I want to prove $$\sum_{k=0}^{n}\left(\sum_{l=0}^{k}\binom{n}{k}\binom{k}{l} l\right)=n \times 3^{n-1}$$ with combinatoric argumentation.
I have tried to understand it by comparing it with similiar problems. One such problem would be to prove the following equation with combinatoric argumentation:
$$\sum_{k=1}^n {{n}\choose{k}}*k = n* 2^{n-1}$$ The proof would be
How can i apply that solution to the given problem?
We have $n$ people, One of them will be the Captain & all the others will be one of $3$ types (A,B or C).
Choose $n-k$ to be of type $A$, now choose $k-l$ out of the remaining $k$ to be of type $B$, now choose one of the remainig $l$ to be the captain and let the final $l-1$ be of type $C$.
Playing algebraically.
$\begin{array}\\ s(n) &=\sum_{k=0}^{n}\left(\sum_{l=0}^{k}\binom{n}{k}\binom{k}{l} l\right)\\ &=\sum_{k=0}^{n}\left(\sum_{l=0}^{k}\dfrac{n!k!}{k!(n-k)!l!(k-l)!}l\right)\\ &=\sum_{k=0}^{n}\left(\sum_{l=0}^{k}\dfrac{n!l}{(n-k)!l!(k-l)!}\right)\\ &=\sum_{k=0}^{n}\sum_{l=0}^{k}\dfrac{n!l}{(n-k)!l!(k-l)!}\\ &=\sum_{l=0}^{n}\sum_{k=l}^{n}\dfrac{n!l}{(n-k)!l!(k-l)!} \quad\text{(when in doubt, reverse the order of summation)}\\ &=\sum_{l=0}^{n}\dfrac{n!l}{l!}\sum_{k=l}^{n}\dfrac{1}{(n-k)!(k-l)!}\\ &=\sum_{l=0}^{n}\dfrac{n!l}{l!}\sum_{k=0}^{n-l}\dfrac{1}{(n-(k+l))!(k+l-l)!}\\ &=\sum_{l=0}^{n}\dfrac{n!l}{l!}\sum_{k=0}^{n-l}\dfrac{1}{(n-k-l)!k!}\\ &=\sum_{l=0}^{n}\dfrac{n!l}{l!(n-l)!}\sum_{k=0}^{n-l}\dfrac{(n-l)!}{(n-k-l)!k!}\\ &=\sum_{l=0}^{n}l\binom{n}{l}\sum_{k=0}^{n-l}\binom{n-l}{k}\\ &=\sum_{l=0}^{n}l\binom{n}{l}2^{n-l}\\ &=\sum_{l=0}^{n}(n-l)\binom{n}{n-l}2^{l}\\ &=\sum_{l=0}^{n}n\binom{n}{n-l}2^{l}-\sum_{l=0}^{n}l\binom{n}{l}2^{l}\\ &=n3^n-\sum_{l=0}^{n}l\binom{n}{l}2^{l}\\ &=n3^n-2n3^{n-1} \qquad (*)\\ &=n \times 3^{n-1}\\ \end{array} $
(*) $(1+x)^n =\sum_{k=0}^n \binom{n}{k}x^k $ so $n(1+x)^{n-1} =\sum_{k=1}^n \binom{n}{k}kx^{k-1} $ so, with $x = 2$, $n3^{n-1} =\sum_{k=1}^n \binom{n}{k}k2^{k-1} $ or $2n3^{n-1} =\sum_{k=1}^n \binom{n}{k}k2^{k} $
