University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.1 - Sequences - Exercises - Page 487: 18



Work Step by Step

We can see that to the subsequent terms' numerator, $2$ is added and the denominator is derived by the formula $n(n+1)$ with $a_1=\frac{-3}{2}$, so $a_n=\frac{2n-5}{n(n+1)}$ for all $n$ natural numbers.
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