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?
 A: 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*}
A: 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 $$\sum_{k=-1}^n\binom{n+1}{k+1}a^{k+1}=(a+1)^{n+1}$$
A: Note that
$$\sum_{k=1}^n2^{k+1}{n+1\choose k+1}=\sum_{m=2}^{n+1}2^m{n+1\choose m}$$
So
$$\sum_{k=1}^n\frac{2^{k+1}}{k+1}{n\choose k}=\frac{1}{n+1}\sum_{m=2}^{n+1}2^m{n+1\choose m}=\frac{1}{n+1}\left((1+2)^{n+1}-1-2(n+1)\right)=\frac{3^{n+1}-3-2n}{n+1}$$
A: The trick is to get rid of the factors $1/(k+1)$.
You do so by absorbing $k+1$ in $k!$, and
$$\frac{\binom nk}{k+1}=\frac{n!}{(k+1)k!(n-k)!}=\frac{n!}{(k+1)!(n+1-k-1)!}=\frac{\binom{n+1}{k+1}}{n+1}.$$
Now,
$$S=\sum\limits_{k=1}^{n}\frac{2^{k+1}}{k+1}{n \choose k}=\frac1{n+1}\sum\limits_{k=1}^{n}2^{k+1}{n+1 \choose k+1}.$$
The summation is $3^{n+1}$ from wich the terms $k=-1$ and $0$ have been removed. Hence
$$S=\frac{3^{n+1}-1-2(n+1)}{n+1}.$$
