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 a representation which is free of sums, binomial coefficients, or any other hypergeometric functions).

I derived the following simple inequality (confirmed numerically): for any $0 < s < 1/2$,
$$
\sum\limits_{i = 0}^p {{k \choose i}{k + p - i \choose p - i}}  \le \frac{1}{{(1 - 2s)^k }}\bigg(\frac{{1 - s}}{s}\bigg)^p.
$$
Given $0 < s < 1/2$, let $X$ be a binomial$(k,s)$ random variable, and $Y$ a binomial$(k+p-X,t)$ random variable, where $t=s/(1-s) \, (\in (0,1))$. Then, by the law of total probability,
$$
{\rm P}(X + Y = p) = \sum\limits_{i = 0}^p {{\rm P}(X + Y = p|X = i){\rm P}(X = i)}  = \sum\limits_{i = 0}^p {{\rm P}(Y = p - i|X = i){\rm P}(X = i)}.
$$
Noting that 
$$
{\rm P}(X = i) = {k \choose i}s^i (1-s)^{k-i}
$$
and
$$
{\rm P}(Y = p - i|X = i) = {k + p - i \choose p - i}t^{p - i} (1 - t)^k ,
$$
and using
$$
\frac{s}{{(1 - s)t}} = 1, 
$$
we get
$$
{\rm P}(X + Y = p) = (1 - s)^k (1 - t)^k t^p \sum\limits_{i = 0}^p {{k \choose i}{k + p - i \choose p - i}} .
$$
Finally, from ${\rm P}(X + Y = p) \leq 1$ and
$$
(1 - s)^k (1 - t)^k t^p = (1 - 2s)^{k} \bigg(\frac{s}{{1 - s}}\bigg)^p ,
$$
it follows that
$$
\sum\limits_{i = 0}^p {{k \choose i}{k + p - i \choose p - i}}  \le \frac{1}{{(1 - 2s)^k }}\bigg(\frac{{1 - s}}{s}\bigg)^p.
$$
