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I know, indeed, and can conceive of no pursuit so antagonistic to the cultivation of the oratorical faculty ... as the study of Mathematics. An eloquent mathematician must, from the nature of things, ever remain as rare a phenomenon as a talking fish, and it is certain that the more anyone gives himself up to the study of oratorical effect the less will he find himself in a fit state to mathematicize.

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# Sums of Powers of Positive Integers

## Archimedes (287-212 BCE), Greece-Italy

The Greek mathematician, scientist, and inventor Archimedes spent most of his life in Syracuse in southern Italy.  Considered the greatest mathematician of antiquity, he is famous for a number of mathematical and scientific discoveries and also for a number of inventions, many of them machines of war used in his city’s defense against Roman armies.  Archimedes knew the Pythagorean “formula” for the sum of the first n positive integers.  He almost certainly knew also a “formula” for the sum of the squares of the first n positive integers.  His Lemma to Proposition 2 in On Conoids and Spheroids (Heath, p. 57) and his Proposition 10 in On Spirals (Heath, pp. 68-69), translated to our symbols, say that

$$(n + 1)n^2 + (1 + 2 + 3 + \cdots + n) = 3(1^2 + 2^2 + 3^2 + \cdots + n^2 )\quad (*),$$

as illustrated in Figure 3 (see also Nelsen, p. 77, and Stein, pp. 46-49).

Figure 3 shows literally that

$$3(1^2 + 2^2 + 3^2 + 4^2 ) = 5 \cdot 4^2 + (1 + 2 + 3 + 4).$$

Using the Pythagorean formula to replace $$1 + 2 + 3 + · · · ­+ n$$ by  $${{n(n + 1)} \over 2}$$ in Archimedes' equation

$$(n + 1)n^2 + (1 + 2 + 3 + \cdots + n) = 3(1^2 + 2^2 + 3^2 + \cdots + n^2 )\quad (*),$$

we have

$$(n + 1)n^2 + {\frac{n(n + 1)}{2}} = 3(1^2 + 2^2 + 3^2 + \cdots + n^2 ).$$

Solving for the sum of squares, we get

$$1^2 + 2^2 + 3^2 + \cdots + n^2 = {\frac{(n + 1)n^2}{3}} + {\frac{n(n + 1)}{6}},$$

or  $$1^2 + 2^2 + 3^2 + \cdots + n^2 = {{n(n + 1)(2n + 1)} \over 6}$$

for all positive integers n.

Since Archimedes concluded in a Corollary to his Lemma to Proposition 2 (Heath, p. 57) that

$$3(1^2 + 2^2 + 3^2 + \cdots + n^2 ) > n^3 > 3(1^2 + 2^2 + 3^2 + \cdots + (n - 1)^2 ),$$

presumably, he, too, replaced $$1 + 2 + 3 + · · · ­+ n$$ by  $${{n(n + 1)} \over 2}$$ in his equation (*), and deduced that

$$1^2 + 2^2 + 3^2 + \cdots + n^2 = {{n(n + 1)(2n + 1)} \over 6}.$$

Again, Archimedes would have described these equations and inequalities in words rather than symbols.

Exercise 2: Use cubes (or squares) to represent the triangular numbers and to illustrate the identity

$$1 + 2 + 3 + \cdots + n = {{n(n + 1)} \over 2}$$

and the identity from Exercise 1.

Exercise 3:  Use 60 or 120 cubes to illustrate the identity

$$1 + 3 + 6 + \cdots + {{n(n + 1)} \over 2} = {{n(n + 1)(n + 2)} \over 6}$$

for n = 3 or n = 4.  Hint:  One way to construct a pyramid representing the sum of the first four triangular numbers, 1, 3, 6, and 10, is to place a single cube representing 1 atop a configuration of three cubes, which, in turn, is placed atop a configuration of 6 cubes, as shown in Figure 4.  Just one more layer of 10 cubes will result in a pyramid containing 1 + 3 + 6 + 10 cubes.  How many such pyramids should you construct in order to form a 4-cube by 5-cube by 6-cube rectangular solid?  (Nelsen, p. 95)

Figure 4. Place the single cube atop the shaded portion of the layer of three cubes.  Place the resulting pyramid of 1 + 3 cubes atop the shaded portion of the layer of six cubes.  Only tops of cubes are shown.

Solutions to these exercises can be found here.

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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.