How to sum up this series? How to sum up this series :
$$2C_o + \frac{2^2}{2}C_1 +  \frac{2^3}{3}C_2 + \cdots +  \frac{2^{n+1}}{n+1}C_n$$
Any hint that will lead me to the correct solution will be highly appreciated.
EDIT: Here $C_i = ^nC_i $
 A: For an elementary method which does not use calculus:
Notice that $\displaystyle \dfrac{{n \choose k}}{k+1} = \dfrac{1}{n+1} {n+1 \choose k+1}$
Thus your sum is 
$$\sum_{k=0}^{n} \dfrac{1}{n+1} {n+1 \choose k+1} 2^{k+1} = \dfrac{\sum_{k=0}^{n+1} {n+1 \choose k} 2^k -1}{n+1} = \dfrac{3^{n+1} - 1}{n+1}$$
A: Hint: Use binomial expansion for $(1+x)^n$ and integrate once. Then choose an appropriate value for $x$.
A: Let's assume $C_i=\binom ni$. I'll give a solution that is not precalculus level. Consider first the equality
$$ (1+x)^n=C_0+xC_1+x^2C_2+\dots+x^nC_n. $$
This is the binomial theorem.
Integrate from 0 to t. On the left hand side we get $\frac{(1+t)^{n+1}-1}{n+1}$ and on the right hand side $\sum \frac1{i+1}t^{i+1}C_i$. 
Now set $t=2$, and a bit of algebra gives you the answer you want.
Pretty sure there is an elementary approach as well. 
A: REMARK $\ $ The various approaches are all equivalent. Namely, suppose that we desire to prove without calculus the identity arising from integrating the binomial formula, viz.  $$\rm (1 + x)^{n+1}\ =\ 1 + \sum_{k=1}^{n+1}\: \frac{n+1}{k+1} {n\choose k}\ x^{k+1}$$
Comparing coefficients reduces it to the identity
$$\rm \quad\quad\  {n+1 \choose k+1}\ =\ \frac{n+1}{k+1} {n\choose k} $$
which is precisely the identity employed in Moron's "calculus free" approach.
