Using the method of induction to show How can I use the method of induction to show for any real number $r$ does not equal $1$ and any positive integer $n$
show that
$$1+r+r^2+\cdots+r^n=\frac{1r^{n+1}-1}{r-1}$$
for $n=1$ it seems to work
$$1+r+\cdots+r^n=(1+r)$$
then $\dfrac{r^2-1}{r-1}$ for the right side
$$\frac{(r-1)(r+1)}{r-1}=1+r$$
Thus the formula is true for $n=1$ then assume the formula is true $n=k$ ,$k$ is an integer greater than $1$.
$$1+r+\cdots+r^k=\frac{r^{k+1}-1}{r-1}$$
then
$$1+r+\cdots+r^k+r^{k+1}$$
then
$$\frac{1r^{k+1}-1}{r-1}  + \frac{r^{k+1}}{1}$$
$$\frac{r^{k+1}+r-1}{r-1}=1+r+\cdots+r^k+r^{k+1}$$
would this be correct method of induction?

Assuming $r\ne1$
Let $\displaystyle F(n): 1+r+r^2+....+r^n=\frac{r^{n+1}-1}{r-1}$ holds true for $n=m $
$\displaystyle\implies 1+r+r^2+....+r^m=\frac{r^{m+1}-1}{r-1}$
$\displaystyle \implies 1+r+r^2+....+r^m+r^{m+1}=\frac{r^{m+1}-1}{r-1}+r^{m+1}=\frac{r^{m+1}-1+r^{m+2}-r^{m+1}}{r-1}$
$\displaystyle \implies 1+r+r^2+....+r^m+r^{m+1}=\frac{r^{(m+1)+1}-1}{r-1}$
So, $f(n)$ will hold true for $n=m+1$ if it holds true for $n=m$
Establish the base case $n=1$
