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?

HINT:
Like Determine: $S = \frac{1}{2}{n \choose 0} + \frac{1}{3}{n \choose 1} + \cdots + \frac{1}{n+2}{n \choose n}$,
$$\sum_{k=1}^n\dfrac{a^{k+1}}{k+1}\binom nk=\dfrac1{n+1}\sum_{k=1}^n\binom{n+1}{k+1}a^{k+1}$$
Now $$\sum_{k=-1}^n\binom{n+1}{k+1}a^{k+1}=(a+1)^{n+1}$$
