Finding the sum $\sum_{k=1}^rk^2\binom {n-k}{r-k}$ I was stuck while finding the given summation.
$$\sum_{k=1}^rk^2\binom {n-k}{r-k}$$
Since $n$ and $r$ are both constants, so I have first converted the above summation into this:
$$\sum_{k=1}^rk^2\binom {n-k}{n-r}$$
I have no idea how to proceed next. Any help will be appreciated.

We have
$$\sum_{k=1}^r k^2 {n-k\choose r-k}
= \sum_{k=0}^r k^2 [z^{r-k}] (1+z)^{n-k}
\\ = [z^r] \sum_{k=0}^r k^2 z^k (1+z)^{n-k}
= [z^r] (1+z)^n \sum_{k=0}^r k^2 z^k (1+z)^{-k}.$$
There is no contribution to the coefficient extractor
when $k\gt r$ and we may continue with
$$[z^r] (1+z)^n \sum_{k\ge 0} k^2 z^k (1+z)^{-k}
= [z^r] (1+z)^n
\frac{(z/(1+z))(1+z/(1+z))}{(1-z/(1+z))^3}
\\ = [z^r] (1+z)^n
z (1+z)^2 (1+z/(1+z))
= [z^r] (1+z)^n  z(1+z)(1+2z)
\\ = [z^{r-1}] (1+z)^{n+1} (1+2z).$$
This is for $r\ge 2$
$$\bbox[5px,border:2px solid #00A000]{
{n+1\choose r-1} + 2 {n+1\choose r-2}.}$$
Alternatively,
$$\frac{n-r+3}{r-1} {n+1\choose r-2} + 2{n+1\choose r-2}
= \frac{n+r+1}{r-1} {n+1\choose r-2}.$$
Here we have used the fact that with formal power series
$$\sum_{k\ge 0} k^2 w^k  = \frac{w(1+w)}{(1-w)^3}.$$
