Finding the Closed Form of: $\sum\limits_{i=1}^n k\cdot{n-2 \choose k-2}$ I am stuck with this example in the textbook: find a closed form of:
$$\sum\limits_{k=1}^n k\cdot{n-2 \choose k-2}.$$
I haven't found anything helpful on the web. Thanks for any advice. 
 A: We have $$\left(1+x\right)^{n}=\sum_{k=0}^{n}\dbinom{n}{k}x^{k}=1+\sum_{k=1}^{n}\dbinom{n}{k}x^{k}
 $$ then if we derive twice $$ n\left(n-1\right)\left(1+x\right)^{n-2}=n\left(n-1\right)\sum_{k=1}^{n}\frac{\left(n-2\right)!}{\left(k-2\right)!\left(n-k\right)!}x^{k-2}
 $$ hence $$x^{2}\left(1+x\right)^{n-2}=\sum_{k=1}^{n}\dbinom{n-2}{k-2}x^{k}
 $$ and if we derive again $$2x\left(1+x\right)^{n-2}+\left(n-2\right)x^{2}\left(1+x\right)^{n-3}=\sum_{k=1}^{n}\dbinom{n-2}{k-2}kx^{k-1}
 $$ now if we put $x=1
 $ we get $$2^{n-3}\left(n+2\right)=\sum_{k=1}^{n}\dbinom{n-2}{k-2}k.
 $$
A: Hint: Using
$$
k\binom{n}{k}=k\,\frac{n!}{k!\,(n-k)!}=n\,\frac{(n-1)!}{(k-1)!\,(n-k)!}=n\binom{n-1}{k-1}
$$
and
$$
\sum_{k=0}^n\binom{n}{k}=(1+1)^n=2^n
$$
Try breaking up
$$
\sum_{k=2}^nk\,\binom{n-2}{k-2}=\sum_{k=2}^n(k-2)\,\binom{n-2}{k-2}+\sum_{k=2}^n2\,\binom{n-2}{k-2}
$$
A: First note that
$$\sum_{k=1}^n k\binom{n-2}{k-2}=\sum_{k=2}^nk\binom{n-2}{k-2}=\sum_{k=0}^{n-2}(k+2)\binom{n-2}k\;.$$
Call this $f(n-2)$, so that
$$f(n)=\sum_{k=0}^n(k+2)\binom{n}k\;.$$
This is not hard to evaluate combinatorially. 
I have $n$ men and $2$ women. From that group I will select an all-male team, possibly empty. I will then select a team captain, who must either be one of the women or be on the selected team. For $k=0,\ldots,n$ there are $\binom{n}k$ ways to select an all-male team of size $k$, and there are then $k+2$ ways to choose the captain. Thus, there are altogether $f(n)$ ways to choose the team and captain.
I could get the same effect by first choosing a captain and then choosing the team. There are $2$ ways to choose a woman as captain, and for each there are $2^n$ ways to choose the all-male team; thus, there are $2^{n+1}$ ways to make the choices and get a woman as captain. There are $n$ ways to choose a male captain, after which there are $2^{n-1}$ ways to choose the rest of the all-male team, so there are $2^{n-1}n$ ways to make the choices and get a man as captain. Thus,
$$f(n)=2^{n+1}+2^{n-1}n=2^{n-1}(n+4)\;.$$
The original $f(n-2)$ is then $2^{n-3}(n+2)$.
