<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] 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

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

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