How do I find a terminating series representation of the hypergeometric function $_2 F_1(p, n+1, n+2, x)$, for real $p \in \mathbb{R}$ but $n \in \mathbb{Z}$, $n\geq0$?
Mathematica gives (Hypergeometric2F1[p, 1 + n, 2 + n, x] /. n -> 0, etc)
\begin{align}
n=0:&\qquad -\frac{(1-x)^{-p} \left((1-x)^p+x-1\right)}{(p-1) x}\\
n=1:&\qquad \frac{2 (1-x)^{-p} \left(-p x^2+(1-x)^p+p x+x^2-1\right)}{(p-2)
(p-1) x^2}\\
\vdots\end{align}
I am getting stuck because in principle $p$ and $n$ can all be positive, so that the standard definition
$$\sum_{k=0}^\infty \frac{(n+1)_k(p)_k}{{(n+2)}_k k! } x^k$$
doesn't terminate. Please help!
Recall $\displaystyle\;\frac{(\gamma)_k}{(\gamma+1)_k} = \frac{\gamma}{\gamma+k}\;$ and assume $p - 1 \ne 0, 1, \ldots, n$ with $x \ne 0$, we have
$$\begin{align} {}_2F_1(p,n+1;n+2;x) &= \sum_{k=0}^\infty \frac{(n+1)_k}{(n+2)_k} \frac{(p)_k x^k}{k!} = \sum_{k=0}^\infty \frac{n+1}{n+k+1}\frac{(p)_k x^k}{k!} \\ &= \frac{n+1}{x^{n+1}}\int_0^x \left(\sum_{k=0}^\infty\frac{(p)_k z^k}{k!} \right) z^n dz\\ &= \frac{n+1}{x^{n+1}}\int_0^x \frac{z^n}{(1-z)^p} dz = \frac{n+1}{x^{n+1}}\int_0^x \frac{(1 - (1-z))^n}{(1-z)^p} dz\\ &= \frac{n+1}{x^{n+1}} \sum_{k=0}^n (-1)^k \binom{n}{k} \int_0^x \frac{dz}{(1-z)^{p-k}}\\ &= \frac{n+1}{x^{n+1}} \sum_{k=0}^n \frac{(-1)^k}{p-k-1}\binom{n}{k} \left[ \frac{1}{(1-x)^{p-k-1}} - 1\right] \end{align} $$