the given expression.

219 Medium

219 of 468

Application +4 | -1

x

is so small such that its square and higher powers may be neglected, then the value of

\frac{(8+3 x)^{2 / 3}}{(2+3 x)(4-5 x)^{1 / 2}}

View answer and explanation

Correct option: C

1-\frac{5}{8} x

Option C (

1-\frac{5}{8} x

) is correct because it is the simplified form of the given expression when higher powers of x are neglected.

220 Medium

220 of 468

Application +4 | -1

An examiner is supposed to give 25 marks from 8 equations. If he has to give at least 2 marks in a question, in how many ways can he do so?

View answer and explanation

Correct option: D

\left(\begin{array}{c}16 \\ 7\end{array}\right)

Option D

\left(\begin{array}{c}16 \\ 7\end{array}\right)

is correct because it represents the number of ways to distribute the marks with the condition that each equation receives at least 2 marks.

221 Medium

221 of 468

Application +4 | -1

If the coefficient of

x^8

\left(a x^2+\frac{1}{b x}\right)^{13}

is equal to the coefficient of

x^{-8}

\left(a x-\frac{1}{b x^2}\right)^{13}

, then

a

and

b

will satisfy the relation

View answer and explanation

Correct option: A

a b+1=0

Option A (

a b+1=0

) is correct because equating the coefficients of

x^8

and

x^{-8}

leads to the relation

ab = -1

222 Medium

222 of 468

Analyzing +4 | -1

\mathrm{\alpha } = \frac{5}{{2!3}} + \frac{{5.7}}{{3!{3^2}}} + \frac{{5.7.9}}{{4!{3^2}}} + . \ldots \ldots ..,\;

if then

{\mathrm{\alpha }^2} + 4\mathrm{\alpha }

equal to

View answer and explanation

Correct option: B

Option B (23) is correct because, after comparing the given series with the binomial expansion and solving for α, we find that α^2 + 4α equals 23.

223 Medium

223 of 468

Application +4 | -1

The coefficient of

x^3

in the infinite series expansion of

\frac{2}{(1-x)(2-x)}

, for

|x| < 1

View answer and explanation

Correct option: B

\frac{15}{8}

Option B (

\frac{{15}}{{8}}

) is correct because it is the result of calculating the coefficient of

x^3

in the series expansion of the given function.

224 Medium

224 of 468

Analyzing +4 | -1

{\left( {1 + x} \right)^n} = \sum \limits_{r = 0}^n {a_r}{x^r}

and

{b_r} = 1 + \frac{{{a_r}}}{{{a_r} - 1}}

and

\mathop \Pi \limits_{r = 1}^n {b_r} = \frac{{{{\left( {101} \right)}^{100}}}}{{100!}}

, then

n

View answer and explanation

Correct option: B

100

Option B (100) is correct because by comparing the given product with the binomial expansion, we can deduce that n = 100.

225 Medium

225 of 468

Application +4 | -1

The sum of coefficients of integral power of

x

in the binomial expansion

(1-2 \sqrt{x})^{50}

is :

View answer and explanation

Correct option: C

\frac{1}{2}\left(3^{50}+1\right)

Option C (

\frac{1}{2}(3^{50}+1)

) is correct because it uses the property of binomial expansions to find the sum of coefficients of integral powers of x.

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226 Medium

226 of 468

Application +4 | -1

The ratio of the coefficient of

x^{10}

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

and the term independent of

x

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

View answer and explanation

Correct option: B

1: 32

Option B (1: 32) is correct because the ratio of the coefficient of x¹⁰ in (1+x²)¹⁰ to the term independent of x in (x+2/x)¹⁰ is 1:32.

227 Medium

227 of 468

Application +4 | -1

{\left( {1 + 2\;x + {x^2}} \right)^n} = \sum \limits_{r = 0}^{2\;n} a,\;{x^r},

Then

{a_r} =

View answer and explanation

Correct option: C

{\;^{2n}}{C_r}

Option C (

\;^{2n}{C_r}

) is correct because simplifying the expression and applying the binomial theorem leads to this result.

228 Medium

228 of 468

Analyzing +4 | -1

The greatest integer less than

(\sqrt{2}+1)^6

is equal to

View answer and explanation

Correct option: B

197

The correct answer is B (197) because it is the greatest integer less than

(\sqrt{2}+1)^6

, calculated by considering

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

and the fact that

(\sqrt{2}-1)^6

is between 0 and 1.

229 Medium

229 of 468

Application +4 | -1

The value of

\sum\limits_{{i=0}}^{n-r}{ }^{n-i} C_r, r \in[1, n]

is equal to

View answer and explanation

Correct option: D

{ }^{n+1} C_{r+1}

Option D (

{ }^{n+1} C_{r+1}

) is correct because the summation represents a hockey stick identity, which simplifies to

{ }^{n+1} C_{r+1}

230 Medium

230 of 468

Analyzing +4 | -1

p

is prime number and

n < p < 2 n

, if

N={ }^{2 n} C_n

, then

View answer and explanation

Correct option: A

p

divides

N

Option A is correct because since p is a prime number between n and 2n, it will divide the numerator of the binomial coefficient but not the denominator, thus dividing N.

231 Medium

231 of 468

Application +4 | -1

There are numerous balls of red, white, black and green colour. In how many ways 11 balls can be chosen from these?

View answer and explanation

Correct option: B

120

The correct answer is 120 because it represents the number of ways to choose 11 balls from the four colors.

232 Medium

232 of 468

Application +4 | -1

\sum\limits_{{r=1}}^n r(r+1) \frac{C_r}{C_{r-1}}=77

, then

n

equals

View answer and explanation

Correct option: B

Option B (6) is correct because substituting n=6 into the simplified summation formula yields 77, satisfying the given relation.

233 Medium

233 of 468

Analyzing +4 | -1

The value of

{ }^n C_0,{ }^{2 n} C_n-{ }^n C_1,{ }^{2 n-2} C_n +{ }^n C_2 \cdot{ }^{2 n-4} C_n-\ldots \text { is }

View answer and explanation

Correct option: C

2^n

Option C (

2^n

) is correct because the expression simplifies to the coefficient of

x^n

(2x+x^2)^n

, which is

2^n

234 Medium

234 of 468

Understanding +4 | -1

The expansion of

\frac{1}{\sqrt[5]{7-3 x}}

in exponents of

x

is possible, if

View answer and explanation

Correct option: A

|x| < \frac{7}{3}

Option A (

|x| < \frac{7}{3}

) is correct because the binomial expansion is valid only when

|\frac{3x}{7}| < 1

235 Medium

235 of 468

Application +4 | -1

In the binomial expansion of

{\left( {a - b} \right)^n},\;n \ge 5,

the sum of 5th and 6th terms is zero, then

\frac{a}{b}

equals

View answer and explanation

Correct option: D

\frac{{n - 4}}{5}

Option D (

\frac{{n - 4}}{5}

) is correct because setting the sum of the 5th and 6th terms of the binomial expansion to zero and solving for a/b yields this result.

236 Medium

236 of 468

Application +4 | -1

The coefficient of

x^3

\left(\sqrt{x^5}+\frac{3}{\sqrt{x^3}}\right)^6

View answer and explanation

Correct option: D

540

Option D (540) is correct because after finding the correct term in the expansion, its coefficient is calculated to be 540.

237 Medium

237 of 468

Application +4 | -1

A set contains

(2 n+1)

elements. If the number of subsets of this set which contain at most

\mathrm{n}

elements is 4096 , then the value of

\mathrm{n}

View answer and explanation

Correct option: A

Option A (6) is correct because by using the properties of binomial coefficients and the given condition, we can derive that n = 6.

238 Medium

238 of 468

Application +4 | -1

Coefficient of constant term in the expansion of

\left(x^3+5(3)^{-\log \sqrt{3}} \sqrt{x^3}\right)^4

View answer and explanation

Correct option: D

150

Option D (150) is correct. The constant term is found by determining the term independent of x in the binomial expansion.

239 Medium

239 of 468

Application +4 | -1

The absolute difference of the coefficients of

x^{10}

and

x^{7}

in the expansion of

\left(2 x^{2}+\frac{1}{2 x}\right)^{11}

is equal to

View answer and explanation

Correct option: D

12^{3}-12

Option D (

12^3 - 12

) is correct because it is the absolute difference between the coefficients of

x^{10}

and

x^{7}

in the binomial expansion.

240 Medium

240 of 468

Application +4 | -1

The last two digit of number

3^{400}

are

View answer and explanation

Correct option: D

Option D (1) is correct because

3^{400}

can be expressed as

81^{100}

, which is equivalent to

(1+80)^{100}

. When expanded using the binomial theorem, all terms except the first one are divisible by 100, leaving a remainder of 1.

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