Sum of the series $\sum_{k=0}^{r}(-1)^k.(k+1).(k+2).\binom{n}{r-k} $ for $n>3$, The sum of the series  $\displaystyle \sum_{k=0}^{r}(-1)^k.(k+1).(k+2).\binom{n}{r-k} = $
where $\displaystyle \binom{n}{r} = \frac{n!}{r!.(n-r)!}$
My try:: I have expand the expression
$\displaystyle 1.2.\binom{n}{r}-2.3.\binom{n}{r-1}+3.4.\binom{n}{r-2}+........+(-1)^r.(r+1).(r+2).\binom{n}{0}$
Now after that how can i calculate it
Thanks

We shall use the combinatorial identity
$$\sum_{j=0}^{k}{(-1)^j\binom{n}{j}}=(-1)^k\binom{n-1}{k}$$
This can be proven easily by induction, and there is also probably some combinatorial argument why it holds. We shall use the equivalent form
$$\sum_{j=0}^{k}{(-1)^{k-j}\binom{n}{j}}=\binom{n-1}{k}$$
Now $(r-k+1)(r-k+2)=k(k-1)-(2r+2)k+(r^2+3r+2)$, so
\begin{align}
& \sum_{k=0}^{r}{(-1)^k(k+1)(k+2)\binom{n}{r-k}} \\
&=\sum_{k=0}^{r}{(-1)^{r-k}(r-k+1)(r-k+2)\binom{n}{k}} \\
& =\sum_{k=2}^{r}{(-1)^{r-k}k(k-1)\binom{n}{k}}-(2r+2)\sum_{k=1}^{r}{(-1)^{r-k}k\binom{n}{k}}+(r^2+3r+2)\sum_{k=0}^{r}{(-1)^{r-k}\binom{n}{k}} 
\end{align}
We have 
\begin{align}
\sum_{k=2}^{r}{(-1)^{r-k}k(k-1)\binom{n}{k}} & =\sum_{k=2}^{r}{(-1)^{(r-2)-(k-2)}n(n-1)\binom{n-2}{k-2}} \\
& =n(n-1)\sum_{k=0}^{r-2}{(-1)^{(r-2)-k}\binom{n-2}{k}} \\
& =n(n-1)\binom{n-3}{r-2} \\
& =\frac{r(r-1)(n-r)}{n-2}\binom{n}{r}
\end{align}
\begin{align}
\sum_{k=1}^{r}{(-1)^{r-k}k\binom{n}{k}} & =\sum_{k=1}^{r}{(-1)^{(r-1)-(k-1)}n\binom{n-1}{k-1}} \\
& =n\sum_{k=0}^{r-1}{(-1)^{(r-1)-k}\binom{n-1}{k}} \\
& =n\binom{n-2}{r-1} \\
& =\frac{r(n-r)}{n-1}\binom{n}{r}
\end{align}
\begin{align}
\sum_{k=0}^{r}{(-1)^{r-k}\binom{n}{k}} & =\binom{n-1}{r} \\
& =\frac{n-r}{n}\binom{n}{r}
\end{align}
Thus 
\begin{align}
& \sum_{k=0}^{r}{(-1)^k(k+1)(k+2)\binom{n}{r-k}} \\
& =\frac{r(r-1)(n-r)}{n-2}\binom{n}{r}-(2r+2)\frac{r(n-r)}{n-1}\binom{n}{r}+(r^2+3r+2)\frac{n-r}{n}\binom{n}{r} \\
& =\binom{n}{r}\frac{(n-r)(r(r-1)n(n-1)-(2r+2)rn(n-2)+(r^2+3r+2)(n-1)(n-2))}{n(n-1)(n-2)} \\
& =\binom{n}{r}\frac{(n-r)(2r^2+(6-4n)r+(2n^2-6n+4))}{n(n-1)(n-2)} 
\end{align}
