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 - Page A 38: 39


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

Work Step by Step

$\sum_{i=1}^{n}(1+i)^4-i^4$ $= (2^4-1^4)+(3^4-2^4)+(4^4-3^4)+...+[(n+1)^4-n^4]$ $= (n+1)^4-1^4$ $= n^4+4n^3+6n^2+4n$ Also: $\sum_{i=1}^{n}(1+i)^4-i^4$ $= \sum_{i=1}^{n}(4i^3+6i^2+4i+1)$ $= 4\sum_{i=1}^{n}i^3+ 6\sum_{i=1}^{n}i^2+ 4\sum_{i=1}^{n}i+ \sum_{i=1}^{n}1$ $= 4S+6(\frac{2n^2+3n^2+n}{6})+ 4(\frac{n^2+n}{2})+ n$ $= 4S+(2n^3+3n^2+n)+ (2n^2+2n)+ (n)$ $= 4S+2n^3+5n^2+4n$ We can equate these two expressions: $4S+2n^3+5n^2+4n = n^4+4n^3+6n^2+4n$ $4S = n^4+2n^3+n^2$ $S = \frac{n^2(n^2+2n+1)}{4}$ $S = \frac{n^2(n+1)^2}{4}$ $S = [\frac{n(n+1)}{2}]^2$ Therefore: $\sum_{i=1}^{n}i^3 = [\frac{n(n+1)}{2}]^2$
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