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!
 A: 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 $$\frac{2^{k+2}+(2-3)(-1)^k}{3\cdot 2^{k+1}}=\frac{2^{k+2}+(-1)^{k+1}}{3\cdot 2^{k+1}}.$$
A: Try writing your sum
$$\sum_{j = 0}^{k} \left(-\frac{1}{2}\right)^k + \left(-\frac{1}{2}\right)^{k + 1} $$
$$ =\frac{2^{k+1} + (-1)^k}{3 \times 2^k} + \frac{(-1)^{k+1}}{2^{k+1}} $$
$$ = \frac{2^{k+2} + 2(-1)^k}{3 \times 2^{k+1}} + \frac{(-1)^{k+1}}{2^{k+1}} \;$$ 
$$ = \frac{2^{k+2} + 2(-1)^k}{3 \times 2^{k+1}} + \frac{3(-1)^{k+1}}{3\times 2^{k+1}}$$
$$ = \quad\quad?$$
A: 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 $2(-1)^k+2(-1)^{k+1}=0$.
A: OK let's tackle the one giving you grief. Just add the fractions and massage it:
$$
\begin{align*}
\frac{2^{k + 1} + (-1)^k}{3 \cdot 2^k} + \left(- \frac{1}{2}\right)^{k + 1}
  &= \frac{2^{k + 1} + (-1)^k}{3 \cdot 2^k} + \frac{(-1)^{k + 1}}{2^{k + 1}} \\
  &= \frac{2^{k + 2} + 2 \cdot (-1)^k + 3 \cdot (-1)^{k + 1}}{3 \cdot 2^{k + 1}} \\
  &= \frac{2^{k + 2} + (2 - 3) \cdot (-1)^k}{3 \cdot 2^{k + 1}} \\
  &= \frac{2^{k + 2} + (-1)^{k + 1}}{3 \cdot 2^{k + 1}}
\end{align*}
$$
A: I'm not sure this is a very 'illuminating' answer, but this is clearly Geometric series:
$$
\sum_{k=0}^{n} a_k = \frac{2}{3} + \frac{1}{3}\bigg(-\frac{1}{2} \bigg)^{n}
$$
ans Wolfram Alpha agrees with me) 
A: This answer does not use induction, but perhaps another method might be helpful.
Let$$
s_n=\sum_{j=0}^n\left(-\frac12\right)^j
$$
Then,
$$
\begin{align}
s_n+\frac12s_n
&=\sum_{j=0}^n\left(-\frac12\right)^j-\sum_{j=1}^{n+1}\left(-\frac12\right)^j\\
&=1-\left(-\frac12\right)^{n+1}
\end{align}
$$
because all but the first term of the left sum and last term of the right sum cancel.
Multiplying both sides by $\frac23$ then simplifying the fraction
$$
\begin{align}
s_n
&=\frac23\left(1-\left(-\frac12\right)^{n+1}\right)\\
&=\frac{2^{n+1}-(-1)^{n+1}}{3\cdot2^n}\\
&=\frac{2^{n+1}+(-1)^n}{3\cdot2^n}\\
\end{align}
$$
