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
 A: Here is a combinatorial approach. Consider the set
$$\{ 1,2,3,4,\ldots,n-1,n,n+1,n+2\}$$
We will pick $m+2$ distinct numbers. This can be done randomly in $\binom{n+2}{m+2}$ ways.
Another way to do the same would be prioritizing the second smallest element.
If second smallest number is $2$, we can choose $1$ as the smallest element and select $m$ out of remaining $n$ numbers. $\binom{n}{m}$ ways.
If second smallest number is $3$, we can choose the smallest in two ways out of $1,2$ and select $m$ out of remaining $n-1$ in $\binom{n-1}{m}$ ways.
In general, if second smallest choice is $i+1$, the smallest can be picked in $i$ ways and rest in $\binom{(n+2)-(i+1)}{m}=\binom{n-i+1}{m}$ ways.
Summing over all valid values of $i$, LHS=RHS is proved.
A: 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.
