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Saturday, April 12, 2008

Sum of consecutive cubics

Show that ...
\center{(\sum 1 + 2 + ... + n)^2 = 1^3 + 2^3 + ... + n^3\\<br />\begin{matrix}<br />n & sum\\ <br />1 & 1 = 1\\ <br />2 & 9 = 1 + 8\\ <br />3 & 36 = 1 + 8 + 27\\ <br />4 & 100 = 1 + 8 + 27 + 64\\<br />\end{matrix}}

It is of course possible to prove this identity by induction but that doesn't make you understand why the above is true. Proofs by induction generally don't contribute to understanding a problem.

( From Getaltheorie voor beginners )
Posted by at 10:58:00 PM
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