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!

You have $$\frac{2^{k+1}+(-1)^k}{3\cdot 2^k}+\left(-\frac{1}{2}\right)^{k+1}.$$Getting a common denominator and combining, we have $$\frac{2^{k+2}+2(-1)^k+3\cdot(-1)^{k+1}}{3\cdot 2^{k+1}}.$$Now we can factor out $(-1)^k$ in the part of the numerator which has it and we get $$\frac{2^{k+2}+(2-3)(-1)^k}{3\cdot 2^{k+1}}=\frac{2^{k+2}+(-1)^{k+1}}{3\cdot 2^{k+1}}.$$
