<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] So far so good. Let $q = -\frac{1}{3}$. Thus, following your lines, you have shown
$$ \sum_{j=1}^{k+1} q^j = \sum_{j=1}^{k} q^j + q^{k+1} = \frac{1-q^{k+1}}{1-q} + q^{k+1}.$$
To finish the proof, note that
$$q^{k+1} = \frac{1-q}{1-q} q^{k+1}.$$
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