## Calculus 8th Edition

Published by Cengage

# Appendix E - Sigma Notation - E Exercises - Page A38: 38

#### Answer

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

#### Work Step by Step

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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