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: 20

Answer

2, $\frac{3}{2}$, $\frac{8}{3}$, $\frac{15}{2}$

Work Step by Step

To find the first four terms of the sequence whose general term is $a_{n}$ = $\frac{(n+1)!}{n^{2}}$, we replace n in the formula with 1,2,3, and 4. n=1, $a_{1}$ = $\frac{(1+1)!}{1^{2}}$= $\frac{2*1}{1}$ n=2, $a_{2}$ =$\frac{(2+1)!}{2^{2}}$ = $\frac{3*2*1}{2*2}$ = $\frac{3}{2}$ n=3, $a_{3}$ = $\frac{(3+1)!}{3^{2}}$ = $\frac{4*3*2*1}{3*3}$ = $\frac{8}{3}$ n=4,$a_{4}$ =$\frac{(4+1)!}{4^{2}}$ = $\frac{5*4*3*2*1}{4*4}$ = $\frac{15}{2}$ The first four terms are 2, $\frac{3}{2}$, $\frac{8}{3}$, $\frac{15}{2}$.
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