Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 11 - 11.1 - Sequences and Series - 11.1 Exercises - Page 777: 26



Work Step by Step

$a_n=\frac{4n^2-n+3}{n(n-1)(n+2)}$ $a_{13}=\frac{4(13)^2-13+3}{13(13-1)(13+2)}=\frac{4(169)-10}{13(12)(15)}=\frac{676-10}{2340}=\frac{666\div18}{2340\div18}=\frac{37}{130}$
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