Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.2 Series - 11.2 Exercises - Page 756: 50

Answer

$16$

Work Step by Step

$a_{1} = 1$, $a_{n} = (5-n)a_{n-1}$ $a_{2} = (5-2)a_{1}$ $=3a_{1}$ $=3(1)$ $=3$ $a_{3} = (5-3)a_{2}$ $=(2)(3)$ $=6$ $a_{4}=(5-4)a_{3}$ $=6$ $a_{5}=(5-5)a_{4}=0$ $a_{6}=(5-6)a_{4}=0$ And all the other terms are zero Hence $\Sigma^{\infty}_{n=1}a_{n} = 1+ 3 +6 +6=16$
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