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Binomial Theorem Part 3

Section: Mathematics

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Question : 12 of 53

Marks: +1, -0

The sum of all possible values of

n \in N

, so that the coefficients of

x, x^2

x^{3}

in the expansion of

(1+x^{2})^{2}(1+x)^{n}

, are in arithmetic progression is :

[23 Jan 2026 Shift 1]

12
9
3
7

Solution:

Concept:

Coefficients of

x

,

x^2

,

x^3

in the expansion of

(1+x^2)^2(1+x)^n

are derived using binomial theorem and set in arithmetic progression.

Explanation:

Expand

(1+x^2)^2 = 1 + 2x^2 + x^4

.

Coefficient of

x

: from

(1+x)^n

only,

= \binom{n}{1}

.

Coefficient of

x^2

:

\binom{n}{2} + 2\cdot\binom{n}{0} = \binom{n}{2} + 2

.

Coefficient of

x^3

:

\binom{n}{3} + 2\cdot\binom{n}{1} = \binom{n}{3} + 2n

.

For AP:

2(\binom{n}{2} + 2) = \binom{n}{1} + \binom{n}{3} + 2n

.

Substitute

\binom{n}{1}=n

,

\binom{n}{2}=n(n-1)/2

,

\binom{n}{3}=n(n-1)(n-2)/6

:

2\cdot\frac{n(n-1)}{2} + 4 = n + \frac{n(n-1)(n-2)}{6} + 2n

\Rightarrow n(n-1) + 4 = 3n + \frac{n(n-1)(n-2)}{6}

.

Multiply by

6

:

6n(n-1) + 24 = 18n + n(n-1)(n-2)

.

Rearrange:

n^3 - 9n^2 + 26n - 24 = 0

.

Factor:

(n-2)(n-3)(n-4)=0

, so

n=2,3,4

.

Sum

= 2+3+4 = 9

.

Answer:

9

, option B.

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