Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

APPENDIX E - Sigma Notation - E Exercises: 47

Answer

$\sum\limits_{i =1}^{n}ar^{i-1}=a+ar+ar^{2}+....+ar^{n-1}=\frac{a(r^{n}-1)}{r-1}$

Work Step by Step

Consider $\sum\limits_{i =1}^{n}ar^{i-1}=a+ar+ar^{2}+....+ar^{n-1}$ $r\times\sum\limits_{i =1}^{n}ar^{i-1}=r\times (a+ar+ar^{2}+....+ar^{n-1})$ $r\times\sum\limits_{i =1}^{n}ar^{i-1}=ar+ar^{2}+....+ar^{n-1}+ar^{n}$ $r\sum\limits_{i =1}^{n}ar^{i-1}-\sum\limits_{i =1}^{n}ar^{i-1}=ar^{n}-a$ $(r-1)\sum\limits_{i =1}^{n}ar^{i-1}=a(r^{n}-1)$ $\sum\limits_{i =1}^{n}ar^{i-1}=\frac{a(r^{n}-1)}{r-1}$ Hence, $\sum\limits_{i =1}^{n}ar^{i-1}=a+ar+ar^{2}+....+ar^{n-1}=\frac{a(r^{n}-1)}{r-1}$
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