<s>[INST] 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! 
 [/INST] 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
$$
\begin{align}
\sum_{n=0}^\infty\binom{6+n}{n}x^n(6+n)
&=\sum_{n=0}^\infty\binom{6+n}{n}x^n(7+n)-\sum_{n=0}^\infty\binom{6+n}{n}x^n\\
&=7\sum_{n=0}^\infty\binom{7+n}{n}x^n-\sum_{n=0}^\infty\binom{6+n}{n}x^n\\
&=7\sum_{n=0}^\infty\binom{-8}{n}(-x)^n-\sum_{n=0}^\infty\binom{-7}{n}(-x)^n\\[3pt]
&=\frac7{(1-x)^8}-\frac1{(1-x)^7}\\[6pt]
&=\frac{6+x}{(1-x)^8}
\end{align}
$$
</s>