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Nicomachus’ theorem∗ PrimeFan† 2013-03-22 1:08:25 Theorem (Nicomachus). The sum of the cubes of the first n integers is equal to the square of the nth triangular number. To put it algebraically, n X 3 i = i=1 n2 + n 2 2 . Proof. There are several formulas for the triangular numbers. Gauss figured out that to compute n X i, i=1 one can, instead of summing the numbers one by one, pair up the numbers thus: 1 + n, 2 + (n − 1), 3 + (n − 2), etc., and each of these sums has the same result, namely, n + 1. Since there are n of these sums, carrying this all the way through to the end, we are in effect squaring n + 1, which is (n + 1)2 = (n + 1)(n + 1) = n2 + n. But this is redundant, since it includes both 1 + n and n + 1, both 2 + (n − 1) and (n − 1) + 2, etc., in effect, each of these twice. Therefore, n n2 + n X = i. 2 i=1 As Sir Charles Wheatstone proved, we can rewrite i3 as i X (2ij + i). j=1 That sum can always be rewritten as a sum of odd terms, namely i X (i2 + i + 2k). k=1 ∗ hNicomachusTheoremi created: h2013-03-2i by: hPrimeFani version: h40668i Privacy setting: h1i hTheoremi h11A25i † This text is available under the Creative Commons Attribution/Share-Alike License 3.0. You can reuse this document or portions thereof only if you do so under terms that are compatible with the CC-BY-SA license. 1 Thus, the sum of the first n cubes is in fact also n−1 X (2i + 1). i=0 The sum of the first n − 1 odd numbers is n2 , and therefore n X n X 3 i = i=1 !2 i , i=1 as the theorem states. For example, the sum of the first four cubes is 1 + 9 + 27 + 64 = 100. This is also equal to 1 + 3 + 5 + 7 + 9 + 11 + 13 + 17 + 19 = 100. The square root of 100 is 10, the fourth triangular number, and indeed 10 = 1 + 2 + 3 + 4. 2