JEE Main 2021 · Mathematics · Binomial Theorem · hard

The coefficient of x^3 in the expansion of (2x^2-1/x)^9 is

Single correct option

The coefficient of

x^{3}

in the expansion of

\left(2x^{2}-\dfrac1x\right)^{9}

Show answer and solution

Answer: (C) $-2016$

The general term is

T_{k+1}=\binom9k(2x^2)^{9-k}\left(-\dfrac1x\right)^k=\binom9k2^{9-k}(-1)^kx^{18-3k}

. Setting

18-3k=3

gives

k=5

. The coefficient is

\binom95\cdot2^{4}\cdot(-1)^5=126\cdot16\cdot(-1)=-2016

. Watch the sign: the odd power of

-1

makes it negative.

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