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.

We will leave out the $r=0$ term since it is $0$.
$$
\begin{align}
\sum_{k=1}^n\sum_{r=1}^kr\binom{n}{r}
&=\sum_{r=1}^n\sum_{k=r}^nr\binom{n}{r}\\
&=\sum_{r=1}^n(n-r+1)r\binom{n}{r}\\
&=\sum_{r=1}^n(n-r)r\binom{n}{r}+\sum_{r=1}^nr\binom{n}{r}\\
&=\sum_{r=1}^nn(n-1)\binom{n-2}{r-1}+\sum_{r=1}^nn\binom{n-1}{r-1}\\[6pt]
&=n(n-1)2^{n-2}+n2^{n-1}\\[15pt]
&=n(n+1)2^{n-2}
\end{align}
$$
