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Q.
In the expansion of $\left(x^3-\frac{1}{x^2}\right)^n, n \in N$, if the sum of the coefficients of $x^5$ and $x^{10}$ is 0 , then $n$ is :
Binomial Theorem
Solution:
$\left(x^3-\frac{1}{x^2}\right)^n$
General term $=\frac{n !}{r !(n-r) !}(-1)^{n-r} x^{5 r-2 n}$
If $5 r-2 n=5$, then $5 r=2 n+5 \Rightarrow r=\frac{2 n}{5}+1$
If $5 r-2 n=10$, then $5 r=2 n+10 \Rightarrow r=\frac{2 n}{5}+2$
Let $n=5 k$
Now $\frac{5 k !}{(2 k+1) !(3 k-1) !}-\frac{5 k !}{(2 k+2) !(3 k-2) !}=0$
$\Rightarrow \frac{1}{3 k -1}-\frac{1}{2 k +2}=0$
$\Rightarrow k =3 \Rightarrow n =15$