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

Felix Marin's approach is nice and short, but I thought I'd add another approach that uses standard Binomial tools: Vandermonde's Identity, the symmetry of Pascal's Triangle, and $\binom{n}{k}=\frac nk\binom{n-1}{k-1}$.
$$
\hspace{-12pt}\begin{align}
&\phantom{={}}\sum_{k=0}^r(-1)^k(k+1)(k+2)\binom{n}{r-k}\\
&=\sum_{k=0}^r(-1)^{r-k}(r-k+1)(r-k+2)\binom{n}{k}\tag{1a}\\
&=\sum_{k=0}^r(-1)^{r-k}[k(k-1)-2(r+1)k+(r+1)(r+2)]\binom{n}{k}\tag{1b}\\
&=\scriptsize n(n-1)\sum_{k=0}^r\binom{-1}{r-k}\binom{n-2}{k-2}-2(r+1)n\sum_{k=0}^r\binom{-1}{r-k}\binom{n-1}{k-1}+(r+1)(r+2)\sum_{k=0}^r\binom{-1}{r-k}\binom{n}{k}\tag{1c}\\
&=n(n-1)\binom{n-3}{r-2}-2(r+1)n\binom{n-2}{r-1}+(r+1)(r+2)\binom{n-1}{r}\tag{1d}\\
&=\left(n(n-1)\frac{r-1}{n-2}-2(r+1)n+(r+1)(r+2)\frac{n-1}r\right)\binom{n-2}{r-1}\tag{1e}\\
&=\frac{2(n-r-1)(n-r-2)}{r(n-2)}\binom{n-2}{n-r-1}\tag{1f}\\
&=\frac{2(n-3)}{r}\binom{n-4}{n-r-3}\tag{1g}\\
&=2\binom{n-3}{r}\tag{1h}\\
\end{align}
$$
Explanation:
$\text{(1a):}$ substitute $k\mapsto r-k$
$\text{(1b):}$ $(r-k+1)(r-k+2)=k(k-1)-2(r+1)k+(r+1)(r+2)$
$\text{(1c):}$ $\binom{n}{k}=\frac{n(n-1)}{k(k-1)}\binom{n-2}{k-2}$ and $\binom{n}{k}=\frac nk\binom{n-1}{k-1}$
$\phantom{\text{(1c):}}$ $\binom{-1}{r-k}=(-1)^{r-k}[k\le r]$ (Iverson Brackets)
$\text{(1d):}$ Vandermonde's Identity
$\text{(1e):}$ $\binom{n-3}{r-2}=\frac{r-1}{n-2}\binom{n-2}{r-1}$ and $\binom{n-1}{r}=\frac{n-1}{r}\binom{n-2}{r-1}$
$\text{(1f):}$ simplify the rational function
$\phantom{\text{(1f):}}$ $\binom{n-2}{r-1}=\binom{n-2}{n-r-1}$ (symmetry of Pascal's Triangle)
$\text{(1g):}$ $\binom{n-2}{n-r-1}=\frac{(n-2)(n-3)}{(n-r-1)(n-r-2)}\binom{n-4}{n-r-3}$
$\text{(1h):}$ $\binom{n-4}{n-r-3}=\binom{n-4}{r-1}$ (symmetry of Pascal's Triangle)
$\phantom{\text{(1h):}}$ $\binom{n-3}{r}=\frac{n-3}r\binom{n-4}{r-1}$
