The coefficient of x^2 in the expansion of (2x^2+1x)^10, x0, is:

Question

The coefficient of $x^2$ in the expansion of $\left(2x^2+\dfrac{1}{x}ight)^{10}$, $x
e0$, is:

Accepted Answer

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

For coefficient of $x^2$: $20-3r=2\Rightarrow r=6$.

$$T_7=\binom{10}{6}2^{4}=210\times16=3360$$

Solutions · 1