Finding a simple expression - Binomial Theorem How does one find a simple expression of the one below applying the binomial theorem:
$$\sum_{k=1}^n k \cdot 2^k{ n \choose k}$$
Edit:
$\frac{d}{dx}(x^k)=kx^{k-1}$
$(1 + x)^n = \sum_{k=0}^n { n \choose k}x^k = \sum_{k=0}^n { n \choose k}\frac{d}{dx}(x^k) = \sum_{k=0}^n { n \choose k}kx^{k-1} $
$n(1+x)^{n-1} = \sum_{k=1}^n k{ n \choose k}x^{k-1}$
Edit 2:
$nx(1+x)^{n-1} = \sum_{k=1}^n k{ n \choose k}2^{k}$
if substitute x = 2
$2n(1+2)^{n-1} = \sum_{k=1}^n k{ n \choose k}2^{k} $
Is this correct?
 A: Since this looks like homework, I won't tell you how to do it with the binomial theorem and will tell you how to do it bijectively instead.  How many ways are there to choose a committee of $k$ people out of $n$ people, split the committee into Democrats and Republicans, then elect a President from that committee, for some $k$?
On the one hand, the answer is $\displaystyle \sum_{k=1}^{n} k \cdot 2^k {n \choose k}$.  On the other hand, one can first choose the President, then split the non-Presidents into Democrats, Republicans, and people not on the committee, then choose a party for the President, which can be done in $2n \cdot 3^{n-1}$ ways.  

As for how to do it through the binomial theorem, here is a different but related application.  Recall the geometric series formula
$$\frac{1}{1 - x} = \sum_{k=0}^{\infty} x^k.$$
What happens if we take the derivative of both sides?  We get
$$\frac{1}{(1 - x)^2} = \sum_{k=1}^{\infty} kx^{k-1}.$$
Substituting $x = \frac{1}{2}$, we then conclude that
$$\sum_{k=1}^{\infty} \frac{k}{2^{k-1}} = 4.$$
This problem requires essentially the same trick.
