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Sums of Powers of Positive Integers

Solutions to Exercises 1-3

Exercise 1.

Figure 12.  The sum of the 4th and 5th triangular numbers is the 5th square number.

Figure 12 illustrates that the sum of the 4th and 5th triangular numbers is the 5th square number (or the square of side length 5), or that 10 + 15 = 25 = 52.  That 6 + 10 = 16 = 42 and 15 + 21 = 36 = 62 can be illustrated in similar fashion.  The general relationship can be expressed as Tn + Tn+1 = (n+1)2, where Tn is the nth triangular number, or, since Tn = n(n + 1)/2, as $${{n(n + 1)} \over 2} + {{(n + 1)(n + 2)} \over 2} = (n + 1)^2.$$

The latter equation can be checked easily using algebra.

Exercise 2.

Figure 13.  The first four triangular numbers represented using squares

Figure 14.  The identity $$1 + 2 + 3 + ... + n = \frac{n(n + 1)}{2}$$ for n = 4 (left),
and the identity from Exercise 1 for n = 3 (right)

Exercise 3.  Six pyramids constructed of 1 + 3 + 6 + 10 = 20 cubes each can be fitted together to form a rectangular solid of dimensions 4 x 5 x 6, or 120 cubes in all, as shown in Figure 15.  This construction illustrates that 6(1 + 3 + 6 + 10) = 4 x 5 x 6 or that $$1 + 3 + 6 + 10 = \frac{4\times 5\times 6}{6}.$$  In general, we have $$6\left( {1 + 3 + 6 + \cdots + {{n(n + 1)} \over 2}} \right) = n(n + 1)(n + 2)$$ or $$1 + 3 + 6 + \cdots + {{n(n + 1)} \over 2} = {{n(n + 1)(n + 2)} \over 6}.$$

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Beery, Janet, "Sums of Powers of Positive Integers," Loci (February 2009), DOI: 10.4169/loci003284

Karaji's solution

I just wonder if the Karaji Generalized his method of solutions? Is there any information about that. For instance did he found sum of fifth powers?

Al-Karaji and higher powers by Janet Beery (posted: 01/20/2010 )
My understanding is that there is no evidence that al-Karaji derived a "formula" for the sum of the fourth, fifth, or higher powers. His justification for his formula for the sum of the cubes was ingenious but al-Haytham's idea seems more readily generalizable.

On the symmetry of the resulting polynomials

== Conjecture == It would be interesting to know that * sums of odd powers can be factored as a polynomial on (n Ã¢â€¹â€¦ (n + 1) / 2) and is therefore symmetric with the center at (Ã¢Ë†â€™Ã‚Â½); and that * sums of even powers can be factored to a function like above multiplied by (2 Ã¢â€¹â€¦ n + 1) and therefore antisymmetric with the center at (Ã¢Ë†â€™Ã‚Â½). It would be interesting to know whether any rule concerning the polynomials thus obtained.

Replies:

Re: On the symmetry of the resulting polynomials by Rick Mabry (posted: 12/30/2011 )
Christopher, see the following article, especially section 3 (Faulhaber Polynomials). Although my browsers don't let me view the code you pasted, I think the article confirms some of what you're getting at (some of which might be gleaned from page 8 of the article we're viewing). A. F. Beardon, Sums of Powers of Integers, American Mathematical Monthly, Vol. 103, No. 3 (Mar., 1996), pp. 201-213.

More about symmetry: by Janet Beery (posted: 01/03/2013 )
The most succinct, beautiful, and perhaps even historically plausible explanation I've seen for the symmetry of the Faulhaber polynomials about -1/2 was in a recent article by Reuben Hersh in the College Math Journal 43:4 (Sept. 2012), pp. 322-4. His approach? Extend the polynomials to negative values of n.