Search

## Search Loci: Convergence:

Keyword

Random Quotation

Progress imposes not only new possibilities for the future but new restrictions.

The Human Use of Human Beings.

See more quotations

# Sums of Powers of Positive Integers

## Aryabhata (b. 476), northern India

The northern Indian mathematician and astronomer, Aryabhata, born in 476, wrote one of the earliest known Indian mathematics and astronomy books, the Aryabhatiya, in 499 (Katz, p. 212).  In Section II, Stanza 22, of the Aryabhatiya, he wrote:

The sixth part of the product of three quantities consisting of the number of terms, the number of terms plus one, and twice the number of terms plus one is the sum of the squares.  The square of the sum of the (original) series is the sum of the cubes. (Katz, 217)

The first sentence gives the formula from pages 1 and 3, above, for the sum of the squares; the second says, in our notation, that

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

If we replace $$1 + 2 + 3 + · · · ­+ n$$ by  $${{n(n + 1)} \over 2}$$, we obtain

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

It seems likely that mathematicians discovered this formula for the sum of the cubes early on by taking note of examples, such as

$$1^3 = 1 = 1^2,$$

$$1^3 + 2^3 = 9 = 3^2 = (1 + 2)^2,$$

$$1^3 + 2^3 + 3^3 = 36 = 6^2 = (1 + 2 + 3)^2,$$ and

$$1^3 + 2^3 + 3^3 + 4^3 = 100 = 10^2 = (1 + 2 + 3 + 4)^2.$$

We would then generalize to

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

and hence to

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

for all positive integers n.

Exercise 4: Use 84 or 180 cubes to demonstrate the identity $$1^2 + 2^2 + 3^2 + \cdots + n^2 = {{n(n + 1)(2n + 1)} \over 6}$$

for n = 3 or n = 4, as described by the Indian mathematician Nilakantha (c. 1445-1545).  A member of the famous Kerala school of mathematics and astronomy in southern India, Nilakantha wrote a commentary on the Aryabhatiya in which he gave the following proof of the identity above.  He explained that six copies of the sum $$1^2 + 2^2 + 3^2 + \cdots + n^2$$

form an n by (n + 1) by (2n + 1) rectangular solid as follows.  Start on the outside with a floor and three walls consisting of 6n2 cubes, then work from the outside inwards, lining the inside of the existing shell with a floor and three walls consisting of 6(n – 1)2 cubes, then 6(n – 2)2 cubes, . . . , then 6  22  cubes, and finally 6 12 cubes.  The outside shell has height n, depth n + 1, and length 2n + 1, with one long side open and the top open.  If n = 3, this shell has dimensions 3, 4, and 7, and consists of 6 32  = 54 cubes.  Construct this shell, then, inside of it, a shell having dimensions 2, 3, and 5 and consisting of 6 22  = 24 cubes, and then, inside of the double shell, a 1 x 2 x 3  layer consisting of 6 12  = 6 cubes.  The result should be a 3 x 4 x 7 rectangular solid.  This construction illustrates that

$$6 \cdot 1^2 + 6 \cdot 2^2 + 6 \cdot 3^2 = 3 \cdot 4 \cdot 7 \quad {\rm or} \quad 1^2 + 2^2 + 3^2 = {{3 \cdot 4 \cdot 7} \over 6},$$

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

for any positive integer n.

That the outside shell contains 6n2 cubes can be seen by noting that the shell can be built using an n x 2n slab for the floor, another n x 2n slab for the back wall, and n x (n - 1) and n x (n + 1) slabs for the side walls.  Remember that the outside shell has height n, depth n + 1, and length 2n + 1, with one long side open and the top open, and that all inside shells should be constructed with dimensions like those of the outside shell.

(See Katz, Victor, 2009, A History of Mathematics:  An Introduction (3rd ed.), Boston: Addison-Wesley, pp. 251-252, or Katz, Victor (editor), 2007, The Mathematics of Egypt, Mesopotamia, China, India, and Islam:  A Sourcebook, Princeton University Press, pp. 493-496.)

Exercise 5:  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, as Nilakantha may have done.  Our strategy will be to demonstrate that six copies of the sum $$1 + 3 + 6 + \cdots + {{n(n + 1)} \over 2}$$ form an n by (n + 1) by (n + 2) rectangular solid, as follows.  Start on the outside with an (n + 1) by (n + 2) floor and two adjoining walls of height n, consisting of $$6 \cdot \frac{n(n+1)}{2}$$ cubes all together, then work from the outside inwards, lining the inside of the existing shell with a floor and two adjoining walls consisting of $$6\cdot \frac{(n-1)n}{2}$$ cubes, . . . , 6 3 cubes, and finally 6 1  cubes.  For n = 3, the 3 x 4 x 5 outside shell consists of $$6\cdot \frac{3\cdot4}{2} = 36$$ cubes, the 2 x 3 x 4 shell inside of this one $$6 \cdot {\frac{2\cdot3}{2}} = 18$$ cubes, and the 1 x 2 x 3 inner shell or layer $$6\cdot \frac{1\cdot2}{2} = 6$$ cubes.  This construction illustrates that

$$6 (1 + 3 + 6) = {{3\cdot 4\cdot 5}}$$

or

$$1 + 3 + 6 = {{3\cdot 4\cdot 5} \over 6}$$

or, more generally,

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

for any integer n.

That the outside shell contains $$6\cdot\frac{n(n+1)}{2}=3n(n+1)$$ cubes can be seen by noting that the shell can be built using an n x (n + 1) slab for the floor and two adjoining wall slabs with dimensions n x n and n x (n + 2).  This will result in an outside shell with height n, length n + 1, and width n + 2.  The inside shells should be constructed with dimensions like those of the outside shell.

(See Joseph, George Gheverghese, 2000, The Crest of the Peacock, Princeton University Press, pp. 295-296, and the references given for Exercise 4.)

Exercise 6:  How might early mathematicians have discovered the identity from Exercise 1?  Provide at least four examples and write the identity in terms of n, where n is a positive integer.

For solutions to these exercises, click here.

Pages: | 1 |  2 |  3 |  4 |  5 |  6 |  7 |  8 |  9 |  10 |  11 |  12 |  13 |  14 |  15 |  16 |  17 |  18 |  19 |  20 |  21 |

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.