Book edition 7th

Author(s) Kenneth H. Rosen

Pages 808 pages

ISBN 9780073383095

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Short Answer

What is the sum of the terms of the geometric progression a+ar+...+arⁿ when $r \neq 1$ .

The sum of the terms of the geometric progression is $S_{n} = \frac{a (1 - r^{n + 1})}{1 - r}$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Geometric progression

Given:

$S_{n} = a + ar + . . . + {ar}^{n}$ with $r \neq 1$

Step 2: Finding sum of geometric progression

Let us determine ${rS}_{n}$ :

${rS}_{n} = ar + {ar}^{2} + . . . + {ar}^{n + 1}$

Next, we determine the difference between $S_{n}$ and ${rS}_{n}$

$S_{n} - r S_{n} = (a + a r + \dots + a r^{n}) - (a r + a r^{2} + \dots + a r^{n + 1}) = a + a r + \dots + a r^{n} - a r - a r^{2} - \dots - a r^{n + 1} = a - a r^{n + 1}$

Let us next use the distributive property on the obtained equality

$S_{n} - {rS}_{n} = a - {ar}^{n + 1} S_{n} (1 - r) = a (1 - r^{n + 1})$

Divide each side by 1-r

$S_{n} = \frac{a (1 - r^{n + 1})}{1 - r}$

Thus, the sum of the terms of the geometric progression is $\frac{a (1 - r^{n +!})}{1 - r}$

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Discrete Mathematics and its Applications

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Short Answer

What is the sum of the terms of the geometric progression a+ar+...+arⁿ when $r \neq 1$ .

Step by Step Solution

Step 1: Geometric progression

Step 2: Finding sum of geometric progression

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What is the sum of the terms of the geometric progression a+ar+...+arn when r≠1.

Step by Step Solution

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What is the sum of the terms of the geometric progression a+ar+...+arⁿ when $r \neq 1$ .