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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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211 Medium

211 of 951

Analyzing +4 | -1

x

is nearly equal to one, then the approximate value of

\frac{m x^m-n x^n}{m-n}

View answer and explanation

Correct option: C

x^{m+n}

Option C (

x^{m+n}

) is correct because using the approximation

x = 1 + h

and neglecting higher powers of

h

simplifies the expression.

212 Medium

212 of 951

Application +4 | -1

The greatest term in the expansion of

(1+x)^{10}

, when

x=\frac{2}{3}

, is

View answer and explanation

Correct option: A

210\left(\frac{2}{3}\right)^4

Option A (210(2/3)⁴) is correct because it is found by determining the greatest term in the binomial expansion using the ratio test and calculating the corresponding term.

213 Easy

213 of 951

Application +4 | -1

Let

P_n

denote the product of all the coefficients in the expansion of

(1+x)^n

. If

(20) ! P_{n+1}=21^{20} P_n

, then

n

is equal to

View answer and explanation

Correct option: B

Option B (20) is correct. By comparing the product of coefficients in consecutive binomial expansions, the value of n is determined.

214 Medium

214 of 951

Application +4 | -1

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

\left(1+x+x^2+x^3+\ldots+x^n\right)=a_0+a_1 x+a_2 x^2+\ldots+a_m x^m

then

\sum_{r=0}^m a_r=

View answer and explanation

Correct option: D

n

Option D (

n

!) is correct because substituting x=1 into the given equation results in the factorial of n.

215 Medium

215 of 951

Understanding +4 | -1

1+\frac{1}{3} x+\frac{1 \times 4}{3 \times 6} x^2+\frac{1 \times 4 \times 7}{3 \times 6 \times 9} x^3+\ldots

is equal to

View answer and explanation

Correct option: D

(1-x)^{-1 / 3}

Option D (

(1-x)^{-1 / 3}

) is correct because the given series is the binomial expansion of

(1-x)^{-1 / 3}

216 Easy

216 of 951

Understanding +4 | -1

m

and

n

are positive integers, then value of

{ }^m C_0{ }^n C_k+{ }^m C_1{ }^n C_{k-1}+\ldots+{ }^m C_k{ }^n C_0

View answer and explanation

Correct option: A

{ }^{m+n} C_k

Option D (None of these) is correct because the value of

\arg(z) + \arg(\bar{z})

depends on whether z is a negative real number or not. If z is a negative real number, the sum is

2\pi

, otherwise it is 0.

217 Easy

217 of 951

Application +4 | -1

If the coefficient of

rth

\left( {r + 1} \right)th

and

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

terms in the binomial expansion of

{(1 + y)^m}

are in AP, then

m

and

r

satisfy the equation

View answer and explanation

Correct option: B

{m^2} - m\left( {4r + 1} \right) + 4{r^2} - 2 = 0

Option B (

{m^2} - m\left( {4r + 1} \right) + 4{r^2} - 2 = 0

) is correct because it is derived from the condition that the coefficients of the given terms are in Arithmetic Progression (AP).

218 Medium

218 of 951

Application +4 | -1

The coefficient of

x^6 \cdot y^{-2}

in the expansion of

\left(\frac{x^2}{y}-\frac{y}{x}\right)^{12}

View answer and explanation

Correct option: C

Option C (0) is correct because the conditions for the powers of x and y to be x⁶ and y^-2 cannot be simultaneously satisfied.

219 Easy

219 of 951

Application +4 | -1

n

is even, then the middle term in the expansion of

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

is 924

{x^6}

, then

n

is equal to

View answer and explanation

Correct option: B

Option B (12) is correct because the middle term in the expansion of

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

is given as 924

{x^6}

, and solving for n yields n=12.

220 Medium

220 of 951

Analyzing +4 | -1

\left(3+x^{2008}+x^{2009}\right)^{2010}=a_0+a_1 x+a_2 x^2

+\ldots+a_n x^n

, the value of

a_0-\frac{a_1}{2}-\frac{a_2}{2}+a_3-\frac{a_4}{2}-\frac{a_5}{2}+a_6-\ldots

View answer and explanation

Correct option: B

2^{2010}

Option B (

2^{2010}

) is correct because substituting complex cube roots of unity into the given polynomial and summing the results leads to the desired expression.

221 Easy

221 of 951

Application +4 | -1

The ninth term of the expansion

{\left( {3x - \frac{1}{{2x}}} \right)^8}

View answer and explanation

Correct option: D

\frac{1}{{256{x^8}}}

Option D (

\frac{1}{{256{x^8}}}

) is correct because substituting r=8 into the general term of the binomial expansion and simplifying leads to this result.

222 Easy

222 of 951

Analyzing +4 | -1

In the expansion of

2 \log _e x-\log _e(x+1)-\log _e(x-1)

, the coefficient of

x^{-4}

View answer and explanation

Correct option: A

\frac{1}{2}

Option A (

\frac{{1}}{{2}}

) is correct because the coefficient of

x^{-4}

in the expansion is found by using logarithmic properties and series expansion, which leads to the coefficient being

\frac{{1}}{{2}}

223 Easy

223 of 951

Application +4 | -1

If the coefficient of 4 th term in the expansion of

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

is 20 , then the respective values of

\alpha

and

n

are

View answer and explanation

Correct option: D

2,6

The correct answer is D (2,6) because by equating the power of x to zero and comparing the constant term, we can find the values of alpha and n.

224 Medium

224 of 951

Application +4 | -1

If the

r^{\text {th }}

term in the expansion of

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

contains

x^4

, then the value of

r

View answer and explanation

Correct option: A

Option A (3) is correct because the

r^{\text {th }}

term in the expansion of

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

containing

x^4

leads to the equation

13-3r = 4

, which gives

r=3

225 Easy

225 of 951

Application +4 | -1

The value of

\left\{\frac{3^{2003}}{28}\right\}

, where

\{ .\}

denotes the fractional part function is

View answer and explanation

Correct option: B

19 / 28

Option B (19/28) is correct because the fractional part of

\frac{3^{2003}}{28}

can be found using modular arithmetic and binomial theorem.

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226 Easy

226 of 951

Application +4 | -1

The coefficient of

x^m

(1+x)^m+(1+x)^{m+1}+\ldots+(1+x)^n, m \leq n

, is

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 binomial coefficients can be simplified to a single binomial coefficient using the hockey-stick identity.

227 Medium

227 of 951

Application +4 | -1

\left(\begin{array}{c} 20 \\ 0 \end{array}\right)-\left(\begin{array}{c} 20 \\ 1 \end{array}\right)+\left(\begin{array}{c} 20 \\ 2 \end{array}\right)-\left(\begin{array}{c} 20 \\ 3 \end{array}\right)+\ldots+\left(\begin{array}{l} 20 \\ 10 \end{array}\right)=

View answer and explanation

Correct option: B

\frac{1}{2}\left(\begin{array}{l}20 \\ 10\end{array}\right)

Option B is correct because by substituting x = -1 into the binomial expansion and using the property that C(n,r) = C(n, n-r), the given sum simplifies to half of C(20, 10).

228 Easy

228 of 951

Application +4 | -1

The value of the sum

\sum\limits_{{j=0}}^8 \frac{1}{(j+1)(j+2)}\left(\begin{array}{l} 8 \\ j \end{array}\right)

View answer and explanation

Correct option: B

\frac{1013}{90}

Option B (

\frac{1013}{90}

) is correct because by rewriting the sum using binomial coefficient identities and then evaluating, we find the value to be

\frac{1013}{90}

229 Medium

229 of 951

Application +4 | -1

\left(1+x-2 x^2\right)^8=a_0+a_1 x+a_2 x^2+\ldots+a_{16} x^{16}

, then the sum

a_1+a_3+a_5+\ldots+a_{15}

is equal to

View answer and explanation

Correct option: A

-2^7

Option A (-2⁷) is correct because it calculates the sum of odd-powered coefficients using the given polynomial and substituting x=1 and x=-1.

230 Easy

230 of 951

Application +4 | -1

The number of irrational terms in the expansion of

\left(2^{1 / 5}+3^{1 / 10}\right)^{55}

View answer and explanation

Correct option: C

Option C (50) is correct because the number of irrational terms is found by subtracting the number of rational terms from the total number of terms.

231 Medium

231 of 951

Analyzing +4 | -1

Let

n

be a positive integer such that

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

, then

a_0+a_1+a_2+\ldots+a_{n-1}

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because it utilizes the symmetry of the coefficients and the total sum to isolate the desired partial sum.

232 Easy

232 of 951

Application +4 | -1

The number of distinct terms in the expansion of

(x+y-z)^{16}

View answer and explanation

Correct option: B

153

Option B (153) is correct because the number of distinct terms in the expansion is the sum of integers from 1 to n+1, where n is the power.

233 Easy

233 of 951

Application +4 | -1

In the expansion of

{(2 - 3{x^3})^{20}},

if the ratio of 10th term to 11th term is

\frac{{45}}{{22}}

, then

x

is equal to

View answer and explanation

Correct option: A

- \frac{2}{3}

Option A (

- \frac{2}{3}

) is correct because setting up the ratio of the 10th and 11th terms and solving for x gives

x = - \frac{2}{3}

234 Easy

234 of 951

Application +4 | -1

Coefficient of

x^4

in the expansion of

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

View answer and explanation

Correct option: B

\frac{405}{256}

Option B (405/256) is correct. The coefficient of x⁴ is found by determining the correct term in the binomial expansion.

235 Easy

235 of 951

Application +4 | -1

The total number of terms in the expansion of

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

after simplification will be

View answer and explanation

Correct option: B

Option B (51) is correct because only the even power terms remain after simplification, resulting in half the number of terms plus one.

236 Medium

236 of 951

Application +4 | -1

Let

m

and

n

be the coefficients of seventh and thirteenth terms respectively in the expansion of

\left(\frac{1}{3} x^{\frac{1}{3}}+\frac{1}{2 x^{\frac{2}{3}}}\right)^{18}

. Then

\left(\frac{n}{m}\right)^{\frac{1}{3}}

View answer and explanation

Correct option: C

\frac{9}{4}

Option C (

\frac{9}{4}

) is correct because it is the simplified ratio of the coefficients of the seventh and thirteenth terms in the binomial expansion, raised to the power of

\frac{1}{3}

237 Easy

237 of 951

Analyzing +4 | -1

The value of

\frac{1}{{{{81}^n}}} - \frac{{10}}{{{{81}^n}}}{\;^{2n}}{C_1}\;\frac{{{{10}^2}}}{{{{81}^n}}}{\;^{2n}}{C_2} - \frac{{{{10}^3}}}{{{{81}^n}}}2{\;^{2n}}{C_3} + ... + \frac{{{{10}^{2n}}}}{{{{81}^n}}}\;\mathrm{is}

View answer and explanation

Correct option: D

The correct answer is D (1) because the expression can be recognized as a binomial expansion, which simplifies to 1.

238 Easy

238 of 951

Understanding +4 | -1

The middle term in the expansion of

{\left( {x - a} \right)^8}

View answer and explanation

Correct option: B

{\;^8}{C_4}{x^4}{a^4}

Option B (

{\;^8}{C_4}{x^4}{a^4}

) is correct because the middle term in the expansion of

{\left( {x - a} \right)^8}

is the 5th term, which is

{\;^8}{C_4}{x^4}{\left( { - a} \right)^4} = {\;^8}{C_4}{x^4}{a^4}

239 Medium

239 of 951

Analyzing +4 | -1

The value of

\frac{{{{18}^3} + {7^3} + 3.18.7.25}}{{{3^6} + 6.243.2 + 15.181.4 + 20.27.8 + 15.9.16 + 6.3.32 + 64}},

View answer and explanation

Correct option: B

The correct answer is B (1) because both the numerator and denominator can be simplified using the binomial theorem, resulting in the same expression.

240 Medium

240 of 951

Analyzing +4 | -1

Larger of

99^{50}+100^{50}

and

101^{50}

View answer and explanation

Correct option: A

101^{50}

Option A (

101^{50}

) is correct because

101^{50}

is larger than

99^{50} + 100^{50}

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