Answer
See proof below.
Work Step by Step
$1. \ (\mathrm{b}\mathrm{a}\mathrm{s}\mathrm{e})$ Check that the formula holds for $n=2$.
We are given by the problem text that
$LHS=\displaystyle \frac{2}{3}$
$RHS=1-\displaystyle \frac{1}{3^{1}}=\frac{3-1}{3}=\frac{2}{3}=LHS \qquad $
Yes, this is true when $n=$1.
$2. \ (\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{p})$ Prove that IF the formula holds for any positive integer $n=k$, THEN it also holds for the next integer, $n=k+1$.
Let us assume that for $n=k$,
$\displaystyle \frac{2}{3}+\frac{2}{3^{2}}+...+\frac{2}{3^{k}}=1-\frac{1}{3^{k}}$
(we call this the step hypothesis)
For $n=k+1$,
the LHS equals $\displaystyle \frac{2}{3}+\frac{2}{3^{2}}+...+\frac{2}{3^{k}}+\frac{2}{3^{k+1}},\quad $
which we rewrite as
$LHS= (\displaystyle \frac{2}{3}+\frac{2}{3^{2}}+...+\frac{2}{3^{k}})+\frac{2}{3^{k+1}}$
apply the hypothesis on the first set of parentheses
$=1-\displaystyle \frac{1}{3^{k}}+\frac{2}{3^{k+1}}$
$=1-(\displaystyle \frac{1}{3^{k}}-\frac{2}{3^{k+1}})$
$=1-\displaystyle \frac{3-2}{3^{k+1}}$
$=1-\displaystyle \frac{1}{3^{k+1}}=RHS$ for $n=k+1$
which makes the formula true for $n=k+1.$
By mathematical induction, the formula is true for any integer $n$.
