<s>[INST] Prove that $\sum\limits_{k=0}^{m}\binom{m}{k}\binom{n+k}{m}=\sum\limits_{k=0}^{m}\binom{n}{k}\binom{m}{k}2^k$ Prove that 
$$\sum_{k=0}^{m}\binom{m}{k}\binom{n+k}{m}=\sum_{k=0}^{m}\binom{n}{k}\binom{m}{k}2^k$$
What should I do for this equation? Should I focus on proving $\binom{m}{k}\binom{n+k}{m}=\binom{n}{k}\binom{m}{k}2^k$?
 [/INST] This can also be done using a basic complex variables technique.
Suppose we seek to verify that
$$\sum_{k=0}^m {m\choose k} {n+k\choose m}
= \sum_{k=0}^m {m\choose k} {n\choose k} 2^k.$$
Introduce the two integral representations
$${n+k\choose m}
= \frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{1}{z^{m+1}} (1+z)^{n+k} \; dz$$
and 
$${n\choose k}
= \frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{1}{z^{k+1}} (1+z)^n \; dz$$
This gives the following integral for the sum on the LHS
$$\frac{1}{2\pi i} 
\int_{|z|=\epsilon} 
\sum_{k=0}^m {m\choose k}
\frac{1}{z^{m+1}} (1+z)^{n+k} \; dz
\\ = \frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^{m+1}}
\sum_{k=0}^m {m\choose k} 
(1+z)^k \; dz
\\ = \frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^{m+1}} (2+z)^m \; dz.$$
We get the following integral for the sum on the RHS
$$\frac{1}{2\pi i} 
\int_{|z|=\epsilon} 
\sum_{k=0}^m {m\choose k} 2^k
\frac{1}{z^{k+1}} (1+z)^n \; dz
\\ = \frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{(1+z)^n}{z}
\sum_{k=0}^m {m\choose k} 2^k
\frac{1}{z^k} \; dz
\\ = \frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{(1+z)^n}{z}
\left(1 + \frac{2}{z}\right)^m \; dz
\\ = \frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{(1+z)^n}{z}
\frac{(z+2)^m}{z^m} \; dz.$$
We can stop here without  further evaluation because the integrals for
LHS and RHS are seen to be the same.

A trace as to when this method appeared on MSE and by whom starts at this
MSE link.
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