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).

Lets examine the first sum.  I can't seem to find a closed form, but there is something very nice with the generating series.  They are simple, and symmetrical with respect to the variables $p$ and $k$. 
Result:
Your sum is the $k^{th}$ coefficient of $\frac{(1+x)^{p}}{\left(1-x\right)^{p+1}},$ and also the $p^{th}$ coefficient of $\frac{(1+x)^{k}}{\left(1-x\right)^{k+1}}.$
The Generating Series for the variable $p$
Consider
$$F(x)=\sum_{p=0}^{\infty}\sum_{i=0}^{p}\binom{k}{i}\binom{k+p-i}{p-i}x^{p}.$$
Changing the order of summation, this becomes
$$F(x)=\sum_{i=0}^{\infty}\binom{k}{i}\sum_{p=i}^{\infty}\binom{k+p-i}{p-i}x^{p},$$
and then shifting the second sum we have
$$F(x)=\sum_{i=0}^{\infty}\binom{k}{i}x^{i}\sum_{p=0}^{\infty}\binom{k+p}{p}x^{p}.$$ Since the rightmost sum is $\frac{1}{(1-x)^{k+1}}$ we see that the generating series is
$$F(x)=\frac{1}{(1-x)^{k+1}}\sum_{i=0}^{\infty}\binom{k}{i}x^{i}=\frac{\left(1+x\right)^{k}}{(1-x)^{k+1}}$$ 
by the binomial theorem.
The Generating Series for the variable $k$:
Lets consider the other generating series with respect to the variable $k$.  Let  
$$G(x)=\sum_{k=0}^{\infty}\sum_{i=0}^{p}\binom{k}{i}\binom{k+p-i}{p-i}x^{k}.$$
Then
$$G(x)=\sum_{i=0}^{p}\sum_{k=i}^{\infty}\binom{k}{i}\binom{k+p-i}{p-i}x^{k}=\sum_{i=0}^{p}x^{i}\sum_{k=0}^{\infty}\binom{k+i}{i}\binom{k+p}{p-i}x^{k}.$$
Splitting up the binomial coefficients into factorials, this is
$$=\sum_{i=0}^{p}x^{i}\sum_{k=0}^{\infty}\frac{(k+i)!}{k!i!}\frac{(k+p)!}{(k+i)!(p-i)!}x^{k}=\sum_{i=0}^{p}\frac{x^{i}p!}{i!\left(p-i\right)!}\sum_{k=0}^{\infty}\frac{\left(k+p\right)!}{k!p!}x^{k}.$$
Consequently,
$$G(x)=\frac{(1+x)^{p}}{\left(1-x\right)^{p+1}}.$$
Comments:  I am not sure why the generating series has this symmetry.  Perhaps you can use this property to tell you more about the sum/generating series.
Hope that helps,
