JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Binomial Theorem

The sum of all possible values of

n \in N

, so that the coefficients of

x, x^{2}

and

x^{3}

in the expansion of

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

, are in arithmetic progression is:

Answer: (D)

Solution

Expanding

(1 + x^{2})^{2} (1 + x)^{n} = (1 + 2 x^{2} + x^{4}) (\sum_{k = 0}^{n} (k n) x^{k})

.

Coefficient of

x

(1 n) = n

Coefficient of

x^{2}

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

Coefficient of

x^{3}

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

(for

n \geq 3

)

For AP:

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

Simplifying:

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

, which factors to give

n = 2, 3, 4

.

Verifying

n = 2

: coefficients are

2, 3, 4

(in AP ✓).

Sum:

2 + 3 + 4 = 9

JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Single correct option