Prove geometric sum with induction I'm not certain how to complete the proof:
Question:
Prove by induction that $1 − \frac{1}{3} + \frac{1}{9} − · · · + (\frac{−1}{3})^n = (\frac{3}{4})[1 − (\frac{−1}{3})^{n+1}]$, for every non negative integer $n$.
Solution
Base Step:
Verify that:
$LHS = 1 = (\frac{3}{4})[1 − (\frac{−1}{3})^{0+1}] = RHS$.
$RHS = (\frac{3}{4})[1 − (\frac{−1}{3})^1]\\
= (\frac{3}{4})[1 − (\frac{−1}{3})]\\
= (\frac{3}{4})(1 + \frac{1}{3})\\
= (\frac{3}{4})(\frac{4}{3}) = 1 = LHS.$
Inductive Step:
Assume that:
$1 − \frac{1}{3} + \frac{1}{9} − · · · + (\frac{−1}{3})^k = (\frac{3}{4})[1 − (\frac{−1}{3})^{k+1}]$, for some integer k.
We try to deduce that:
$1 − \frac{1}{3} + \frac{1}{9} − · · · + (\frac{−1}{3})^{k+1} = (\frac{3}{4})[1 − (\frac{−1}{3})^{k+2} ]$.
$LHS
= 1 − \frac{1}{3} + \frac{1}{9} − · · · + (\frac{−1}{3})^{k+1} \\
= 1 − \frac{1}{3} + \frac{1}{9} − · · · + (\frac{−1}{3})^{k} + (\frac{−1}{3})^{k+1}\\
= \frac{3}{4}[1-(\frac{-1}{3})^{k+1}] + (\frac{-1}{3})^{k+1}\\
... $
Lost from this point onwards.

After a small amount of work, my final solution is:
\begin{align}
LHS &= 1 − \frac{1}{3} + \frac{1}{9} − · · · + (\frac{−1}{3})^{k+1} \\
&= 1 − \frac{1}{3} + \frac{1}{9} − · · · + (\frac{−1}{3})^{k} + (\frac{−1}{3})^{k+1}\\
&= \frac{3}{4}[1-(\frac{-1}{3})^{k+1}] + (\frac{-1}{3})^{k+1}\\
&= \frac{3}{4}[1-(\frac{-1}{3})^{k+1}] + \frac{3}{4}[\frac{4}{3}(\frac{-1}{3})^{k+1}]\\
&= \frac{3}{4}[1-(\frac{-1}{3})^{k+1} + \frac{4}{3}(\frac{-1}{3})^{k+1}]\\
&= \frac{3}{4}[1+(\frac{4}{3}-1)(\frac{-1}{3})^{k+1}]\\
&= \frac{3}{4}[1 - (\frac{-1}{3})^{k+2}]\\
&= RHS
\end{align}
