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Q.
If in the expansion of $\left(x^{3}-\frac{2}{\sqrt{x}}\right)^{n}$ a term like $x^{2}$ exists and $n$ is a double digit number, then least value of $n$ is
Binomial Theorem
Solution:
Let the term is $T_{r+1}={ }^{n} C_{r}\left(x^{3}\right)^{n-r}\left(\frac{-2}{\sqrt{x}}\right)^{r}$
Term like $x^{2}$ exists
$\therefore 3 n-3 r-r / 2=2$
$\Rightarrow 3 n=2+\frac{7 r}{2}$
$\Rightarrow n=\frac{4+7 r}{6}$
Hence, least value of $n$ which is double digit is $10$.