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 808: 80

Answer

$89964$

Work Step by Step

I know that $\sum_{k=1}^{n} (k^3)=\left[\frac{n(n+1)}{2}\right]^2.$ Hence $\sum_{k=4}^{24} (k^3)=\sum_{k=1}^{24} (k^3)-\sum_{k=1}^{3} (k^3)=\left[\frac{24(24+1)}{2}\right]^2-\left[\frac{3(3+1)}{2}\right]^2=90000-36=89964$
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