Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 9 - Section 9.6 - Conic Sections in Polar Coordinates - Exercise Set - Page 1032: 61


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Work Step by Step

Given $a_j=j^2+1$, we have: $a_1=1^2+1=2$ $a_2=2^2+1=5$ $a_3=3^2+1=10$ $a_4=4^2+1=17$ $a_5=5^2+1=26$ $a_6=6^2+1=37$ The sum is $\sum_1^6 a_j=2+5+10+17+26+37=97$
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