Infinite Sum with Combination I am trying to figure out what the following sum converges to:
$$\sum_{n=0}^\infty {6+n\choose n}x^n(6+n),\qquad\qquad0<x<1$$
An answer would be great, but if you have an explanation, that'd be better! 
 A: Since
$$
7\binom{7+n}{n}=(7+n)\binom{6+n}{n}
$$
and using negative binomial coefficients,
$$
\binom{k+n}{n}=(-1)^n\binom{-k-1}{n}
$$
we get
$$
\begin{align}
\sum_{n=0}^\infty\binom{6+n}{n}x^n(6+n)
&=\sum_{n=0}^\infty\binom{6+n}{n}x^n(7+n)-\sum_{n=0}^\infty\binom{6+n}{n}x^n\\
&=7\sum_{n=0}^\infty\binom{7+n}{n}x^n-\sum_{n=0}^\infty\binom{6+n}{n}x^n\\
&=7\sum_{n=0}^\infty\binom{-8}{n}(-x)^n-\sum_{n=0}^\infty\binom{-7}{n}(-x)^n\\[3pt]
&=\frac7{(1-x)^8}-\frac1{(1-x)^7}\\[6pt]
&=\frac{6+x}{(1-x)^8}
\end{align}
$$
A: Related techniques: (I), (II). Follow the steps:
1) simplify $(n+6){ n+6\choose n} $ as

$$ (n+6){ n+6\choose n} = \frac{1}{6!}(n+1)(n+2)(n+3)(n+4)(n+5)(n+6)^2. $$

2) use the series identity
$$ \sum_{n=0}^{\infty} x^{n+1}=\frac{x}{1-x} \longrightarrow (*) $$
3) Applying the operators $D(xD)(x^2D)^5 $ to both sides of $(*)$  , where $D=\frac{d}{dx}$, gives
$$ \sum_{n=0}^{\infty}(n+1)(n+2)(n+3)(n+4)(n+5)(n+6)^2 x^{n+5}=D(xD)(x^2D)^5 \frac{x}{1-x}  $$

$$ \implies \frac{1}{6!}\sum_{n=0}^{\infty}(n+1)(n+2)(n+3)(n+4)(n+5)(n+6)^2 x^{n}\\
=\frac{1}{6!\,x^5}D(xD)(x^2D)^5 \frac{x}{1-x} . $$

Note: The operator $(x^2 D)^5$ means

$$ (x^2 D)^5 = (x^2D)(x^2D)(x^2D)(x^2D)(x^2D). $$

A: Starting fom the previous answers to this post, it seems to me (assuming no mistakes o my side) that $$\sum_{n=0}^\infty {6+n\choose n}x^n(6+n)=\frac{x+6}{(1-x)^8}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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\begin{align}
&\bbox[10px,#ffd]{\ds{\left.\sum_{n = 0}^{\infty}{6 + n \choose n}x^{n}\pars{6 + n}
\,\right\vert_{\ 0\ <\ x\ <\ 1}}} =
{1 \over x^{5}}\sum_{n = 0}^{\infty}\bracks{{-7 \choose n}\pars{-1}^{n}}
x^{n + 5}\pars{n + 6}
\\[5mm] = &\
{1 \over x^{5}}\,\partiald{}{x}
\sum_{n = 0}^{\infty}{-7 \choose n}\pars{-x}^{n + 6} =
{1 \over x^{5}}\,\partiald{}{x}\bracks{\pars{-x}^{6}
\sum_{n = 0}^{\infty}{-7 \choose n}\pars{-x}^{n}} =
{1 \over x^{5}}\,\partiald{}{x}\bracks{x^{6}\pars{1 - x}^{\,-7}}
\\[5mm] = &\
{1 \over x^{5}}\bracks{6x^{5}\pars{1 - x}^{-7} + 7x^{6}\pars{1 - x}^{-8}} =
{6\pars{1 - x} + 7x \over \pars{1 - x}^{8}} =
\bbx{6 + x \over \pars{1 - x}^{8}}
\end{align}
