<s>[INST] Finding the coefficient of $x^r$ in an expansion. Suppose that the summation of the infinite series $$1+nx+\frac{n(n-1)}{2} x^2+\cdots+\frac{n(n-1)\cdots(n-r+1)}{r}x^r+\cdots$$ is equal to $(1+x)^n$ for $|x|<1$.
Show that the coefficient of $x^r$ in the expansion of $\frac{1+x+x^2}{(1-x)^2}$  is $3r $. 
Hence show that $(217)^\frac{1}{3} \simeq 6.0092$
My attempt :
$$\frac{1+x+x^2}{(1-x)^2}=\frac{(1+x+x^2)(1-x)}{(1-x)^3}$$
$$\frac{1+x+x^2}{(1-x)^2}=\frac{1-x^3}{(1-x)^3}$$
$$=\frac{1}{(1-x)^3}-\frac{x^3}{(1-x)^3}$$
$$=\frac{1}{\left(1+(-x)\right)^3}+\frac{1}{\left(1+\left(-\frac{1}{x}\right)\right)^3}$$
$$=(1+(-x))^{-3}+\left(1+\left(-\frac{1}{x}\right)\right)^{-3}$$
How can I proceed after this ? Is there another method ? Is my method correct ? 
 [/INST] In another approach, as inquired in the OP, we write the term of interest as
$$\begin{align}
\frac{1+x+x^2}{(1-x)^2}&=1-\frac{3}{1-x}+\frac{3}{(1-x)^2}\\\\
&=1-3\sum_{n=0}^\infty x^n+3\frac{d}{dx}\sum_{n=0}^\infty x^n\\\\
&=1-3\sum_{n=0}^\infty x^n+3\sum_{n=1}^\infty nx^{n-1}\\\\
&=1-3\sum_{n=0}^\infty x^n+3\sum_{n=0}^\infty (n+1)x^n\\\\
&=1+3\sum_{n=0}^\infty nx^n
\end{align}$$
Hence, the coefficient on the $n$'th term is indeed $3n$ as was to be shown!
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