<s>[INST] Binomial coefficients identity : $\sum_{k=1}^{n-m+1} k\binom{n-k+1}{m}=\binom{n+2}{m+2}$ For any positive integer m&n.$n\ge m$ , let $\left( {\begin{array}{*{20}{c}}
n\\
m
\end{array}} \right) = {}^n{C_m}$. Prove that $\left( {\begin{array}{*{20}{c}}
n\\
m
\end{array}} \right) + 2\left( {\begin{array}{*{20}{c}}
{n - 1}\\
m
\end{array}} \right) + 3\left( {\begin{array}{*{20}{c}}
{n - 2}\\
m
\end{array}} \right) + .. + \left( {n - m + 1} \right)\left( {\begin{array}{*{20}{c}}
m\\
m
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{n + 2}\\
{m + 2}
\end{array}} \right)$
My approach is as follow
$n=4, m=2$
$\left( {\begin{array}{*{20}{c}}
4\\
2
\end{array}} \right) + 2\left( {\begin{array}{*{20}{c}}
3\\
2
\end{array}} \right) + 3\left( {\begin{array}{*{20}{c}}
2\\
2
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
6\\
4
\end{array}} \right) = 15$
$6 + 6 + 3 = 15 \Rightarrow LHS = RHS$
But not able to solve it via property
 [/INST] Starting from
$$\sum_{k=1}^{n-m+1} k {n-k+1\choose m}$$
we re-write as
$$\sum_{k=1}^{n-m+1} k {n-k+1\choose n-m+1-k}
= [z^{n-m+1}] (1+z)^{n+1}
\sum_{k=1}^{n-m+1} k \frac{z^k}{(1+z)^k}$$
Here the coefficient extractor enforces the upper limit of the sum and
we obtain
$$[z^{n-m+1}] (1+z)^{n+1}
\sum_{k\ge 1} k \frac{z^k}{(1+z)^k}
\\ = [z^{n-m+1}] (1+z)^{n+1}
\frac{z/(1+z)}{(1-z/(1+z))^2}
= [z^{n-m+1}] (1+z)^{n+1}
z (1+z)
\\ = [z^{n-m}] (1+z)^{n+2} = {n+2\choose n-m}
= {n+2\choose m+2}$$
as claimed.
</s>