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!

OK let's tackle the one giving you grief. Just add the fractions and massage it:
$$
\begin{align*}
\frac{2^{k + 1} + (-1)^k}{3 \cdot 2^k} + \left(- \frac{1}{2}\right)^{k + 1}
  &= \frac{2^{k + 1} + (-1)^k}{3 \cdot 2^k} + \frac{(-1)^{k + 1}}{2^{k + 1}} \\
  &= \frac{2^{k + 2} + 2 \cdot (-1)^k + 3 \cdot (-1)^{k + 1}}{3 \cdot 2^{k + 1}} \\
  &= \frac{2^{k + 2} + (2 - 3) \cdot (-1)^k}{3 \cdot 2^{k + 1}} \\
  &= \frac{2^{k + 2} + (-1)^{k + 1}}{3 \cdot 2^{k + 1}}
\end{align*}
$$
