The coefficient of $x^{25}$ in $(1+x)^5+(1+x)^6+(1+x)^7+\ldots+(1+x)^{50}$ is

$\left(\begin{array}{l}50 \\ 25\end{array}\right)$

$\left(\begin{array}{l}51 \\ 25\end{array}\right)-1$

$\left(\begin{array}{l}46 \\ 25\end{array}\right)$

The value of ${1^2}.{C_1} + {3^2}.{C_3} + {5^2}.{C_5} + ...$ is

$n{\left( {n - 1} \right)^{n - 2}} + n{.2^{n - 1}}$

$n{\left( {n - 1} \right)^{n - 2}}$

$n\left( {n - 1} \right){.2^{n - 3}}$

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JEE • MATHEMATICS • BINOMIAL THEOREM

Binomial theorem practice · Medium

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271 Medium

271 of 468

Application +4 | -1

In how many ways can 15 objects be placed in 3 boxes? (Each box can hold atmost 15 objects).

View answer and explanation

Correct option: D

136

Option D (136) is correct because it represents the number of ways to distribute 15 identical objects into 3 distinct boxes, which can be calculated using stars and bars method.

272 Medium

272 of 468

Analyzing +4 | -1

Let

n \geq 5

be an integer. If

9^n-8 n-1=64 \alpha

and

6^n-5 n-1=25 \beta

, then

\alpha-\beta

is equal to

View answer and explanation

Correct option: C

^n{C_3}(8 - 5){ + ^n}{C_4}\left( {{8^2} - {5^2}} \right) + \ldots { + ^n}{C_n}\left( {{8^{n - 2}} - {5^{n - 2}}} \right)

Option C is correct because it represents the difference between the series expansions of (1+8)^n and (1+5)^n, after accounting for the initial terms and dividing by 64 and 25 respectively.

273 Medium

273 of 468

Application +4 | -1

\left(2+\frac{x}{3}\right)^{55}

is expanded in the ascending powers of

x

and the coefficients of powers of

x

in two consecutive terms of the expansion are equal, then these terms are:

View answer and explanation

Correct option: A

7^{\text {th }}

and

8^{\text {th }}

Option A (7th and 8th) is correct because the coefficients of x in the 7th and 8th terms are equal after solving the binomial expansion.

274 Medium

274 of 468

Application +4 | -1

The coefficient of

{x^5}

in the expansion of

{\left( {2 - x + 3{x^2}} \right)^6},

View answer and explanation

Correct option: D

- 5052

Option D (

- 5052

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

{x^5}

by considering all possible combinations of terms in the multinomial expansion.

275 Medium

275 of 468

Analyzing +4 | -1

If in the expansion of

(1+x)^n, a, b

and

c

are three consecutive coefficients, then

n

is equal to

View answer and explanation

Correct option: B

\frac{2 a c+a b+b c}{b^2-a c}

Option D (None of these) is correct because the given expression for

n

does not hold true when tested with simple values.

276 Medium

276 of 468

Analyzing +4 | -1

The term independent of

x

in the binomial expansion of

\left(1-\frac{1}{x}+3 x^5\right)\left(2 x^2-\frac{1}{x}\right)^8

is:

View answer and explanation

Correct option: A

400

Option A (400) is correct because it is the sum of the constant terms obtained by multiplying the given expression and considering the powers of x that cancel out.

277 Medium

277 of 468

Application +4 | -1

The last digit of

(2137)^{754}

View answer and explanation

Correct option: D

Option D (9) is correct because the last digit is found by looking at the pattern of the last digit of powers of 7, and 754 leaves a remainder of 2 when divided by 4.

278 Medium

278 of 468

Analyzing +4 | -1

The coefficient of the

(r+1)

th term of

\left(x+\frac{1}{x}\right)^{20}

when expanded in the descending powers of

x

, is equal to the coefficient of the 6 th term of

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

when expanded in ascending powers of

x

. The value of

r

View answer and explanation

Correct option: D

None of the above

Option A (5) is the closest to the correct value of r, which is derived by equating the coefficients of the binomial expansions and solving for r.

279 Medium

279 of 468

Application +4 | -1

The coefficient of

x^4

in the expansion of

\left(1+x+x^2+x^3\right)^{11}

View answer and explanation

Correct option: A

990

Option A (990) is correct because expanding the expression and finding the coefficient of x⁴ results in 990.

280 Medium

280 of 468

Application +4 | -1

Evaluate

\frac{1}{\sqrt[3]{6-3 x}}

View answer and explanation

Correct option: B

6^{-1 / 3}\left[1+\frac{x}{6}+\frac{2 x^2}{6^2}+\ldots\right]

Option B (

6^{-1 / 3}\left[1+\frac{x}{6}+\frac{2 x^2}{6^2}+\ldots\right]

) is correct because it is the binomial expansion of the given expression.

281 Medium

281 of 468

Understanding +4 | -1

The sum

{ }^{20} C_0+{ }^{20} C_1+{ }^{20} C_2+\ldots .+{ }^{20} C_{10}

is equal to

View answer and explanation

Correct option: D

None of these

None of these is the correct answer because the sum is equal to

2^{19} + \frac{{ }^{20}C_{10}}{2}

282 Medium

282 of 468

Application +4 | -1

The coefficient of

x^{25}

(1+x)^5+(1+x)^6+(1+x)^7+\ldots+(1+x)^{50}

View answer and explanation

Correct option: B

\left(\begin{array}{l}51 \\ 26\end{array}\right)

Option B is correct because the coefficient of

x^{25}

is found by summing a geometric progression and then extracting the relevant coefficient.

283 Medium

283 of 468

Application +4 | -1

If the middle term in the expansion of

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

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 is used to find the value of n, which is determined to be 12.

284 Medium

284 of 468

Application +4 | -1

\alpha \in[-1,1]

and

I(\alpha)

denotes the term independent of

x

in the expansion of

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

then

|I(\alpha)|

cannot exceed

View answer and explanation

Correct option: C

{ }^{10} C_5 \frac{\pi^{10}}{2^{20}}

Option C (

{ }^{10} C_5 \frac{\pi^{10}}{2^{20}}

) is correct because the term independent of x is maximized when

\sin^{-1} \alpha = \cos^{-1} \alpha = \frac{\pi}{4}

285 Medium

285 of 468

Application +4 | -1

The coefticient of

x^n

in the expansion of

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

View answer and explanation

Correct option: A

\frac{2^n}{n !}

Option A (

\frac{2^n}{n !}

) is correct. The coefficient of

x^n

in the expansion of the given expression is equivalent to the coefficient of

x^n

e^{2x}

, which is

\frac{2^n}{n !}

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286 Medium

286 of 468

Application +4 | -1

n !\left(\frac{1}{n !}+\frac{1}{2 !(n-2) !}+\frac{1}{4 !(n-4) !}+\ldots \ldots+\frac{1}{n !}\right)

is equal to -

View answer and explanation

Correct option: B

2^{n-1}

Option B (

2^{n-1}

) is correct because the given expression simplifies to the sum of even-indexed binomial coefficients, which equals

2^{n-1}

287 Medium

287 of 468

Analyzing +4 | -1

The value of

{1^2}.{C_1} + {3^2}.{C_3} + {5^2}.{C_5} + ...

View answer and explanation

Correct option: D

None of these

The correct answer is D (None of these) because the given options do not match the derived expression for the sum of the series.

288 Medium

288 of 468

Application +4 | -1

Sum of the series

\sum\limits_{{r=0}}^{10}(-1)^r\left({ }^{10} C_r\right)\left[\frac{1}{2^r}+\frac{3^r}{2^{2 r}}\right]

View answer and explanation

Correct option: A

\frac{2^{10}+1}{2^{20}}

Option A (

\frac{2^{10}+1}{2^{20}}

) is correct. The sum is simplified using the binomial theorem.

289 Medium

289 of 468

Analyzing +4 | -1

The integer just greater than

(\sqrt{3}+1)^{2 m}

contains

View answer and explanation

Correct option: B

2^{m+1}

as a factor

Option B (

2^{{m+1}}

as a factor) is correct because by expressing

(\sqrt{{3}}+1)^{{2 m}}

I+F

and using its conjugate, we can show that

2^{{m+1}}

is a factor of the integer just greater than

(\sqrt{{3}}+1)^{{2 m}}

290 Medium

290 of 468

Application +4 | -1

If the numerical coefficient of the

p

th term in the expansion of

(2 x+3)^6

is 4860 , then the value(s) of

p

is/are

View answer and explanation

Correct option: D

Option D (5) is correct because substituting p=5 into the general term formula yields the given numerical coefficient of 4860.

291 Medium

291 of 468

Application +4 | -1

C_0^2+\frac{1}{2} C_1^2+\frac{1}{3} C_2^2+\cdots+\frac{1}{n+1} C_n^2

equals

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because the given sum can be transformed using combinatorial identities and then evaluated using the Vandermonde's identity.

292 Medium

292 of 468

Application +4 | -1

x+y=1

, then value of

\sum\limits_{{r=0}}^n(r)\left({ }^n C_r\right) x^{n-r} y^r

View answer and explanation

Correct option: D

n y

Option D (ny) is correct because differentiating the binomial expansion and substituting x+y=1 simplifies the expression to ny.

293 Medium

293 of 468

Analyzing +4 | -1

n

is an even integer and

a, b

and

c

are distinct numbers, then the number of distinct terms in the expansion of

(a+b+c)^n+(a+b-c)^n

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the number of distinct terms in the expansion is

\frac{(n+2)^2}{2}

294 Medium

294 of 468

Application +4 | -1

If the 7 th term in the binomial expansion of

\left(\frac{3}{\sqrt[3]{84}}+\sqrt{3} \ln x\right)^{9}, x>0

, is equal to 729 , then

x

can be :

View answer and explanation

Correct option: B

e

Option B (e) is correct because solving the equation derived from the 7th term of the binomial expansion leads to x = e.

295 Medium

295 of 468

Analyzing +4 | -1

The term independent of

x

in the expansion of

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

View answer and explanation

Correct option: A

0 , if

n

is odd

Option A (0, if

n

is odd) is correct because when

n

is odd, the term independent of

x

in the expansion is 0.

296 Medium

296 of 468

Application +4 | -1

x

is so small that

{x^3}

and higher powers of

x

may be neglected,then

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

may be approximated as

View answer and explanation

Correct option: B

- \frac{3}{8}{x^2}

Option B (

\frac{{ - 3}}{8}{x^2}

) is correct because by using binomial expansion and neglecting terms of

x^3

and higher, the expression simplifies to this value.

297 Medium

297 of 468

Analyzing +4 | -1

The coefficient of

x^n

in the expansion of

\frac{1}{(1-x)(1-2 x)(1-3 x)}

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because partial fraction decomposition and series expansion are used to find the coefficient of

x^n

298 Medium

298 of 468

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: C

\frac{1}{2} n

Option C (

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

) is correct because by using the property

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

and summing the series, we can find that

t_n = \frac{n}{2} s_n

, thus

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

299 Medium

299 of 468

Application +4 | -1

The

\operatorname{sum} \sum\limits_{{r=1}}^n r \cdot{ }^{2 n} C_r

is equal to

View answer and explanation

Correct option: A

n \cdot 2^{2 n-1}

Option A (

n \cdot 2^{2 n-1}

) is correct because the sum can be simplified using combinatorial identities to

n \cdot 2^{2 n-1}

300 Medium

300 of 468

Analyzing +4 | -1

x+\frac{1}{x}=1

and

p=x^{4000}+\frac{1}{x^{4000}}

and

q

is the digit at unit place in the number

2^{2^n}+1, n \in N

and

n>1

, then

p+q

, is

View answer and explanation

Correct option: B

Option B (6) is correct because solving the given equation leads to x being a complex cube root of unity, allowing calculation of p, and q is found using the properties of powers of 2.

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