Answer
See the proof below.
Work Step by Step
$1. \ (\mathrm{b}\mathrm{a}\mathrm{s}\mathrm{e})$ Check that the formula holds for n = 1.
LHS$=1+r^{1},\qquad $
RHS$=\displaystyle \frac{1-r^{1+1}}{1+r}=\frac{1-r^{2}}{1-r}=\frac{(1-r)(1+r)}{1-r}$
We recognized a difference of squares in the numerator.
We now cancel the common factor,
$RHS=1+r=LHS\qquad $... yes, it 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 $n=k$,
$1+r^{1}+r^{2}+...+r^{k}=\displaystyle \frac{1-r^{k+1}}{1-r}$
(we call this the step hypothesis)
For $n=k+1$,
the LHS equals $1+r^{1}+r^{2}+...+r^{k}+r^{k+1}$
which we rewrite as
$LHS=(1+r^{1}+r^{2}+...+r^{k})+r^{k+1}=\qquad $
use the hypothesis:
$=\displaystyle \frac{1-r^{k+1}}{1-r}+r^{k+1}$
$=\displaystyle \frac{1-r^{k+1}+r^{k+1}(1-r)}{1-r}$
$=\displaystyle \frac{1-r^{k+1}+r^{k+1}-r^{k+2}}{1-r}$
$=\displaystyle \frac{1-r^{k+2}}{1-r}$
$=\displaystyle \frac{1-r^{(k+1)+1}}{1-r}=RHS,\quad $
which makes the formula true for $n=k+1.$
By mathematical induction, the formula is true for any integer $n.$
