Question

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

Answer 1

Step $\left(1\right):$ Assume given statement
Let the given statement be $P\left(n\right)$, i.e,
$P\left(n\right):a+ar+a{r}^2+\cdots +a{r}^{n-1}=\frac{a(r^n-1)}{r-1}$

Step $\left(2\right):$ Checking statement $P\left(n\right)$ for $n=1$
Put $n=1$ in $P\left(n\right)$, we get
$P\left(1\right):a=\frac{a(r^1-1)}{r-1}$
$\Rightarrow a=a$
Thus $P\left(n\right)$ is true for $n=1$.

Step $\left(3\right):$ $P\left(n\right)$ for $n=K$
Put $n=K$ in $P\left(n\right)$ and assume this is true for some natural number $K$ i.e,
$P\left(K\right):a+ar+a{r}^2+\cdots +a{r}^{K-1}=\frac{a(r^K-1)}{r-1}\cdots \left(1\right)$

Step $\left(4\right):$ Checking statement $P\left(n\right)$ for $n=K+1$
Now we shall prove that $P(K+1)$ is true whenever $P\left(K\right)$ is true.
Now, we have
$a+ar+a{r}^2+\cdots +a{r}^{K-1}+a{r}^K$
$=(a+ar+a{r}^2+\cdots +a{r}^{K-1})+a{r}^K$
$=\frac{a(r^K-1)}{r-1}+a{r}^K$ (using $1$)
$=\frac{a{r}^K-a+a{r}^K(r-1)}{r-1}$
$=\frac{a{r}^K-a+a{r}^{K+1}-a{r}^K}{r-1}$
$=\frac{a{r}^{K+1}-a}{r-1}$
$=\frac{a(r^{K+1}-1)}{r-1}$
Thus, $P(K+1)$ is true whenever $P\left(K\right)$ is true.
Final answer :
Therefore, by the principle of mathematical induction statement $P\left(n\right)$ is true for all $n\in N$.