<s>[INST] 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
 [/INST] Suppose we seek to evaluate
$$\sum_{k=0}^r (-1)^k (k+2)(k+1) {n\choose r-k}.$$
Start from
$${n\choose r-k}
= \frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{1}{z^{r-k+1}} (1+z)^n \; dz.$$
This yields the following expression for the sum
$$\frac{1}{2\pi i} 
\int_{|z|=\epsilon}
\sum_{k=0}^r (-1)^k (k+2)(k+1)
\frac{1}{z^{r-k+1}} (1+z)^n \; dz
\\ = \frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^{r+1}}
\sum_{k=0}^r (-1)^k (k+2)(k+1) z^k \; dz.$$
Observe that the defining integral of the binomial coefficient is zero
when $k > r.$ This condition  is precisely $r-k+1 < 1$ which makes the
integrand into an entire function.  Therefore we may extend the sum to
infinity.

We thus have
$$\frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^{r+1}}
\sum_{k=0}^\infty (-1)^k (k+2)(k+1) z^k \; dz
\\ = \frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^{r+1}}
\left(\frac{1}{1+z}\right)'' \; dz
\\ = \frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^{r+1}}
\frac{2}{(1+z)^3} \; dz
\\ = \frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{2(1+z)^{n-3}}{z^{r+1}} \; dz$$
Therefore the value of the sum is given by
$$[z^r] 2 (1+z)^{n-3} = 2{n-3\choose r}.$$
This may be re-written as
$$2{n\choose r} \frac{(n-r)(n-r-1)(n-r-2)}{n(n-1)(n-2)}$$
which is indeed the same as the earlier results, namely
$${n\choose r} \frac{(n-r)(2r^2 + (6-4n)r + 2n^2-6n+4}{n(n-1)(n-2)}.$$
A trace as to when this method appeared on MSE and by whom starts at this
MSE link.
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