Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 11 - Section 11.3 - Series - Exercise Set: 18

Answer

$1+\frac{41}{330} = \frac{371}{330}$

Work Step by Step

We sum from i = low bound to i = high bound. $\frac{6-2}{6+2} + \frac{6-3}{6+3} + \frac{6-4}{6+4} + \frac{6-5}{6+5} = \frac{4}{8} + \frac{3}{9} + \frac{2}{10} + \frac{1}{11}$ $=\frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{11} = \frac{341}{330} + \frac{30}{330} = \frac{371}{330} = 1+\frac{41}{330}$
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