As part of another problem I'm working on, I find myself needing to prove the following:

$$\sum\_{k=0}^n\binom{2k+1}{k}\binom{m-(2k+1)}{n-k} = \sum\_{k=0}^{n}\binom{m+1}{k}$$

where $n\leq m$. I've checked it computationally for all $n\leq m\leq 16$. 

A few thoughts: this looks like a binomial convolution, but the $k$'s show up in the top of the binomial coefficients which disqualifies it from Vandermonde-esque identities I've found. Further, it uses weird binomial coefficients where the top is less than the bottom and the top can be negative - seems strange to me. 

Some references I've found [(for example)](http://www.kurims.kyoto-u.ac.jp/EMIS/journals/EJC/Volume_17/PDF/v17i1n24.pdf) have similar looking sums of products, but the $2k$ instead of $k$ seems to hurt. Another ("Some Generalizations of Vandermonde's Convolution" by H. W. Gould) reveals to me that 
$$\sum\_{k=0}^n\binom{2k+1}{k}\binom{m-(2k+1)}{n-k} = 
\sum\_{k=0}^n\binom{2k+1+j}{k}\binom{m-(2k+1)-j}{n-k} $$
where $j$ can be any integer. Not sure if this can help.

I see from [this question](https://math.stackexchange.com/questions/69532/partial-sum-of-rows-of-pascals-triangle) and elsewhere that partial sums of Pascal triangle rows don't really have closed forms. I can't think of how to use a generating function here (I'm trying to show a sum is equal to a sum), and the terms in each sum seem completely different. I'm not really sure how to proceed, any help/advice would be much appreciated!