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 have that $$\begin{align} \sum^{n}_{k=1}\sum^{k}_{r=0}r\binom{n}{r}&= n\sum_{k=1}^{n}\sum_{r=1}^{k}\binom{n-1}{r-1}=n\sum_{r=1}^{n}\binom{n-1}{r-1}\sum_{k=r}^{n}1\\ &=n\sum_{r=1}^{n}\binom{n-1}{r-1}(n-(r-1))= n\sum_{k=0}^{n-1}\binom{n-1}{k}(n-k)\\ &=n^2\sum_{k=0}^{n-1}\binom{n-1}{k}-n(n-1)\sum_{k=1}^{n-1}\binom{n-2}{k-1} \\ &=n^22^{n-1}-n(n-1)2^{n-2}= n(n+1)2^{n-2}. \end{align}$$
$$\begin{eqnarray*}\sum_{k=1}^{n}\sum_{r=0}^{k}r\binom{n}{r}&=&\sum_{k=1}^{n}\sum_{r=1}^{k}r\binom{n}{r}=n\sum_{k=1}^{n}\sum_{r=1}^{k}\binom{n-1}{r-1}\\&=&n\sum_{r=1}^{n}r\binom{n-1}{r-1}=n\sum_{r=1}^{n}\left[1+(r-1)\right]\binom{n-1}{r-1}\\&=&n2^{n-1}+n(n-1)\sum_{r=2}^{n}\binom{n-2}{r-2}\\&=&n2^{n-1}+n(n-1)2^{n-2}=\color{red}{n(n+1)2^{n-2}.}\end{eqnarray*}$$
$$ E:=n\sum^{n}_{k=1}\bigg[\binom{n-1}{0}+\binom{n-1}{1}+\cdots \cdots +\binom{n-1}{k-1}\bigg]$$ $$ = n\bigg[n\binom{n-1}{0}+(n-1)\binom{n-1}{1}+(n-2)\binom{n-1}{2}\cdots \cdots +1\binom{n-1}{n-1}\bigg] $$
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: $$ E = n[(n-1)2^{n-2}+2^{n-1}] = n2^{n-2}(n-1+2) = n(n+1)2^{n-2}$$
We will leave out the $r=0$ term since it is $0$. $$ \begin{align} \sum_{k=1}^n\sum_{r=1}^kr\binom{n}{r} &=\sum_{r=1}^n\sum_{k=r}^nr\binom{n}{r}\\ &=\sum_{r=1}^n(n-r+1)r\binom{n}{r}\\ &=\sum_{r=1}^n(n-r)r\binom{n}{r}+\sum_{r=1}^nr\binom{n}{r}\\ &=\sum_{r=1}^nn(n-1)\binom{n-2}{r-1}+\sum_{r=1}^nn\binom{n-1}{r-1}\\[6pt] &=n(n-1)2^{n-2}+n2^{n-1}\\[15pt] &=n(n+1)2^{n-2} \end{align} $$
