Chapter 9, Problem 67RE

To determine

To proof: The expression ${\displaystyle \sum _{i=0}^{n-1}a{r}^i}=\frac{a\left(1-r^n\right)}{1-r}$ with help of mathematical induction for integers $n\ge 1$ .

Explanation of Solution

Given information:

The expression ${\displaystyle \sum _{i=0}^{n-1}a{r}^i}=\frac{a\left(1-r^n\right)}{1-r}$ .

Formula used:

The steps to prove a statement by mathematical induction are as follows:

Step 1. Show that the statement is true for $n=1$ .

Step 2. Assume that the statement is true for $n=k$ .

Step 3. Prove that the statement is true for $n=k+1$ with help of step 2.

Proof:

Consider expression ${\displaystyle \sum _{i=0}^{n-1}a{r}^i}=\frac{a\left(1-r^n\right)}{1-r}$ .

Denote the expression by $S_n$ .

Recall that the steps to prove a statement by mathematical induction are as follows:

Step 1. Show that the statement is true for $n=1$ .

Step 2. Assume that the statement is true for $n=k$ .

Step 3. Prove that the statement is true for $n=k+1$ with help of step 2.

Substitute $n=1$ in the expression,

  $\begin{array}{c}a{r}^0=\frac{a\left(1-r^1\right)}{1-r}\\ a=\frac{a\left(1-r\right)}{1-r}\\ a=a\end{array}$

The above equation holds true.

Therefore, the statement is true for $n=1$ so $S_1$ is true.

Now, assume that the statement if true for $n=k$ .

  ${\displaystyle \sum _{i=0}^{k-1}a{r}^k}=\frac{a\left(1-r^k\right)}{1-r}$  ...(1)

Therefore, the statement holds true for $n=k$ so $S_k$ is true.

Now, prove it for $n=k+1$ . Write the expression for $n=k+1$ . Group the terms that involve $S_k$ .

  $\begin{array}{c}{\displaystyle \sum _{i=0}^{\left(k+1\right)-1}a{r}^i}={\displaystyle \sum _{i=0}^{k}a{r}^i}\\ ={\displaystyle \sum _{i=0}^{k-1}a{r}^i}+a{r}^k\\ =S_k+a{r}^k\end{array}$

Substitute the value of $S_k$ from equation (1),

  $\begin{array}{c}{\displaystyle \sum _{i=0}^{\left(k+1\right)-1}a{r}^i}=\frac{a\left(1-r^k\right)}{1-r}+a{r}^k\\ =\frac{a\left(1-r^k\right)+a{r}^k\left(1-r\right)}{1-r}\\ =\frac{a-a{r}^k+a{r}^k-a{r}^{k+1}}{1-r}\\ =\frac{a-a{r}^{k+1}}{1-r}\end{array}$

Take the common terms and simplify it further as,

  $\begin{array}{c}{\displaystyle \sum _{i=0}^{\left(k+1\right)-1}a{r}^i}=\frac{a-a{r}^{k+1}}{1-r}\\ =\frac{a\left(1-r^{k+1}\right)}{1-r}\end{array}$

Result obtained is that $S_k$ implies $S_{k+1}$ that is,

  ${\displaystyle \sum _{i=0}^{\left(k+1\right)-1}a{r}^i}=\frac{a\left(1-r^{k+1}\right)}{1-r}$

Hence, by the principle of mathematical induction the statement ${\displaystyle \sum _{i=0}^{n-1}a{r}^i}=\frac{a\left(1-r^n\right)}{1-r}$ is true for all integers $n\ge 1$ .