Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 12 - Sequences; Induction; the Binomial Theorem - 12.1 Sequences - 12.1 Assess Your Understanding - Page 809: 99


See proof

Work Step by Step

In Problem 98, we proved: $u_{n+1}=\dfrac{(n+1)(n+2)}{2}$ Compute $u_{n+1}+u_n$ using the formula: $u_{n+1}+u_n=\dfrac{(n+1)(n+2)}{2}+\dfrac{n(n+1)}{2}$ $=\dfrac{n+1}{2}(n+2+n)$ $=\dfrac{(n+1)(2n+2)}{2}$ $=\dfrac{2(n+1)(n+1)}{2}$ $=(n+1)^2$ We proved: $u_{n+1}+u_n=(n+1)^2$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.