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: 14



Work Step by Step

Need to find the first four terms of $a_n=\dfrac{(n+1)!}{n^2}$ when $n=1,2,3,4$ Thus, $a_1=\dfrac{(1+1)!}{1^2}=2$ $a_2=\dfrac{(2+1)!}{2^2}=\dfrac{3}{2}$ $a_3=\dfrac{(3+1)!}{3^2}=\dfrac{4 \cdot 3 \cdot 2 \cdot 1}{9}=\dfrac{8}{3}$ $a_4=\dfrac{(4+1)!}{4^2}=\dfrac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{16}=\dfrac{30}{4}$ Hence, the first four terms are: $2,\dfrac{3}{2},\dfrac{8}{3},\dfrac{30}{4}$
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