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JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Binomial Theorem

The coefficient of

x^{2}

in the expansion of

(2 x^{2} + \frac{1}{x})^{10}, x \neq = 0

, is :

(A)3240
(B)3360✓ Correct
(C)3480
(D)3600

Answer: (B)

Solution

The general term of

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

is

T_{r + 1} = (r 10) (2 x^{2})^{10 - r} (\frac{1}{x})^{r} = (r 10) 2^{10 - r} x^{20 - 3 r}

. The exponent of

x

equals

2

when

20 - 3 r = 2

, that is

r = 6

. The corresponding coefficient is

(6 10) 2^{4} = 210 \times 16 = 3360

. The coefficient of

x^{2}

is

3360

.

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Single correct option

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