# Yet another sum involving binomial coefficients

Let $k,p$ be positive integers. Is there a closed form for the sums

$$\sum_{i=0}^{p} \binom{k}{i} \binom{k+p-i}{p-i}\text{, or}$$

$$\sum_{i=0}^{p} \binom{k-1}{i} \binom{k+p-i}{p-i}\text{?}$$

(where 'closed form' should be interpreted as usual, i.e. meaning free of sums and hypergeometric functions).

We know that the first sum has generating function $(1+z)^k/(1-z)^{k+1}$, and the second sum has generating function $(1+z)^{k-1}/(1-z)^{k+1}$, but that doesn't help me find a closed form so far.

• I'm sure you already checked, but Mathematica gives $$\sum_{i=0}^p\binom{k}{i}\binom{k+p-i}{p-i} = \binom{k+p}{p}\ _2 F_1(-k, -p;-k-p;-1)$$ So having a closed form for the left side would imply a closed form for the hypergeometric function on the right. – Aeryk Jul 15 '11 at 13:47
• Maple agrees, and also gives for the second sum $$\sum_{i=0}^p {k-1 \choose i} {k + p - i \choose p - i } = {k+p \choose p} {}_2F_1(-k+1,-p; -k-p; -1)$$ – Robert Israel Jul 16 '11 at 0:17

Your first sum is the Delannoy number $D(k,p)$. See OEIS sequence A008288