If ${\left( {1 + x} \right)^n} = {C_0} + {C_1}x + {C_2}{x^2} + ... + {C_n}{x^n}$, then $\begin{array}{*{20}{c}} {\sum \;}\\ {0 \le r < s \le n} \end{array}\;\;\sum \left( {r + s} \right){C_r}{C_s}\;$ is equal to

$2n\left[ {{2^{2n - 1}}{ - ^{2n - 1}}{C_{n - 1}}} \right]$

The coefficient of ${\lambda ^n}{{\rm{\mu }}^n}$ in the expansion of ${\left[ {\left( {1 + \lambda } \right)\left( {1 + {\rm{\mu }}} \right)\left( {\lambda + {\rm{\mu }}} \right)} \right]^n}$ is

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

$\mathop \sum \limits_{r = 0}^n C_r^2$

$\mathop \sum \limits_{r = 0}^n C_{r + 2}^2$

$\mathop \sum \limits_{r = 0}^n C_{r + 3}^2$

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JEE • MATHEMATICS • BINOMIAL THEOREM

Binomial theorem practice · Easy

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361 Easy

361 of 434

Application +4 | -1

The term independent of

x

in the expansion of

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

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the term independent of x does not exist as the power of x cannot be zero.

362 Easy

362 of 434

Analyzing +4 | -1

The total number of terms which depend on the value of

x

in the expansion of

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

View answer and explanation

Correct option: B

2 n

Option B (

2 n

) is correct because the terms independent of x are removed, leaving 2n terms that depend on x.

363 Easy

363 of 434

Application +4 | -1

Constant term in the expansion of

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

View answer and explanation

Correct option: C

-252

Option C (-252) is correct because it is the constant term obtained by finding the term independent of x in the binomial expansion.

364 Easy

364 of 434

Application +4 | -1

\left(1+x+x^2\right)^8=a_0+a_1 x+a_2 x^2+\cdots+

a_{16} x^{16}

, then

a_5

equals

View answer and explanation

Correct option: A

504

Option A (504) is correct because the coefficient of x⁵ in the expansion of (1+x+x²)⁸ is 504.

365 Easy

365 of 434

Application +4 | -1

The coefficient of

a^3 b^4

in the expansion of

(a+2 b)^3(2 a-b)^4

View answer and explanation

Correct option: B

-15

Option B (-15) is correct because it is the coefficient of

a^3b^4

obtained after expanding and multiplying the two binomial expressions.

366 Easy

366 of 434

Application +4 | -1

If,

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

, then

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

equals

View answer and explanation

Correct option: C

\frac{1}{2}n{a_n}

Option C (

\frac{1}{2}n{a_n}

) is correct because by using the property of binomial coefficients and summing the series in reverse, we can derive the given expression.

367 Easy

367 of 434

Application +4 | -1

{49^n} + 16n - 1

is divisible by

View answer and explanation

Correct option: C

Option C (64) is correct because the expression can be rewritten using the binomial theorem, and the resulting terms are divisible by 64.

368 Easy

368 of 434

Application +4 | -1

The coefficient of

{x^7}

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

is equal to

View answer and explanation

Correct option: A

62640

Option A (62640) is correct because it is the coefficient of

{x^7}

obtained by considering all possible combinations of powers of the terms in the expansion of

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

that result in

{x^7}

369 Easy

369 of 434

Analyzing +4 | -1

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

, then

\begin{array}{cccccccccccccccccccc} {\sum \;}\\ {0 \le r < s \le n} \end{array}\;\;\sum \left( {r + s} \right){C_r}{C_s}\;

is equal to

View answer and explanation

Correct option: A

n\left[ {{2^{2n - 1}}{ - ^{2n - 1}}{C_{n - 1}}} \right]

Option A (

n[2^{2n-1} - ^{2n-1}C_{n-1}]

) is correct because it is derived by manipulating the sum of products of binomial coefficients.

370 Easy

370 of 434

Analyzing +4 | -1

{C_r}

stands for

{\;^n}{C_r},

then the sum of first

\left( {n + 1} \right)

terms of the series

a\;{C_0} - \left( {a + d} \right){C_1} + \left( {a + 2d} \right){C_2} - \left( {a + 3d} \right){C_3} + \ldots ,

View answer and explanation

Correct option: C

The correct answer is C (0) because the given series can be expressed as a sum involving binomial coefficients with alternating signs, which evaluates to zero.

371 Easy

371 of 434

Analyzing +4 | -1

The coefficient of

{\lambda ^n}{\mathrm{\mu }^n}

in the expansion of

{\left[ {\left( {1 + \lambda } \right)\left( {1 + \mathrm{\mu }} \right)\left( {\lambda + \mathrm{\mu }} \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 it represents the coefficient of

\lambda ^n \mu^n

in the expansion of the given expression, which simplifies to the sum of cubes of binomial coefficients.

372 Easy

372 of 434

Application +4 | -1

If the

r

th term in the expansion of

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

contains

{x^4},

then

r

is equal to

View answer and explanation

Correct option: B

Option B (3) is correct because the value of r that makes the term contain x⁴ is 3, as found by solving the equation 13 - 3r = 4.

373 Easy

373 of 434

Application +4 | -1

x^2-x+1=0

, then

\sum\limits_{{n=1}}^5\left(x^n+\frac{1}{x^n}\right)=

View answer and explanation

Correct option: A

Option A (8) is correct because it represents the sum of the given expression for the roots of the quadratic equation.

374 Easy

374 of 434

Application +4 | -1

A positive value of

{m}

for which the coefficient of

x^2

in the expansion

(1+x)^m

is 6, is

View answer and explanation

Correct option: A

Option A (4) is correct because the coefficient of

x^2

in the expansion of

(1+x)^m

being 6 leads to the equation

\frac{m(m-1)}{2} = 6

, which gives

m=4

375 Easy

375 of 434

Application +4 | -1

The coefficient of

\frac{1}{x}

in the expansion of

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

View answer and explanation

Correct option: B

{\;^{2n}}{C_{n - 1}}\;

Option B (

\;{\;^{2n}}{C_{n - 1}}\;

) is correct because simplifying the expression and finding the coefficient of 1/x leads to this result.

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376 Easy

376 of 434

Application +4 | -1

If the ratio of the coefficient of

x^{10}

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

and term independent of

x

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

\frac{p}{q}

(where

{p}

and

{q}

are relative prime numbers), then

\frac{{q}}{8}-{p}

View answer and explanation

Correct option: A

Option A (3) is correct because the ratio of the coefficients simplifies to 1/32, leading to p=1, q=32, and (q/8)-p = 3.

377 Easy

377 of 434

Application +4 | -1

\mathrm{\;}{x^{2r}}

occurs in

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

, then

\mathrm{\;}n - 2r

must be of the form

View answer and explanation

Correct option: B

3k

Option B (3k) is correct because the general term of the binomial expansion leads to the conclusion that n-2r must be a multiple of 3.

378 Easy

378 of 434

Application +4 | -1

The sum of the coefficient in the expansion of

{(1 + x - 3{x^2})^{3148}}\;

View answer and explanation

Correct option: C

Option C (1) is correct because the sum of the coefficients is found by substituting x = 1 into the given expression.

379 Easy

379 of 434

Application +4 | -1

The reminder of

3^{2003}

when divided by 28

View answer and explanation

Correct option: A

Option A (19) is correct because after simplifying the expression using modular arithmetic, the remainder is found to be 19.

380 Easy

380 of 434

Application +4 | -1

If the coefficient of

(3 r+4)

th term and

(2 r-2)

th term in the expansion of

(1+x)^{20}

are equal then

r

is equal to:

View answer and explanation

Correct option: A

Option A (4) is correct because if the coefficients of the (3r+4)th and (2r-2)th terms in the expansion of (1+x)²⁰ are equal, then r = 4.

381 Easy

381 of 434

Analyzing +4 | -1

For

r=0,1,2, \ldots, 10

, let

A_r, B_r

and

C_r

denote respectively, the coefficients of

x^r

in the expansion of

(1+x)^{10},(1+x)^{20}

and

(1+x)^{30}, \sum\limits_{r=1}^{10} A_r\left(B_{10} B_r-C_{10} A_r\right)

is equal to

View answer and explanation

Correct option: D

C_{10}-B_{10}

Option D (

C_{10}-B_{10}

) is correct because by substituting the binomial coefficients and simplifying the summation, the expression reduces to

C_{10}-B_{10}

382 Easy

382 of 434

Application +4 | -1

In the expansion of

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

, the sum of the coefficients of odd powers 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}} = {2^{49}}

383 Easy

383 of 434

Application +4 | -1

Coefficient of

{x^{ - 4}}

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

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the calculated coefficient of x^-4 does not match any of the given options. The correct coefficient is

\frac{{3^{12} \times 5}}{{{{2^8}}}}

384 Easy

384 of 434

Application +4 | -1

In what direction a line be drawn through the point

(1,2)

so that it point of intersection with the line

x+y=4

is at a distance

\frac{\sqrt{6}}{3}

from the given point?

View answer and explanation

Correct option: D

75^{\circ}

Option D (75°) is correct, but the provided explanation is insufficient and lacks proper steps. A more detailed solution is needed to confirm this answer.

385 Easy

385 of 434

Application +4 | -1

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

. Then,

{C_0}{C_1} + {C_1}{C_2} + ... + {C_{n - 1}}{C_n}\;

is equal to

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because multiplying the two binomial expansions and comparing coefficients of x^(n-1) leads to this expression.

386 Easy

386 of 434

Application +4 | -1

The coefficient of the middle term in the binomial expansion in powers of

x

{(1 + \mathrm{\alpha }x)^4}

and of

{(1 - \mathrm{\alpha }x)^6}

is the same, if

\alpha

equals

View answer and explanation

Correct option: C

- \frac{3}{{10}}

Option C (

- \frac{3}{{10}}

) is correct because it is the value of

\alpha

that makes the coefficients of the middle terms in the two binomial expansions equal.

387 Easy

387 of 434

Application +4 | -1

The coefficient of

{x^{ - 17}}

in the expansion of

{\left( {{x^4} - \frac{1}{{{x^3}}}} \right)^{15}}\;\mathrm{is}

View answer and explanation

Correct option: C

- {\;^{15}}{C_{11}}\;

Option C (-^15C_11) is correct because the coefficient of x^(-17) in the expansion of (x⁴ - 1/x³)¹⁵ is -^15C_11.

388 Easy

388 of 434

Analyzing +4 | -1

The coefficient of

{x^{15}}

in the product

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

is equal to

View answer and explanation

Correct option: A

{2^{105}} - {2^{121}}

Option A (

{2^{105}} - {2^{121}}

) is correct because after expanding the product and identifying the coefficient of

{x^{15}}

, it simplifies to

{2^{105}} - {2^{121}}

389 Easy

389 of 434

Application +4 | -1

The value of the sum of the series

{ }^{14} {C}_0 \cdot{ }^{15} {C}_1+{ }^{14} {C}_1 \cdot{ }^{15} {C}_2+{ }^{14} {C}_2 \cdot{ }^{15} {C}_3+\ldots+{ }^{14} {C}_{14} \cdot{ }^{15} {C}_{15}

View answer and explanation

Correct option: C

{ }^{29} {C}_{14}

Option C (

{ }^{29} {C}_{14}

) is correct because the given sum represents the coefficient of x¹⁴ in the product of two binomial expansions, (1+x)¹⁴ and (x+1)¹⁵, which simplifies to (1+x)²⁹, and the coefficient of x¹⁴ is

{ }^{29} {C}_{14}

390 Easy

390 of 434

Application +4 | -1

In the expansion of

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

, the 5 th term from the end is

View answer and explanation

Correct option: B

\frac{17010}{x^8}

Option B (

\frac{{17010}}{{x^8}}

) is correct because it is the 5th term from the end in the expansion.

Revision notes and key takeaways

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