Back to BINOMIAL THEOREM

Your Progress

0 / 434 Answered

Correct

Incorrect

Accuracy

JEE • MATHEMATICS • BINOMIAL THEOREM

Binomial theorem practice · Easy

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 331-360 of 434 questions

Preparing practice and saved progress…

331 Easy

331 of 434

Application +4 | -1

The coefficient of

\left(a^3 \cdot b^6 \cdot c^8 \cdot d^9 \cdot e \cdot f\right)

in the expansion of

(a+b+c-d-e-f)^{31}

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the coefficient is calculated using the multinomial theorem, and the provided options do not match the correct calculation.

332 Easy

332 of 434

Application +4 | -1

If the expansion of

{\left( {\frac{{3\sqrt x }}{7} - \frac{5}{{2x\sqrt x }}} \right)^{13n}}

contains a term independent of

x

in 14th term,then

n

should be

View answer and explanation

Correct option: D

Option D (4) is correct because the value of n that makes the 14th term independent of x is found by setting the exponent of x in the 14th term to zero and solving for n.

333 Easy

333 of 434

Application +4 | -1

The term independent of

x

in the expansion of

\left( {1 - {x^2} + 3{x^3}} \right){\left( {\frac{5}{2}{x^3} - \frac{1}{{5{x^2}}}} \right)^{11}},\,\,x \ne 0

is:

View answer and explanation

Correct option: B

\frac{{33}}{{200}}

Option B (33/200) is correct because it is the constant term obtained after expanding the binomial and multiplying by the polynomial.

334 Easy

334 of 434

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.

335 Easy

335 of 434

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.

336 Easy

336 of 434

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.

337 Easy

337 of 434

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.

338 Easy

338 of 434

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.

339 Easy

339 of 434

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.

340 Easy

340 of 434

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.

341 Easy

341 of 434

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.

342 Easy

342 of 434

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.

343 Easy

343 of 434

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.

344 Easy

344 of 434

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}}

345 Easy

345 of 434

Application +4 | -1

If the coefficient of

{x^7}

in the expansion of

{\left( {a{x^2} + {b^{ - 1}}{x^{ - 1}}} \right)^{11}}

is equal to the coefficient of

{x^{ - 7}}

{\left( {ax - {b^{ - 1}}{x^{ - 2}}} \right)^{11}},

, then

ab =

View answer and explanation

Correct option: A

Option A (1) is correct because equating the coefficients of

{x^7}

and

{x^{ - 7}}

leads to the relationship

ab = 1

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

346 Easy

346 of 434

Understanding +4 | -1

The coefficients of

x^{p}

and

x^{q}

in the expansion of

(1+x)^{p+q}

are

View answer and explanation

Correct option: A

equal

Option A is correct because the coefficients of x^p and x^q in the expansion of (1+x)^(p+q) are equal due to the property of binomial coefficients: nCr = nC(n-r).

347 Easy

347 of 434

Analyzing +4 | -1

s_n=\sum\limits_{{r=0}}^n \frac{1}{{ }^n C_r}

and

t_n=\sum\limits_{{r=0}}^n \frac{r}{{ }^n C_r}

, then

\frac{t_n}{s_n}

is equal to

View answer and explanation

Correct option: A

\frac{n}{2}

Option A (

\frac{n}{2}

) is correct because by using the property

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

and manipulating the summations, we find that

\frac{t_n}{s_n} = \frac{n}{2}

348 Easy

348 of 434

Application +4 | -1

Which term in the expansion of the binomial

\left[\sqrt[3]{\left(\frac{a}{\sqrt{b}}\right)}+\sqrt{\left(\frac{b}{\sqrt[3]{a}}\right)}\right]^{21}

contains

a

and

b

to one and the same power?

View answer and explanation

Correct option: C

10 th

Option B (6) is correct because the area of the quadrilateral OABC is 6 times the area of the parallelogram formed by vectors OA and OC.

349 Easy

349 of 434

Application +4 | -1

The coefficient of

{x^7}

in the expansion of

{\left( {x - 2\;{x^2}} \right)^{ - 3}},

View answer and explanation

Correct option: C

67584

Option C is correct. The coefficient of x⁷ is found by identifying the correct term in the expansion and calculating its coefficient.

350 Easy

350 of 434

Application +4 | -1

In the expansion of

\left(\sqrt{\frac{q}{p}}+\sqrt[10]{\frac{p^7}{q^3}}\right)^n

, there is a term similar to

p q

, then that term is equal to

View answer and explanation

Correct option: D

252 p q

Option D (

252 p q

) is correct because when the binomial expansion is simplified and the powers of p and q are set to 1, the resulting coefficient is 252.

351 Easy

351 of 434

Understanding +4 | -1

The largest term in the expansion of

\left(\frac{b}{2}+\frac{b}{2}\right)^{100}

View answer and explanation

Correct option: C

{ }^{100} C_{50}\left(\frac{b}{2}\right)^{100}

Option C (

{ }^{100} C_{50}(\frac{b}{2})^{100}

) is correct because the largest term in a binomial expansion occurs at the middle term when the coefficients are equal.

352 Easy

352 of 434

Application +4 | -1

5^{99}

is divided by 13 , the remainder is

View answer and explanation

Correct option: D

Option D (8) is correct because when

5^{99}

is divided by 13, the remainder can be found using modular arithmetic and binomial theorem.

353 Easy

353 of 434

Analyzing +4 | -1

The term independent of

x

in expansion of

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

View answer and explanation

Correct option: C

210

Option C (210) is correct because after simplifying the expression inside the parentheses and applying the binomial theorem, the term independent of x is found to be 210.

354 Easy

354 of 434

Analyzing +4 | -1

a=99^{50}+100^{50}

and

b=101^{50}

, then

View answer and explanation

Correct option: A

a < b

Option A (

a < b

) is correct because

101^{50}-99^{50}

can be expanded using the binomial theorem, showing that

b

is greater than

a

355 Easy

355 of 434

Application +4 | -1

In the expansion of

{\left( {1 + x} \right)^{50}},

the sum of the coefficient of add power of

x

View answer and explanation

Correct option: B

{2^{49}}

Option B (

{2^{49}}

) is correct because the sum of the coefficients of odd powers of x in the expansion of

{\left( {1 + x} \right)^{50}}

is equal to

{2^{50 - 1}}

356 Easy

356 of 434

Application +4 | -1

5^{40}

is divided by 11 , then remainder is

\alpha

and when

2^{2003}

is divided by 17 , then remainder is

\beta

, then the value of

\beta-\alpha

View answer and explanation

Correct option: C

Option C (7) is correct because after calculating the remainders α and β, their difference is 7.

357 Easy

357 of 434

Application +4 | -1

The number of distinct terms in the expansion of

(x+2 y-3 z+5 w-7 u)^n

View answer and explanation

Correct option: D

None of these

Option D is correct because the number of distinct terms in the expansion is calculated using combinations, and the result does not match the other options.

358 Easy

358 of 434

Analyzing +4 | -1

The remainder, if

1+2+2^2+2^3+\ldots+2^{1999}

is divided by 5 , is

View answer and explanation

Correct option: A

Option A (0) is correct because the sum is a geometric progression that simplifies to a form divisible by 5, leaving no remainder.

359 Easy

359 of 434

Application +4 | -1

If the coefficient of

r th

and

\left( {r + 1} \right)

th terms in the expansion of

{(3 + 7x)^{29}}

are equal, then

r

equals

View answer and explanation

Correct option: B

Option B (21) is correct because it is the value of

r

that makes the coefficients of the

r

th and

(r+1)

th terms equal in the expansion of

{(3 + 7x)^{29}}

360 Easy

360 of 434

Application +4 | -1

If the fifth term in the expansion of

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

does not depend upon

x

, then

{ }^n C_2+{ }^n C_{14}

equals

View answer and explanation

Correct option: A

240

Option A (240) is correct because the value of n is determined by the condition that the fifth term is independent of x, and then the expression is calculated.

Revision notes and key takeaways

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

Showing 331-360 of 434 questions

Your Progress

Binomial theorem practice · Easy

Explore Other Chapters in MATHEMATICS

Continue Your Practice

Your practice