Question

Prove the following by using the principle of mathematical induction for all $n\in N:P\left(n\right):a+ar+a{r}^2+......+a{r}^{n-1}=\frac{a(r^n-1)}{r-1}$

Answer 1

Let the given statement be $P\left(n\right)$ i.e.,
$P\left(n\right):a+ar+a{r}^2+......+a{r}^{n-1}=\frac{a(r^n-1)}{r-1}$
For $n=1$, we have
$P\left(1\right):a=\frac{a(r^1-1)}{(r-1)}=a$, which is true.
Let $P\left(k\right)$ be true for some $k\in N$, i.e
$a+ar+a{r}^2+.......+a{r}^{k-1}=\frac{a(r^k-1)}{r-1}...........\left(i\right)$
We shall now prove that $P(k+1)$ is true.
Consider $a+ar+a{r}^2+...........a{r}^{k-1}-1$
$=\frac{a(r^k-1)}{r-1}+a{r}^k$ [Using (i)]
$=\frac{a(r^k-1)+a{r}^{k+1}-a{r}^k}{r-1}$
$=\frac{a{r}^k-a+a{r}^{k+1}-a{r}^k}{r-1}$
$=\frac{a{r}^{k+1}-a}{r-1}$
$=\frac{a(r^{k+1}-1)}{r-1}$
Thus $P(k+1)$ is true whenever $P\left(k\right)$ is true.
Hence by the principle of mathematical induction, statement $P\left(n\right)$ is true for all natural numbers i.e., $n$