Evaluate $\sum^{n}_{k=1}\sum^{k}_{r=0}r\binom{n}{r}$ 
Evaluate the summation $$\sum^{n}_{k=1}\sum^{k}_{r=0}r\binom{n}{r}$$

$\bf{Attempt:}$ From $$\sum^{n}_{k=1}\sum^{k}_{r=0}r\binom{n}{r} = \sum^{n}_{k=1}\sum^{k}_{r=0}\left[r\cdot \frac{n}{r}\binom{n-1}{r-1}\right] = n\sum^{n}_{k=1}\sum^{k}_{r=0}\binom{n-1}{r-1}$$
So $$ = n\sum^{n}_{k=1}\bigg[\binom{n-1}{0}+\binom{n-1}{1}+\cdots +\binom{n-1}{k-1}\bigg]$$
Could some help me to solve it, thanks.
 A: We have that
$$\begin{align}
\sum^{n}_{k=1}\sum^{k}_{r=0}r\binom{n}{r}&=
n\sum_{k=1}^{n}\sum_{r=1}^{k}\binom{n-1}{r-1}=n\sum_{r=1}^{n}\binom{n-1}{r-1}\sum_{k=r}^{n}1\\
&=n\sum_{r=1}^{n}\binom{n-1}{r-1}(n-(r-1))=
n\sum_{k=0}^{n-1}\binom{n-1}{k}(n-k)\\
&=n^2\sum_{k=0}^{n-1}\binom{n-1}{k}-n(n-1)\sum_{k=1}^{n-1}\binom{n-2}{k-1}
\\
&=n^22^{n-1}-n(n-1)2^{n-2}= n(n+1)2^{n-2}.
\end{align}$$
A: $$\begin{eqnarray*}\sum_{k=1}^{n}\sum_{r=0}^{k}r\binom{n}{r}&=&\sum_{k=1}^{n}\sum_{r=1}^{k}r\binom{n}{r}=n\sum_{k=1}^{n}\sum_{r=1}^{k}\binom{n-1}{r-1}\\&=&n\sum_{r=1}^{n}r\binom{n-1}{r-1}=n\sum_{r=1}^{n}\left[1+(r-1)\right]\binom{n-1}{r-1}\\&=&n2^{n-1}+n(n-1)\sum_{r=2}^{n}\binom{n-2}{r-2}\\&=&n2^{n-1}+n(n-1)2^{n-2}=\color{red}{n(n+1)2^{n-2}.}\end{eqnarray*}$$
A: $$ E:=n\sum^{n}_{k=1}\bigg[\binom{n-1}{0}+\binom{n-1}{1}+\cdots \cdots +\binom{n-1}{k-1}\bigg]$$
$$ = n\bigg[n\binom{n-1}{0}+(n-1)\binom{n-1}{1}+(n-2)\binom{n-1}{2}\cdots \cdots +1\binom{n-1}{n-1}\bigg] $$
Since we have How can I solve $\sum\limits_{i = 1}^k i \binom{k}{i-1}$:
$$(n-1)2^{n-2}=\sum\limits_{i=0}^{n-1} i\binom{n-1}{i} = \sum\limits_{i=0}^{n-1} i\binom{n-1}{n-i-1} = S - \sum\limits_{i=0}^{n-1}\binom{n-1}{i}=S-2^{n-1}$$
the finally result is:
$$ E = n[(n-1)2^{n-2}+2^{n-1}] = n2^{n-2}(n-1+2) = n(n+1)2^{n-2}$$
A: We will leave out the $r=0$ term since it is $0$.
$$
\begin{align}
\sum_{k=1}^n\sum_{r=1}^kr\binom{n}{r}
&=\sum_{r=1}^n\sum_{k=r}^nr\binom{n}{r}\\
&=\sum_{r=1}^n(n-r+1)r\binom{n}{r}\\
&=\sum_{r=1}^n(n-r)r\binom{n}{r}+\sum_{r=1}^nr\binom{n}{r}\\
&=\sum_{r=1}^nn(n-1)\binom{n-2}{r-1}+\sum_{r=1}^nn\binom{n-1}{r-1}\\[6pt]
&=n(n-1)2^{n-2}+n2^{n-1}\\[15pt]
&=n(n+1)2^{n-2}
\end{align}
$$
