Binomial TheoremMedium

Question

The coefficient of $x^{2}$ in the expansion of $\left( 2x^{2} + \frac{1}{x} \right)^{10},x \neq 0$, is :

Options

A.3240
B.3360
C.3480
D.3600

Solution

Sol. [Key Concept: General term has $x$-power $20 - 3r$; set it to $2$ (so $r = 6$); the coefficient is $10C_{6}{\ 2}^{4}$.]

$${T_{r + 1} = \ ^{10}C_{r}\left( 2x^{2} \right)^{10 - r}(1/x)^{r} }{= \ ^{10}C_{r}2^{10 - r}x^{20 - 2r - r} }{= \ ^{10}C_{r}2^{10 - r}x^{20 - 3r} }{\Rightarrow 20 - 3r = 2 }{r = 6 }$$So required coefficient is $\ ^{10}C_{6}2^{4}$

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