Chapter 12.4, Problem 30AYU

Use mathematical induction to prove that if $r\ne 1$, then

$a\text{ + }ar\text{ + }a{r}^{\text{2}}\text{ + }\cdot \cdot \cdot \text{ + }a{r}^{n-1}\text{ = }a\frac{1-r^n}{1-r}$

To determine

To prove: The given statement $a + ar + a{r}^2 + \cdots + a{r}^{n - 1} = a\frac{1 - r^n}{1 - r}$ if $r \ne 1$ is true for all natural numbers $n$ using the Principle of Mathematical Induction.

Answer to Problem 30AYU

The statement is true for the natural number $\left($ k + 1$\right)$ terms, it is true for all natural numbers by the theorem of mathematical induction.

Explanation of Solution

Given:

Statements says the $a + ar + a{r}^2 + \cdots + a{r}^{n - 1} = a\frac{1 - r^n}{1 - r}$ if $r \ne 1$ is true for all natural number.

Formula used:

The Principle of Mathematical Induction

Suppose that the following two conditions are satisfied with regard to a statement about natural numbers:

CONDITION I: The statement is true for the natural number 1.

CONDITION II: If the statement is true for some natural number $k$, it is also true for the next natural number $k + 1$. Then the statement is true for all natural numbers.

Proof:

Consider the statement

$a + ar + a{r}^2 + \cdots + a{r}^{n - 1} = a\frac{1 - r^n}{1 - r}\text{ if }r \ne 1$    -----(1)

Step 1: Show that statement (1) is true for the natural number $n = 1$.

That is $a = a\frac{1 - r^1}{1 - r^1} = a$. Hence the statement is true for the natural number $n = 1$.

Step 2: Assume that the statement is true for some natural number $k$.

That is $a + ar + a{r}^2 + \cdots + a{r}^{k - 1} = a\frac{1 - r^k}{1 - r}\text{ if }r \ne 1$    -----(2)

Step 3: Prove that the statement is true for the next natural number $k + 1$.

That is, to prove that $a + ar + a{r}^2 + \cdots + a{r}^{k + 1 - 1} = a\frac{1 - r^{k + 1}}{1 - r}\text{ if }r \ne 1$

Consider $a + ar + a{r}^2 + \cdots + a{r}^{k + 1 - 1}$

$= a + ar + a{r}^2 + \cdots + a{r}^{k - 1} + a{r}^k$

$= a\frac{1 - r^k}{1 - r} + a{r}^k$ (From equation (2))

$= a\left(\frac{1 - r^k + r^{k.} - r.r^k}{1 - r}\right)$

$= \frac{1 - r^{k + 1}}{1 - r}\text{ if }r \ne 1$.

As the statement is true for the natural number $\left($ k + 1$\right)$ terms, it is true for all natural numbers by the theorem of mathematical induction.