simplification of combinatorial sum

Question

Is there a simplified form of the following sum?

$$ \sum_{k=0}^n p^k{n+k\choose k} $$

Answer 1

You should be able to prove, by induction, that at $p = 1$, the series is equal to $\binom{2n+1}{n+1}$.

At $p \neq 1$, according to Wolfram Alpha, you can write this in terms of the hypergeometric function: $$ \frac{1}{\left(1-p\right)^{n+1}}-{}_{2}F_{1}(1,2n+2,n+2;p)\binom{2n+1}{n+1}p^{n+1}. $$ You should be able to prove the above with some patient algebra (you'll also need the definition of the hypergeometric function when its third argument is equal to a positive integer).

Note: At $p = 0$, the expression above agrees with the series under the convention $0^0 = 1$, in which case both are equal to $1$.

That the expression is in terms of a special function should be a good indicator that outside of special cases (such as $p = 1$), you probably will not be able to find a "nice" closed form expression.

Answer 2

Hypergeometric functions are just special ways to write series with certain ratios between terms: $$ \begin{align} \sum_{k=n+1}^\infty\!\!\binom{n+k}{k}p^k &=\binom{2n+1}{n+1}p^{n+1}\left[1+\frac{2n+2}{n+2}p+\frac{2n+2}{n+2}\frac{2n+3}{n+3}p^2+\cdots\right]\\ &=\binom{2n+1}{n+1}p^{n+1}{}_2F_1(1,2n+2;n+2;p)\tag1 \end{align} $$ The first argument of $1$ cancels the $k!$ in $\frac{p^k}{k!}$.

The Binomial Theorem says $$ \begin{align} \sum_{k=0}^\infty\!\binom{n+k}{k}p^k &=\sum_{k=0}^\infty(-1)^k\binom{-n-1}{k}p^k\\ &=(1-p)^{-n-1}\tag2 \end{align} $$ Subtracting $(1)$ from $(2)$ yields $$ \sum_{k=0}^n\!\binom{n+k}{k}p^k =(1-p)^{-n-1}-\binom{2n+1}{n+1}p^{n+1}{}_2F_1(1,2n+2;n+2;p)\tag3 $$