Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

APPENDIX E - Sigma Notation - E Exercises: 38

Answer

$\sum\limits_{i =1}^{n}i^{3}=[\frac{n(n+1)}{2}]^{2}$

Work Step by Step

We need to prove $\sum\limits_{i =1}^{n}i^{3}=[\frac{n(n+1)}{2}]^{2}$ Consider $n=1,k,k+1$ $\sum\limits_{i =1}^{1}i^{3}=[\frac{1(1+1)}{2}]^{2}=1$ $\sum\limits_{i =1}^{k}i^{3}=[\frac{k(k+1)}{2}]^{2}$ $\sum\limits_{i =1}^{k+1}(i)^{3}=[\frac{k(k+1)}{2}]^{2}+(k+1)^{3}$ $\sum\limits_{i =1}^{k+1}(i)^{3}=\frac{1}{4}(k+1)^{2}(k+2)^{2}$ $\sum\limits_{i =1}^{k+1}i^{3}=[\frac{(k+1)((k+1)+1)} {2}]^{2}$ Hence, $\sum\limits_{i =1}^{n}i^{3}=[\frac{n(n+1)}{2}]^{2}$
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