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  4. In the expansion of ((1/x2)-x3)n n ∈ N, if the sum of the coefficients of x5 and x10 is 0, then n is:

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

Binomial Theorem

Solution:

$\left(\frac{1}{x^{2}}-x^{3}\right)^{n}$
$T_{r+1}=\frac{n !}{r !(n-r) !}(-1)^{n-r} x^{5 r-2 n}$
If $5 r-2 n=5,$ then $5 r=2 n+5$
$r=\frac{2 n}{5}+1$
If $ 5 r-2 n=10,$ then $ 5 r=2 n+10$
$r=\frac{2 n}{5}+2 $
Let $n=5 k$
According to question
$\frac{5 k !}{(2 k+1) !(3 k-1) !}-\frac{5 k !}{(2 k+2) !(3 k-2) !}=0$
$\Rightarrow \frac{1}{3 k-1}-\frac{1}{2 k+2}=0$
$\Rightarrow k=3 $
$\Rightarrow n=15$