The coefficient of $x^{60}$ in $(1+x)^{51}\left(1-x+x^2\right)^{50}$ is

$-\left({ }^{50} C_{20}\right)$

$-\left({ }^{51} C_{20}\right)$

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JEE • MATHEMATICS • BINOMIAL THEOREM

Binomial theorem practice · Mixed difficulty

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481 Medium

481 of 951

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.

482 Medium

482 of 951

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.

483 Medium

483 of 951

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 the correct answer because it is the coefficient of

x^3

after expanding and simplifying the given expression.

484 Hard

484 of 951

Analyzing +4 | -1

The coefficient of

{x^n}\;{y^n}

in the expansion of

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

View answer and explanation

Correct option: D

\sum \limits_{r = 0}^n C_r^3

Option D (

\sum \limits_{r = 0}^n C_r^3

) is correct because the coefficient of x^n y^n in the given expansion is the sum of the cubes of the binomial coefficients.

485 Medium

485 of 951

Application +4 | -1

r

and

n

are positive integers

r>1, n>2

and coefficient of

(r+2)^{\text {th }}

term and

3 r^{\text {th }}

term in the expansion of

(1+x)^{2 n}

are equal, then

n

equals)

View answer and explanation

Correct option: C

2 r

Option C (2r) is correct because equating the coefficients and solving for n yields n = 2r.

486 Easy

486 of 951

Application +4 | -1

The coefficient of

{x^r}

in the expansion of

{(1 - x)^{ - 2}}

View answer and explanation

Correct option: B

r + 1

Option B (r + 1) is correct because the coefficient of x^r in the expansion of (1-x)^(-2) is (r+1).

487 Medium

487 of 951

Application +4 | -1

\left(1+x-2 x^2\right)^{20}=a_0+a_1 x+a_2 x^2+a_3 x^3

+\ldots+a_{40} x^{40}

, then

a_1+a_3+a_5+\ldots+a_{39}

is equal to

View answer and explanation

Correct option: A

-2^{19}

Option A (

-2^{19}

) is correct because substituting x=1 and x=-1 into the polynomial and subtracting the results allows us to isolate the sum of odd-powered coefficients.

488 Easy

488 of 951

Application +4 | -1

If middle term in the expansion to

\left(x^{1 / 2}+x^k\right)^{16}

is also independent term

View answer and explanation

Correct option: C

16 x^7

The correct answer is C (

16 x^7

) because it is obtained by finding the value of k that makes the middle term independent of x, and then finding the term independent of x.

489 Easy

489 of 951

Analyzing +4 | -1

1 + \frac{{2.1}}{{3.2}} + \frac{{2.5}}{{3.6}}{\left( {\frac{1}{2}} \right)^2} + \frac{{2.5.8}}{{3.6.9}}.{\left( {\frac{1}{2}} \right)^3} + ...

is equal to

View answer and explanation

Correct option: C

{4^{1/3}}

Option C (

{4^{1/3}}

) is correct because the given series can be recognized as a binomial expansion of the form (1-x)^(-n), which simplifies to

{4^{1/3}}

490 Medium

490 of 951

Application +4 | -1

If the term free from

x

in the expansion of

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

is 405 , then the value of

k

View answer and explanation

Correct option: B

\pm 3

Option B (

\pm 3

) is the correct value of k because it satisfies the condition that the term free from x in the given expansion is 405.

491 Medium

491 of 951

Understanding +4 | -1

The value of

\sum\limits_{0 \le i} {\sum\limits_{ < j < } {\sum\limits_{k < l} {\sum\limits_{ \le n} 2 } } }

is equal to

View answer and explanation

Correct option: B

2^{n+1} C_4

Option B (

2^{n+1} C_4

) is correct because the quadruple summation can be expressed as a combination multiplied by 2.

492 Easy

492 of 951

Application +4 | -1

If the coefficients of

pth

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

and

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

terms in the expansion of

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

are in AP, then

View answer and explanation

Correct option: B

{n^2} - n\left( {4p + 1} \right) + 4{p^2} - 2 = 0

Option B (

{n^2} - n\left( {4p + 1} \right) + 4{p^2} - 2 = 0

) is correct because it is the relationship between n and p when the coefficients of the pth, (p+1)th, and (p+2)th terms are in arithmetic progression.

493 Medium

493 of 951

Application +4 | -1

\sum\limits_{{k=1}}^{n-r} { }^{{n}-{k}} {C}_{{r}}={ }^{{x}} {C}_{{y}}

then

View answer and explanation

Correct option: B

x=n, y=r+1

Option B (

x=n, y=r+1

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

\sum\limits_{{k=1}}^{n-r} { }^{{n}-{k}} {C}_{{r}}={ }^{{n}} {C}_{{r}+1}

494 Easy

494 of 951

Application +4 | -1

If the middle term in the expansion of

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

equals to

7 \frac{7}{8}

then

x

is equal to

(n \in I)

View answer and explanation

Correct option: C

n \pi+(-1)^n \frac{\pi}{6}

Option C (

n \pi+(-1)^n \frac{\pi}{6}

) is correct because it is the general solution for x when the middle term of the binomial expansion equals

7 \frac{7}{8}

495 Medium

495 of 951

Application +4 | -1

If in the expansion of

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

r\mathrm{th}

term is independent of

x

,then value of

r

View answer and explanation

Correct option: A

Option A (6) is correct because it is the value of r that makes the term in the binomial expansion independent of x.

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496 Easy

496 of 951

Analyzing +4 | -1

n

is the degree of the polynomial,

\left[\frac{2}{\sqrt{5 x^3+1}-\sqrt{5 x^3-1}}\right]^8+\left[\frac{2}{\sqrt{5 x^3+1}+\sqrt{5 x^3-1}}\right]^8

and

\mathrm{m}

is the coefficient of

x^n

in it, then the ordered pair

(n, m)

is equal to

View answer and explanation

Correct option: B

\left(12,(20)^4\right)

Option B (

(12, (20)^4)

) is correct because after rationalizing and simplifying, the polynomial's degree is 12 and the coefficient of

x^{12}

(20)^4

497 Medium

497 of 951

Application +4 | -1

The sum of binomial coefficient in the expansion of

\left(2-\frac{3}{2} x+\frac{1}{2 x^2}\right)^{2001}

View answer and explanation

Correct option: C

Option C (1) is correct because substituting

x=1

into the expression gives

(2 - \frac{3}{2} + \frac{1}{2})^{2001} = 1^{2001} = 1

, which is the sum of the binomial coefficients.

498 Medium

498 of 951

Application +4 | -1

If three successive coefficient in the expansion of

(1+x)^n

are in A.P., then

(n+2)

View answer and explanation

Correct option: C

a perfect square

Option C (a perfect square) is correct because the condition that three successive coefficients are in A.P. leads to an expression that can be written as a perfect square.

499 Easy

499 of 951

Application +4 | -1

If the coefficients of second, third and fourth terms in the expansion of

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

are in A.P., then

View answer and explanation

Correct option: B

2\;{n^2} - 9\;n + 7 = 0

Option B (

2\;{n^2} - 9\;n + 7 = 0

) is correct because it is derived from the given condition that the coefficients are in A.P.

500 Easy

500 of 951

Application +4 | -1

{\left( {1 + x - 2\;{x^2}} \right)^6} = 1 + {C_1}x + {C_2}{x^2} + {C_3}{x^3} + \ldots + {C_{12}}{x^{12}},

then the value of

{C_2} + {C_4} + {C_6} + \ldots + {C_{12}},

View answer and explanation

Correct option: C

Option C (31) is correct because it is the sum of the even-indexed coefficients

{C_2} + {C_4} + {C_6} + \ldots + {C_{12}}

obtained by substituting

x = 1

and

x = -1

into the given equation and solving.

501 Easy

501 of 951

Analyzing +4 | -1

The number of rational terms in the expansion of

{\left( {\sqrt[3]{4} + \frac{1}{{\sqrt[4]{6}}}} \right)^{20}},

View answer and explanation

Correct option: A

Option A (3) is correct because there are only 3 terms where both the powers of 2 and 3 are rational numbers, which occurs when r = 8, 16, and 20.

502 Medium

502 of 951

Analyzing +4 | -1

a_{r}

is the coefficient of

x^{10-r}

in the Binomial expansion of

(1+x)^{10}

, then

\sum\limits_{{r=1}}^{10} r^{3}\left(\frac{a_{r}}{a_{r-1}}\right)^{2}

is equal to

View answer and explanation

Correct option: A

1210

Option A (1210) is correct because the summation simplifies to a telescoping series that can be evaluated using standard summation formulas.

503 Easy

503 of 951

Application +4 | -1

(1+x)^n=C_0+C_1 x+\ldots+C_n x^n

, then the value of

\sum\limits_{{r=0}}^n \sum\limits_{{s=0}}^n C_r C_s

is equal to

View answer and explanation

Correct option: B

2^{2 n}

Option B (

2^{2 n}

) is correct because the sum can be simplified to

(2^n)^2

using the binomial theorem.

504 Easy

504 of 951

Analyzing +4 | -1

For

x \in {R}, x \neq-1

, if

(1+x)^{2016}+x(1+x)^{2015}+x^2(1+x)^{2014}+\ldots+x^{2016}=\sum\limits_{{i=0}}^{2016} a_i x^i,

then

a_{17}

is equal to:

View answer and explanation

Correct option: A

\frac{2017 !}{17 ! 2000 !}

Option A (

\frac{2017 !}{17 ! 2000 !}

) is correct because it represents the coefficient of

x^{17}

in the simplified expression after summing the geometric series.

505 Medium

505 of 951

Application +4 | -1

\frac{\mathrm{C}_0}{1}+\frac{\mathrm{C}_2}{3}+\frac{\mathrm{C}_4}{5}+\frac{\mathrm{C}_6}{7}+\ldots \ldots . .=

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because the given series can be expressed as a binomial coefficient sum, which simplifies to

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

506 Easy

506 of 951

Application +4 | -1

The sum of the rational terms in the expansion of

{(\sqrt 2 + {3^{1/5}})^{10}}

View answer and explanation

Correct option: A

The correct answer is A (41) because it represents the sum of the rational terms obtained from the binomial expansion of the given expression.

507 Easy

507 of 951

Analyzing +4 | -1

The coefficient of

x^{60}

(1+x)^{51}\left(1-x+x^2\right)^{50}

View answer and explanation

Correct option: A

{ }^{50} C_{20}

Option A (

{ }^{50} C_{20}

) is correct because

(1+x)^{51}(1-x+x^2)^{50}

simplifies to

(1+x)(1+x^3)^{50}

, and the coefficient of

x^{60}

{ }^{50} C_{20}

508 Medium

508 of 951

Application +4 | -1

If the coefficient of

r

th,

(r+1)

th and

(r+2)

th terms in the expansion of

(1+x)^n

are in A.P., then

n

is a root of the equation

View answer and explanation

Correct option: D

x^2-x(4 r+1)-4 r^2-2=0

Option D (

x^2-x(4r+1)-4r^2-2=0

) is correct because the condition that the coefficients are in A.P. leads to a quadratic equation in n.

509 Easy

509 of 951

Application +4 | -1

In the expansion of

{(1 + x + {x^2} + {x^3})^6},

then coefficient of

{x^{14}}

View answer and explanation

Correct option: B

120

Option B (120) is correct because the coefficient of

{x^{14}}

in the expansion of

{(1 + x + {x^2} + {x^3})^6}

is found by expanding and summing the appropriate terms, resulting in 120.

510 Medium

510 of 951

Application +4 | -1

\frac{T_2}{T_3}

in the expansion of

(a+b)^n

and

\frac{T_3}{T_4}

in the expansion of

(a+b)^{n+3}

are equal, then

n

is equal to

View answer and explanation

Correct option: C

Option C (5) is correct because by equating the ratios of consecutive terms in the binomial expansions, we can solve for n.

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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