Question

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

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

Answer 1

Answer: (C)

$15$

Term of $x^5{=}^n{C}_r{x}^{3r}{\left(\frac{-1}{x^2}\right)}^{n-r}{=}^n{C}_r{x}^{3r}.x^{2r-2n}.{(-1)}^{n-r}$
So $3r+2r-2n=5$
$r=\frac{5+2n}{5}⟶\left(1\right)$
Also for $x^{10}$ be ${(r^1+1)}^{th}$ term
So term with $x^{10}{=}^n{C}_{r^1}{x}^{{3r}^1}{\left(\frac{-1}{x^2}\right)}^{n-r}$
$\Rightarrow 3r^1+2r^1-2n=10$
$r^1=\frac{2n+10}{5}⟶\left(2\right)$
Now we know if $\left|{}^n{C}_r\right|=\left|{}^n{C}_{r^1}\right|\Rightarrow n=r+r^1$
Now we add co oefficient of $x^5\&x^{10}$
${}^n{C}_r{(-1)}^{n-r}{+}^n{C}_{r^1}{(-1)}^{{n-r}^1}=0$
Co oefficient of $x^{5}is{}^n{C}_r{(-1)}^{n-r}$; co oeffiecient of $x^{10}is{}^n{C}_{r^1}{(-1)}^{{n-r}^1}$
${\Rightarrow }^n{C}_r{(-1)}^{n-r}=-\left({}^n{C}_{r^1}{(-1)}^{{n-r}^1}\right)$
$\Rightarrow \left|{}^n{C}_r\right|=\left|{}^n{C}_{r^1}\right|$
So, $n=r+r^1$
$n=\frac{2n+10}{5}+\frac{2n+5}{5}$

$5n=4n+15$
$n=15$