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.