Binomial sum of a sequence I have the following sequence:
$$
\sum\limits_{k=1}^n k\binom{n-1}{k-1}
$$ 
What is the sum of this sequence.
 A: HINT: That $k-1$ in the binomial coefficient is a little ugly, and the index starts at $k=1$, so why not shift it? Then you get
$$\begin{align*}
\sum_{k=1}^n\binom{n-1}{k-1}k&=\sum_{k=0}^{n-1}\binom{n-1}k(k+1)\\
&=\sum_{k=0}^{n-1}\binom{n-1}kk+\sum_{k=0}^{n-1}\binom{n-1}k\;.\tag{1}
\end{align*}$$
You should immediately recognize the second summation in $(1)$. The first isn’t quite so obvious, but if you apply a useful standard identity, you can turn it into something pretty straightforward:
$$k\binom{n}k=\frac{kn!}{k!(n-k)!}=\frac{n!}{(k-1)!(n-k)!}=\frac{n(n-1)!}{(k-1)!(n-k)!}=n\binom{n-1}{k-1}\;.$$
A: We have by the Binomial Theorem
$$(1+x)^{n-1}=\sum_{j=0}^{n-1}\binom{n-1}{j}x^j=\sum_{k=1}^n \binom{n-1}{k-1}x^{k-1}.$$
Let 
$$f(x)=x(1+x)^{n-1}.\tag{$1$}$$
 Then $f(x)=\sum_{k=1}^n \binom{n-1}{k-1}x^{k}$, and therefore
 $$f'(x)=\sum_{k=1}^n k\binom{n-1}{k-1}x^{k-1}.$$
It follows that our sum is $f'(1)$. But $f'(1)$ is easy to find from Expression $(1)$.
A: $\sum_{k=1}^n \binom{n-1}{k-1}\cdot k$
$=\sum_{k=1}^n \binom{n-1}{k-1}+\sum_{k=1}^n \binom{n-1}{k-1}\cdot(k-1)$ 
Now, $ \binom{n-1}{k-1}\cdot(k-1)=\frac{(n-1)!}{\{n-1-(k-1)\}!(k-1)!}\cdot(k-1)$
$=(n-1)\frac{(n-2)!}{\{n-2-(k-2)\}!(k-2)!}=(n-1)\binom{n-2}{k-2}$
$\sum_{k=1}^n \binom{n-1}{k-1}\cdot k$
$=\sum_{k=1}^n \binom{n-1}{k-1}+(n-1)\sum_{k=1}^n\binom{n-2}{k-2}$
$=\sum_{r=0}^{n-1} \binom{n-1}{r}+(n-1)\sum_{s=0}^{n-2}\binom{n-2}{s}$ as $\binom{n}{r}=0$ if $r<0$ or $r>n$
$=(1+1)^{n-1}+(n-1)(1+1)^{n-2}=2^{n-2}(2+n-1)=(n+1)2^{n-2} $
