If in the expansion of ${(x^{3} - \frac{1}{x^{2}})}^{n}, n \in N,$ sum of the coefficient of $x^{5}$ and $x^{10}$ is $0$ , then value of $n$ is

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$5$

$10$

$15$

$20$

answer is C.

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Detailed Solution

$T_{r + 1,}$ and $(r + 1)$ th term in the expansion of ${(x^{3} - \frac{1}{x^{2}})}^{n}$ is
$T_{r + 1} =^{n} C_{r} {(x^{3})}^{n - r} {(- \frac{1}{x^{2}})}^{r} =^{n} C_{r} (- 1)^{r} x^{3 n - 5 r}$

For the coefficient of $x^{5},$ we set $3 n - 5 r = 5 \Rightarrow r = \frac{3 n - 5}{5} = p$ (say) and
for the coefficient of $x^{10}$ , we set $3 n - 5 r = 10 \Rightarrow r = \frac{3 n - 10}{5} = q$ (say)
Note that $p - q = 1$ . We are given

$\begin{array}{l} ^{n} C_{p} (- 1)^{p} +^{n} C_{q} (- 1)^{q} = 0 \\ \Rightarrow^{n} C_{p} (- 1)^{p} +^{n} C_{p - 1} (- 1)^{p - 1} = 0 \\ \Rightarrow^{n} C_{p} =^{n} C_{p - 1} \Rightarrow n - p = p - 1 \\ \Rightarrow n = 2 p - 1 \Rightarrow p = \frac{n + 1}{2} \end{array}$
Thus, $\frac{n + 1}{2} = \frac{3 n - 5}{5} \Rightarrow 5 n + 5 = 6 n - 10$
$\Rightarrow 15 = n$