<s>[INST] Showing ${n + a - 1 \choose a - 1} = \sum_{k = 0}^{\left\lfloor n/2 \right\rfloor} {a \choose n-2k}{k+a-1 \choose a-1}$ Prove that for integers $n \geq 0$ and $a \geq 1$, $${n + a - 1 \choose a - 1} = \sum_{k = 0}^{\left\lfloor n/2 \right\rfloor} {a \choose n-2k}{k+a-1 \choose a-1}.$$
I figured I'd post this question, which was on an assignment I did, since I thought the solution was so nice.
 [/INST] This one can also be done using complex variables.

Suppose we seek to evaluate
$$\sum_{k=0}^{\lfloor n/2\rfloor} 
{a\choose n-2k} {k+a-1\choose a-1}.$$
Introduce the integral representation
$${a\choose n-2k}
=\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{a}}{z^{n-2k+1}} \; dz.$$
Note that this integral is zero when $k>\lfloor n/2\rfloor$
so we may extend the sum to infinity.

This gives for the sum the integral
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{a}}{z^{n+1}} 
\sum_{k\ge 0} {k+a-1\choose a-1} z^{2k}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{a}}{z^{n+1}} 
\frac{1}{(1-z^2)^a} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} 
\frac{1}{(1-z)^a} \; dz.$$
This last one evaluates to
$${n+a-1\choose a-1}$$
by inspection (Newton binomial).
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