Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Section 10.1 - Sequences and Summation Notation - Exercise Set: 19

Answer

1,2, $\frac{3}{2}$ , $\frac{2}{3}$

Work Step by Step

To find the first four terms of the sequence whose general term is $a_{n}$ =$\frac{n^{2}}{n!}$, we replace n in the formula with 1,2,3, and 4. n=1, $a_{1}$ = $\frac{1^{2}}{1!}$ = 1 n=2, $a_{2}$ =$\frac{2^{2}}{2!}$ = $\frac{2*2}{2*1}$ = 2 n=3, $a_{3}$ =$\frac{3^{2}}{3!}$ = $\frac{3*3}{3*2*1}$ = $\frac{3}{2}$ n=4,$a_{4}$ =$\frac{4^{2}}{4!}$ = $\frac{4*4}{4*3*2*1}$ = $\frac{2}{3}$ The first four terms are 1,2, $\frac{3}{2}$ , $\frac{2}{3}$.
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