Calculus: Early Transcendentals 8th Edition

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

Chapter 11 - Section 11.2 - Series - 11.2 Exercises - Page 717: 75

Answer

$c=-\frac{-1-\sqrt 3}{2}$

Work Step by Step

$\Sigma^{\infty}_{n=2}(1+c)^{-n}=\Sigma^{\infty}_{n=2}\frac{1}{(1+c)^{n}}=2$ $a=\frac{1}{(1+c)^{2}}$ and $r=\frac{1}{1+c}$ $\frac{a}{1-r} =2$ $\frac{\frac{1}{(1+c)^{2}}}{1-\frac{1}{1+c}}=2$ $\frac{\frac{1}{(1+c)^{2}}}{\frac{1+c}{1+c}-\frac{1}{1+c}}=2$ $\frac{\frac{1}{(1+c)^{2}}}{\frac{c}{1+c}}=2$ $\frac{1}{(1+c)^{2}} \times \frac{1+c}{c}=2$ $\frac{1}{(1+c)c}=2$ $\frac{1}{c^{2}+c}=2$ $2c^{2} +2c-1=0$ $c=\frac{-2±\sqrt (2^{2} -4(2)(-1))}{2(2)}$ $c=\frac{-2±\sqrt (4+8)}{4}$ $c=\frac{-2±\sqrt 12}{4}$ $c=\frac{-2±2\sqrt 3}{4}$ $c=\frac{-1±\sqrt 3}{2}$ For the series to converge, we need $|r| = |1+c| \lt 1$ Therefore $c=-\frac{-1-\sqrt 3}{2}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.