Chapter 9.5, Problem 21E

Solution

To prove : Using mathematical induction to prove that statement “the sum of the first $n$ terms of a geometric sequence with first term $a_1$ and common ratio $r\ne 1$ is $a_1\left(1-r^n\right)/\left(1-r\right)$ ” holds true for all positive integers.

Given information :

The statement “the sum of the first $n$ terms of a geometric sequence with first term $a_1$ and common ratio $r\ne 1$ is $a_1\left(1-r^n\right)/\left(1-r\right)$ ”.

Formula used :

Follow the steps to prove that the statement is true for any positive integer.

Anchor step: Prove that any statement $P_n$ is true for $n=1$ .

Inductive hypothesis: Assume $P_n$ is true for $n=k$ .

Inductive Step: Prove that $P_n$ is true for $n=k+1$ .

Proof :

Conjecture:

Take $P_n$ to be statement $P_n:a_1+a_{1}r+a_1{r}^2+\dots +a_1{r}^{n-1}=\frac{a_1\left(1-r^n\right)}{1-r}$ .

Anchor step:

Prove that $P_n$ is true for $n=1$ .

For $n=1$ ,

  $\begin{array}{l}{a}_1\stackrel{?}{=}\frac{a_1\left(1-r^1\right)}{1-r}\\ a_1=a_1\end{array}$

Inductive hypothesis:

Assume $P_n$ is true for $n=k$ . Replace $n$ with $k$ in $P_n$ .

  $P_k:a_1+a_{1}r+a_1{r}^2+\dots +a_1{r}^{k-1}=\frac{a_1\left(1-r^k\right)}{1-r}$

Inductive Step:

Prove that $P_n$ is true for $n=k+1$ . This means it is needed to be proved that $P_{k+1}$ must be true.

Proceed with the inductive hypothesis.

  $P_k:a_1+a_{1}r+a_1{r}^2+\dots +a_1{r}^{k-1}=\frac{a_1\left(1-r^k\right)}{1-r}$

Add $a_1{r}^k$ on both sides.

  $\begin{array}{c}{a}_1+a_{1}r+a_1{r}^2+\dots +a_1{r}^{k-1}+a_1{r}^k=\frac{a_1\left(1-r^k\right)}{1-r}+a_1{r}^k\\ =\frac{a_1\left(1-r^k\right)+\left(1-r\right)a_1{r}^k}{1-r}\\ =\frac{a_1\left(1-r^k+\left(1-r\right)r^k\right)}{1-r}\\ =\frac{a_1\left(1-r^k+r^k-r^{k+1}\right)}{1-r}\end{array}$

  $a_1+a_{1}r+a_1{r}^2+\dots +a_1{r}^{k-1}+a_1{r}^k=\frac{a_1\left(1-r^{k+1}\right)}{1-r}$

The statement is same as the statement $P_{k+1}$ , so $P_{k+1}$ is true.

So, if $P_n$ is true for $n=k$ , then it is also true for $n=k+1$ .

Conclusion:

As $P_n$ is true for $n=1$ and $P_k$ further implies $P_{k+1}$ , this means $P_n$ is true for $n=2$ , $n=3$ and so on.

By mathematical induction, “the sum of the first $n$ terms of a geometric sequence with first term $a_1$ and common ratio $r\ne 1$ is $a_1\left(1-r^n\right)/\left(1-r\right)$ ” holds true for all positive integers.