The coefficient of x^2 in the expansion of (2x^2+(1)/(x))^10,x 0, is :

Question

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

Accepted Answer

Using the binomial theorem,

${(2{x}^{2}+\frac{1}{x})}^{10}=\sum _{10}^{r=0}(\frac{10}{r})(2{x}^{2}{)}^{10-r}{(\frac{1}{x})}^{r}$

Now simplify the general term:

${T}_{r+1}=(\frac{10}{r}){2}^{10-r}{x}^{2(10-r)}{x}^{-r}$

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

We need the coefficient of ${x}^{2}$, so set the power of $x$ equal to $2$:

$20-3r=2$

$3r=18$

$r=6$

Now substitute $r=6$:

$=(\frac{10}{6}){2}^{10-6}$

$=(\frac{10}{6}){2}^{4}$

$=210	imes 16$

$=3360$

So, the coefficient of ${x}^{2}$ is

$3360$

Hence, the correct option is B.

Solutions · 1