Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 11 - Review Exercises - Page 840: 14



Work Step by Step

$\sum_{k=1}^{10} 2k^3=2\sum_{k=1}^{10} k^3$ The sum of the cubes of the first $n$ integers can be obtained by the following formula: $\frac{n^2(n+1)^2}{4}.$ Thus the sum:$\frac{10^2(10+1)^2}{4}=\frac{10^211^2}{4}=\frac{12100}{4}=3025$ Thus $2\sum_{k=1}^{10} k^3=2\cdot3025=6050$
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