Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 11 - Section 11.1 - Sequences and Summation Notation - Exercise Set - Page 829: 13



Work Step by Step

Need to find the first four terms of $a_n=\dfrac{n^2}{n!}$ when $n=1,2,3,4$ Thus, $a_1=\dfrac{1^2}{1!}=1$ $a_2=\dfrac{2^2}{2!}=2$ $a_3=\dfrac{3^2}{3!}=\dfrac{9}{3 \cdot 2 \cdot 1}=\dfrac{3}{2}$ $a_4=\dfrac{4^2}{4!}=\dfrac{16}{4 \cdot 3 \cdot 2 \cdot 1}=\dfrac{2}{3}$ Hence, the first four terms are: $1,2,\dfrac{3}{2},\dfrac{2}{3}$
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