Determine: $S = \frac{2^2}{2}{n \choose 1} + \frac{2^3}{3}{n \choose 2} + \frac{2^4}{4}{n \choose 3} + \cdots + \frac{2^{n+1}}{n+1}{n \choose n}$ We are given two hints: consider $(n+1)S$; and use the Binomial Theorem.  But we are not to use calculus.  
My consideration of $(n+1)S$ goes like this:
\begin{align*}
 \sum\limits_{k=1}^{n}\frac{2^{k+1}}{k+1}{n \choose k}
 &= \frac{1}{n+1}\sum\limits_{k=1}^{n}(n+1)\frac{2^{k+1}}{k+1}{n \choose k} \\
 &= 2\frac{1}{n+1}\sum\limits_{k=1}^{n}2^k{n+1 \choose k+1} \\
 &= 2\frac{1}{n+1}\sum\limits_{k=1}^{n}(1+1)^k{n+1 \choose k+1} \\
\end{align*}
Now I think I'm in a position to use the Binomial Theorem, giving
\begin{equation*}
 2\frac{1}{n+1}\sum\limits_{k=1}^{n}\sum\limits_{i=0}^{k}{k \choose i}{n+1 \choose k+1}
\end{equation*}
I don't know if I am on the right track, but I do know that I'm stuck.  Can anyone offer any advice on how to proceed?

Observe that
$$\frac1{k+1}{n\choose k}=\frac{n!}{(k+1)\cdot k!\cdot(n-k)!}=\frac{n!}{(k+1)!\cdot(n-k)!}=\frac{1}{n+1}\cdot{n+1\choose k+1}$$
Then
\begin{align*}\frac{2^2}{2}{n\choose 1}+\frac{2^3}{3}{n\choose 2}+\ldots+\frac{2^{n+1}}{n+1}{n\choose n}&=\frac1{n+1}\left[\sum_{k=1}^n2^{k+1}{n+1\choose k+1}\right]\\
&=\frac1{n+1}\left[(1+2)^{n+1}-2{n+1\choose1}-1\right]
\end{align*}
