Question

In the expansion of ${(\frac{1}{x^2}-x^3)}^n,n\in N$, if the sum of the coefficients of $x^5$ and $x^{10}$ is $0$, then $n$ is:

• A. $25$
• B. $20$
• C. $15$
• D. None of these

Answer 1

The correct option is C $15$

Given, ${(\frac{1}{x^2}-x^3)}^n$

$T_{r+1}{=}^n{C}_r{\left(\frac{1}{x^n}\right)}^{n-r}(-x^3{)}^r$

${=}^n{C}_r{x}^{2r-2n+3r}(-1{)}^r$

${=}^n{C}_r(-1{)}^r{x}^{5r-2n}$

$5r-2n=5$

$\Rightarrow 5r-5=2n$

$\Rightarrow x=\frac{5}{2}(r-1)$

$5r-2n=10$

$\Rightarrow 5r-10=2n$

$n=\frac{5}{2}(r-2)$

$T_5+T_{10}=0$

${}^n{C}_{\frac{5+2n}{5}}(-1){+}^n{C}_{\frac{10+2n}{5}}=0$

$\Rightarrow n=15$