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: 17

Answer

$6+\frac{31}{35} = \frac{241}{35}$

Work Step by Step

We sum from i = low bound to i = high bound. $\frac{4(1)}{1+3} + \frac{4(2)}{2+3} + \frac{4(3)}{3+3} + \frac{4(4)}{4+3} = \frac{4}{4} + \frac{8}{5} + \frac{12}{6} + \frac{16}{7} = 1 + \frac{56}{35} + 2 + \frac{80}{35}$ $= 3+\frac{136}{35} = 6+\frac{31}{35} = \frac{241}{35}$
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