<s>[INST] Can anyone give a combinatorial proof of the identity ${n \choose m} + 2{n-1 \choose m}+3{n-2 \choose m}+...+(n-m+1){m \choose m}={n+2 \choose m+2}$ Can anyone give a combinatorial proof of the identity
$${n \choose m} + 2{n-1 \choose m}+3{n-2 \choose m}+\ldots+(n+1-m){m \choose m}={n+2 \choose m+2}$$
I am finding difficult as $n$ is varying instead of $m$
 [/INST] One way is by induction.  If $n = m$, then we have
$$
\binom{n}{m} = \binom{n+2}{m+2}
$$
which is true, as both sides equal $1$.  Next, suppose that
$$
\binom{m+k}{m} + 2\binom{m+k-1}{m} + 3\binom{m+k-2}{m} + \cdots
               + (k+1)\binom{m}{m} = \binom{m+k+2}{m+2}
$$
for some $k \geq 0$.  Then, for $k+1$, your left-hand side equals
\begin{align}
L & = \binom{m+k+1}{m} + 2\binom{m+k}{m} + 3\binom{m+k-1}{m} + \cdots
                       + (k+2)\binom{m}{m} \\
  & = \binom{m+k+1}{m} + \binom{m+k}{m} + \binom{m+k-1}{m} + \cdots
                       + \binom{m}{m} + \binom{m+k+2}{m+2} \\
  & = \binom{m+k+2}{m+1} + \binom{m+k+2}{m+2}
      \qquad \leftarrow \text{hockey-stick identity} \\
  & = \binom{m+k+3}{m+2}
\end{align}
which establishes the proposition.
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