A proposition $P(n)$ is true for all positive integers $n$ (or natural numbers $n$) if both of the following cases are true:
- Base case: $P(1)$ is true.
- Inductive case: For some positive integer $k \ge 1$, $P(k+1)$ is true if $P(k)$ is true.
$\dsum_{k=1}^n ar^k=\dfrac{ar(r^n-1)}{r-1}$.
[Base case]
For $n=1$,
$\begin{array}{rcl}
\text{LHS} & = & ar \\
\text{RHS} & = & \dfrac{ar(r-1)}{r-1} \\
& = & ar
\end{array}$
$\therefore \text{LHS}=\text{RHS}$.
$\therefore P(1)$ is true.
[Inductive case]
Assume that $P(m)$ is true for some $m\ge 1$.
i.e. $\dsum_{k=1}^m ar^k=\dfrac{ar(r^m-1)}{r-1}$.
For $n=m+1$,
$\begin{array}{cl}
& \text{LHS} \\
= & \dsum_{k=1}^{m+1} ar^k \\
= & \dsum_{k=1}^m ar^k +ar^{m+1} \\
= & \dfrac{ar(r^m-1)}{r-1}+ar^{m+1} & \text{, by using the assumption} \\
= & \dfrac{ar(r^m-1)}{r-1}+\dfrac{ar^{m+1}(r-1)}{r-1} \\
= & \dfrac{ar^{m+1}-ar+ar^{m+2}-ar^{m+1}}{r-1} \\
= & \dfrac{ar(r^{m+1}-1)}{r-1} \\
= & \text{RHS}
\end{array}$
$\therefore P(m+1)$ is also true.
$\therefore$ by the principal of mathematical induction, $P(n)$ is true for all positive integers $n$.
