Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 11 - Sequences; Induction; the Binomial Theorem - Section 11.1 Sequences - 11.1 Assess Your Understanding - Page 828: 79

Answer

$44, 000$

Work Step by Step

Recall the formula: $\displaystyle \sum_{k=1}^{n}k^{3}=\left[\dfrac{n(n+1)}{2}\right]^{2}$ We can see that for the given sequence, the index does not start at 1. So, we will rewrite the given sequence as: $\sum_{k=5}^{20} k^{3}=$ (Terms from 5 to 20) = (Terms from 1 to 20) - (Terms from 1 to 4) We rewrite the sequence as stated above and apply the sum formula: $\displaystyle \sum_{k=5}^{20} k^{3}= \sum_{k=1}^{20}k^{3}-\sum_{k=1}^{4}k^{3} \\= \displaystyle [\dfrac{20(20+1)}{2}]^{2}-[\dfrac{4(4+1)}{2}]^{2} \\=(210)^{2}-(10)^{2} \\= 44, 000$
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.