Chapter 12: Q 30. (page 830)

Use mathematical induction to prove that if $r \neq 1$ 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

Step 1. Given information

$r \neq 1$

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,

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

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

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

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

Short Answer

Step by step solution

Step 1. Given information

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

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Use mathematical induction to prove that if r≠1thenrole="math" localid="1647843083263" a+ar+ar2+...+arn-1=a1-rn1-r

Short Answer

Step by step solution

Step 1. Given information

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

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Theoretical and Mathematical Physics

Statistics

Calculus

Logic and Functions

Decision Maths

Study anywhere. Anytime. Across all devices.

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