Prove that $\sum_{j = 0}^{n} (-\frac{1}{2})^j = \frac{2^{n+1} + (-1)^n}{3 \times 2^n}$ whenever $n$ is a nonnegative integer. I'm having a really hard time with the algebra in this proof. I'm supposed to use mathematical induction (which is simple enough), but I just don't see how to make the algebra work.
$\sum_{j = 0}^{k} (-\frac{1}{2})^k + (-\frac{1}{2})^{k + 1} = \frac{2^{k+1} + (-1)^k}{3 \times 2^k}+(-\frac{1}{2})^{k + 1}$, by adding $(-\frac{1}{2})^{k + 1}$ to both sides.
I want to show that the right side is equal to:
$\frac{2^{k+1+1} + (-1)^{k+1}}{3 \times 2^{k+1}}$
Thank you!

Try writing your sum
$$\sum_{j = 0}^{k} \left(-\frac{1}{2}\right)^k + \left(-\frac{1}{2}\right)^{k + 1} $$
$$ =\frac{2^{k+1} + (-1)^k}{3 \times 2^k} + \frac{(-1)^{k+1}}{2^{k+1}} $$
$$ = \frac{2^{k+2} + 2(-1)^k}{3 \times 2^{k+1}} + \frac{(-1)^{k+1}}{2^{k+1}} \;$$ 
$$ = \frac{2^{k+2} + 2(-1)^k}{3 \times 2^{k+1}} + \frac{3(-1)^{k+1}}{3\times 2^{k+1}}$$
$$ = \quad\quad?$$
