What is the co-efficient of $x^2$ in the expansion of $(x+\frac{1}{x^2})^{20}$?

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What is the co-efficient of $x^2$ in the expansion of $\left(x+\frac{1}{x^2}\right)^{20}$?

The correct answer should be $605\frac{5}{8}$.

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It's $$\frac{(x^3+1)^{20}}{x^{40}}$$ and since $x^{42}=(x^3)^{14}$, we get the answer: $\binom{20}{14}$.

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By the binomial formula, it is the coefficient $\dbinom{20}i$, $i$ being such that $$x^i\frac1{x^{2k}}=x^2\iff\begin{cases}i-2k=2\\i+k=20\end{cases}\iff i=14,~. k=6,$$ so it is $\;\dfrac{20!}{14!\,6!}=38\,760$.