University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Section 5.2 - Sigma Notation and Limits of Finite Sums - Exercises - Page 299: 27



Work Step by Step

Use formula: $\Sigma_{k=1}^n k=\dfrac{n(n+1)}{2}$ and $\Sigma_{k=1}^n k^2=\dfrac{n(n+1)(n+2)}{6}$ Here, we have $\Sigma_{k=1}^{5} \dfrac{k^3}{225}+(\Sigma_{k=1}^{5} k)^2=\dfrac{1}{225}[\Sigma_{k=1}^{5} k^3]+(\dfrac{5 \cdot 6}{2})^3=1+3375=3376$
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