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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 12: Q 30. (page 830)

Use mathematical induction to prove that if $r 鈮� 1$ then
role="math" localid="1647843083263" $a + a r + a r^{2} + . . . + a r^{n - 1} = a (\frac{1 - r^{n}}{1 - r})$

Short Answer

Expert verified

The given statement is shown.

Step by step solution

01

Step 1. Given information

$r 鈮� 1$

02

Step 2. Prove that a+ar+ar2+...+arn-1=a1-rn1-r.

First we consider the above statement for $n = 1$

Since $a = a \frac{1 - r}{1 - r}$ , we say that above statement for $n = 1$

Hence condition holds true.

Now we assume that above statement is true for some $k$

So we have,

$a + a r + a r^{2} + 鈰� + a r^{k - 1} = a \frac{1 - r^{k}}{1 - r}$

Now let us consider the sum of first $k + 1$ terms

$\begin{array}{rcl} a + a r + a r^{2} + 鈰� + a r^{k - 1} + a r^{k + 1 - 1} & = & (a + a r + a r^{2} + 鈰� + a r^{k - 1}) + a r^{k} \\ = & a \frac{1 - r^{k}}{1 - r} + a r^{k} \\ = & a \frac{1 - r^{k + 1}}{1 - r} \end{array}$

This shows that the statement is for $n = k + 1$

Hence condition holds true.

As a result, we say that given statement is true for any natural number $n$

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