<s>[INST] 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.
 [/INST] You just about have it. From your last line, you get
$$\begin{equation}\begin{aligned}
\frac{3}{4}\left[1-\left(\frac{-1}{3}\right)^{k+1}\right] + \left(\frac{-1}{3}\right)^{k+1} & = \frac{3}{4} - \frac{3}{4}\left(\frac{-1}{3}\right)^{k+1} + \left(\frac{-1}{3}\right)^{k+1} \\
& = \frac{3}{4} + \frac{1}{4}\left(\frac{-1}{3}\right)^{k+1} \\
& = \frac{3}{4} + \frac{(-3)}{4}\left(\frac{1}{-3}\right)^{k+2} \\
& = \frac{3}{4}\left[1 - \left(\frac{-1}{3}\right)^{k+2}\right]
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
which is the RHS of what you are trying to deduce, thus completing your induction procedure.
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