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 ? 
 A: 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!
A: If we MUST use the binomial coefficients, this becomes rather tedious:
$$\frac{1+x^2+x}{(1-x)^2}=\frac{(1-x)^2+3x}{(1-x)^2}=1+\frac {3x}{(1-x)^2}$$$$=1-3x(1-x)^{-2}=1-3(1-x)^{-1}+3(1-x)^{-2}$$
$$=1-3\sum\binom{-1}{r}(-1)^rx^r+3\sum\binom{-2}{r}(-1)^rx^r$$
$$=1+3\sum \left(-\binom{-1}{r}+\binom{-2}{r}\right)(-1)^rx^r$$
Let's try to prove that the binomial difference if equal to $(-1)^r\times r$. For $r=0$ the sum is $0$, for $k=1$ the sum is $-1$, assuming this is true for $r=m$, let's check for $r=m+1$: $$-\binom{-1}{m+1}+\binom{-2}{m+1}=-\frac{-1-m}{1+m}\binom{-1}{m}+\frac{-2-m}{1+m}\binom{-2}{m}$$ $$=(-1)m(-1)^m-\frac 1{m+1}\binom{-2}{m}=m(-1)^{m+1}+(-1)^{m+1}$$
So the expansion becomes $$1+\sum_r3rx^r$$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
 \newcommand{\mrm}[1]{\,\mathrm{#1}}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\color{#f00}{{1 + x + x^{2} \over \pars{1 - x}^{2}}} & =
{1 - x^{3} \over \pars{1 - x}^{3}} =
\pars{1 - x^{3}}\sum_{r = 0}^{\infty}{-3 \choose r}\pars{-x}^{r} =
\half\pars{1 - x^{3}}\sum_{r = 0}^{\infty}\pars{r + 2}\pars{r + 1}x^{r}
\end{align}

Note that
  $\ds{{-3 \choose r} = {-\pars{-3} + r - 1 \choose r}\pars{-1}^{r} =
{r + 2 \choose r}\pars{-1}^{r}  =
\half\pars{r + 2}\pars{r + 1}\pars{-1}^{r}}$

\begin{align}
\color{#f00}{{1 + x + x^{2} \over \pars{1 - x}^{2}}} & =
\half\sum_{r = 0}^{\infty}\pars{r + 2}\pars{r + 1}x^{r} -
\half\sum_{r = 0}^{\infty}\pars{r + 2}\pars{r + 1}x^{r + 3}
\\[5mm] & =
\half\sum_{r = 0}^{\infty}\pars{r + 2}\pars{r + 1}x^{r} -
\half\sum_{r = 3}^{\infty}\pars{r - 1}\pars{r - 2}x^{r}
\\[5mm] & =
1 + 3x + 6x^{2} +
\half\sum_{r = 3}^{\infty}\bracks{%
\pars{r + 2}\pars{r + 1} - \pars{r - 1}\pars{r - 2}}x^{r}
\\[5mm] & =
1 + 3x + 6x^{2} + \sum_{r = 3}^{\infty}\pars{3r}x^{r} =
1 + \sum_{r = 1}^{\infty}\pars{3r}x^{r}
\end{align}

$$
\color{#f00}{\bracks{x^{r}}\bracks{{1 + x + x^{2} \over \pars{1 - x}^{2}}}}
=
\color{#f00}{\left\lbrace\begin{array}{rcrcl}
\ds{1} & \mbox{if} & \ds{r} & \ds{=} & \ds{0}
\\[2mm]
\ds{3r} & \mbox{if} & \ds{r} & \ds{=} & \ds{1,2,3,\ldots}
\\[2mm]
\ds{0} &&&& \mbox{otherwise}
\end{array}\right.}
$$
