Question

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

• A. $5$
• B. $10$
• C. $15$
• D. none of these

Answer 1

The correct option is C $15$

$(r+1)$th term in the expansion of
${(x^3-\frac{1}{x^2})}^n$ is
${{}^{n}C}_r{\left(x^3\right)}^{n-r}{(-\frac{1}{x^2})}^r={{}^{n}C}_r{(-1)}^r{x}^{3n-5r}$
For coefficient of $x^5$, we set $3n-5r=5$ $\Rightarrow r=\frac{3n-5}{5}=p$ (say) and for coefficient of $x^{10}$, we set
$3n-5r=10$ $\Rightarrow r=\frac{3n-10}{5}=q$ (say)
Note that $p-q=1$. We are given
${{}^{n}C}_p{(-1)}^p+{{}^{n}C}_q{(-1)}^q=0$
$\Rightarrow {{}^{n}C}_p{(-1)}^p+{{}^{n}C}_{p-1}{(-1)}^{p-1}=0$
$\Rightarrow {{}^{n}C}_p={{}^{n}C}_{p-1}\Rightarrow n-p=p-1$
$\Rightarrow n=2p-1\Rightarrow p=\frac{n+1}{2}$
Thus, $\frac{n+1}{2}=\frac{3n-5}{5}\Rightarrow 5n+5=6n-10$
$\Rightarrow 15=n$