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!

We start from $\frac{2^{k+1}+(-1)^{k}}{3\cdot 2^k} +\frac{(-1)^{k+1}}{2^{k+1}}$. Multiply numerator and denominator of the first term by $2$, and numerator and denominator of the second term by $3$. Now we can add safely and get
$$\frac{2^{k+2}+2(-1)^{k}+3(-1)^{k+1}}{3\cdot 2^{k+1}}.$$
We need to verify that $2(-1)^k+3(-1)^{k+1}=(-1)^{k+1}$. This is clear, since $2(-1)^k+2(-1)^{k+1}=0$.
