The following problem is from MOP 2007.
Prove that$$\sum_{i=0}^n\binom{n}{2i+1}\binom{i}{m}=2^{n-2m-1}\binom{n-m-1}{m}$$
I tried to approach this using generating functions:
Let $f(X),g(X)$ be defined as:$$f(X)=\sum_{n\geq 0}\left[\sum_{i=0}^n\binom{n}{2i+1}\binom{i}{m}\right]X^n$$$$g(X)=\sum_{n\geq 0}\left[2^{n-2m-1}\binom{n-m-1}{m}\right]X^n$$
To evaluate $f$, we can swap the order of summation and separate the binomials to get:$$f(X)=\sum_{i\geq 0}\sum_{n\geq i}\binom{n}{2i+1}\binom{i}{m}X^n=\sum_{i\geq 0}\binom{i}{m}\sum_{n\geq i}\binom{n}{2i+1}X^n$$
However, using the well known generating function $$\sum_{k\geq 0}\binom{k}{m}X^k=\frac{X^m}{(1-X)^{m+1}}$$we get by evaluating the inner summation that$$f(X)=\sum_{i\geq 0}\binom{i}{m}\frac{x^{2i+1}}{(1-X)^{2i+2}}$$
At this point, I'm not sure how to evaluate this sum in closed form (or, if this is even possible). But fortunately we don't need to evaluate it, we just need to show it is equivalent to $g(X)$.
However, I wasn't able to get anywhere with this. I was wondering if we could use the generating function I mentioned earlier to do so. We can write $$g(X)=X^{2m+1}\left(\binom{m}{m}+2^1\binom{m+1}{m}X+2^2\binom{m+2}{m}X^2+2^3\binom{m+3}{m}X^3+\dots\right)$$
Using that generating function I mentioned earlier, this can be written as $$g(X)=\frac{1}{(1-X)^{m+1}}+\left(\frac{X}{(1-X)^{m+1}}-X\right)+\left(\frac{2X^2}{(1-X)^{m+1}}-X^2-\binom{m+1}{m}X^2\right)+\dots$$
Which happens to resemble a geometric series, but with some left over terms being subtracted off. Not sure if there is a clever way to evaluate this.
Or maybe I'm just going the wrong way, any thoughts?
(Also, ideas of a combinatorial proof would be nice to see as well!)
Update (due to Lord Shark):
Using the above generating function again, we can write $f$ in closed form as$$f(x)=\frac{X^{2m+1}}{(1-2X)^{m+1}}$$
Looks like it remains to investigate $g(x)$.
