Evaluation of the expression : $ \sum\limits_{n=0}^{\infty}{{14+2n}\choose{n}}p^{14+n}(1-p)^{n} $, $ 0 \lt p \lt 1$ I've been trying to compute an expression related to a certain gambling system that I'm analysing.
I have arrived that such a probability can be expressed as $$ \sum_{n=0}^{\infty}{{14+2n}\choose{n}}p^{14+n}(1-p)^{n} \ \ \text{where}\ 0 < p < 1$$
Anyone has a hint on how can this be evaluated ?
Thanks!
 A: We seek to evaluate:
$$S(K,p)=p^K\sum_{n=0}^{\infty}{{K+2n}\choose{n}}x^n, \tag1
$$
where $x=p(1-p)$.
Using the identity:
$$
\binom nk=\frac1{2\pi i}\oint_{|z|=1}\frac{(1+z)^n}{z^{k+1}}dz
$$
the expression (1) can be rewritten as:
$$\begin{align}
2\pi i\, S(K,p)&=p^K \sum_{n=0}^{\infty}x^n\oint_{|z|=1}\frac{(1+z)^{K+2n}}{z^{n+1}}dz\\
&=p^K \oint_{|z|=1}\left[\frac{(1+z)^K}z\sum_{n=0}^{\infty}\frac{x^n(1+z)^{2n}}{z^{n}}\right]dz\\
&=p^K \oint_{|z|=1}\left[\frac{(1+z)^K}z\frac1{1-\frac{x(1+z)^{2}}{z}}\right]dz\\
&=p^K \oint_{|z|=1}\left[\frac{(1+z)^K}{z-p(1-p)(1+z)^{2}}\right]dz.\tag2\\
\end{align}
$$
The poles of the integrand are $z_1=\frac p{1-p}$ and $z_2=\frac{1-p}p$. If $p=\frac12$ we have $z_1=z_2=1$ with the pole of second order lying on the 
integration contour. It can be shown that in this case the integral (as well as the original sum) diverges.
If $p\ne\frac12$ one pole is inside the integration contour, and the other one is outside of it. 
Rewriting the integrand as:
$$
f(z)=-\frac{(1+z)^K}{p(1-p)(z-\frac p{1-p})(z-\frac {1-p}p)}
$$
we have by residue theorem:
$$\oint_{|z|=1}f(z)dz=2\pi i\times
\begin{cases}
\operatorname{res}(f(z),z_1)=\frac1{(1-p)^K(1-2p)},& p<\frac12;\\
\operatorname{res}(f(z),z_2)=\frac1{p^K(2p-1)},& p>\frac12,\\
\end{cases}
$$
so that finally:
$$
S(K,p)=\begin{cases}
\frac1{1-2p}\left(\frac{p}{1-p}\right)^K,& p<\frac12;\\
\frac1{2p-1},& p>\frac12.\\
\end{cases}\tag3
$$
