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Q. If $x^{p}$ occurs in the expansion of $\left(x^{2}+\frac{1}{x}\right)^{2 n}$, its coeffi- cient is

Binomial Theorem

Solution:

Let $t_{r+1}$ contains $x^{p} .$
Then, $t_{r+1}={ }^{2 n} C_{r}\left(x^{2}\right)^{2 n-r}\left(\frac{1}{x}\right)^{r}={ }^{2 n} C_{r} x^{4 n-3 r}$
$\therefore 4 n-3 r=p ; $ or $r=\frac{4 n-p}{3}$
Hence, the coefficient of $x^{p}={ }^{2 n} C_{\frac{4 n-p}{3}}$