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A formal manipulator in mathematics often experiences the discomforting feeling that his pencil surpasses him in intelligence.

In Mathematical Circles, Boston: Prindle, Weber and Schmidt, 1969.

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# Sums of Powers of Positive Integers

## Abu Ali al-Hasan ibn al-Hasan ibn al-Haytham (965-1039), Egypt

The mathematician and scientist Abu Ali al-Hasan ibn al-Hasan ibn al-Haytham (965-1039) was born in Basra in what is now southern Iraq, but spent most of his working life in Egypt.  He is most famous for his book Optics and specifically for the reflection problem named after him, “Alhazen’s Problem” (Katz, p. 256).

For his volume computations, al-Haytham needed formulas for the sums of the first n integer cubes and the first n fourth powers (C.H. Edwards, p. 83).  He may have used a diagram like that in Figure 6 (Baron, p. 70) to describe the relationship $$(4 + 1)\sum_{i = 1}^4 i = \sum_{i = 1}^4 {i^2 } + \sum_{p = 1}^4 {\sum_{i = 1}^p i }.$$

Figure 6.  The area of this rectangle can be written in (at least) two different ways (after Baron, p. 70).

The area of the rectangle in Figure 6 can be written in (at least) two different ways.  Length times width gives area $$(4 + 1)(1 + 2 + 3 + 4),$$ whereas adding together the areas of the eight regions gives area $$(1^2 + 2^2 + 3^2 + 4^2) + (1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4)).$$

Since each of these two expressions equals the area of the same rectangle, the two expressions are equal to one another.  Writing them using summation notation, we obtain the equation $$(4 + 1)\sum_{i = 1}^4 i = \sum_{i = 1}^4 {i^2 } + \sum_{p = 1}^4 {\sum_{i = 1}^p i },$$

which can be solved for $$\sum_{i = 1}^4 {i^2 },$$

the sum of the first four squares.  Replacing 4 by n, we have the equation $$(n + 1)\sum_{i = 1}^n i = \sum_{i = 1}^n {i^2 } + \sum_{p = 1}^n {\sum_{i = 1}^p i },$$

which can be solved for $$\sum_{i = 1}^n {i^2 },$$

the sum of the first n squares.

The diagram in Figure 7 illustrates the more general relationship $$(n + 1)\sum_{i = 1}^n {i^k = } \sum_{i = 1}^n {i^{k + 1} } + \sum_{p = 1}^n {\sum_{i = 1}^p {i^k } },$$

where each side of the equation gives the area of the rectangle (C.H. Edwards, p. 84).

Figure 7.  The area of the rectangle in this figure, a generalization of Figure 6, can be written in (at least) two different ways (after C.H. Edwards, p. 84).

If we let k = 1 (see the preceding paragraph and Exercise 10) and use the formula $$1 + 2 + 3 + \cdots + n = {{n(n + 1)} \over 2},$$

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

If we let k = 2 and substitute, using the formula for the sum of the squares, we obtain $$1^3 + 2^3 + 3^3 + \cdots + n^3 = \left( {{{n(n + 1)} \over 2}} \right)^2$$

(see Exercise 11).  If we let k = 3, we obtain $$(n + 1)\sum_{i = 1}^n i^3 = \sum_{i = 1}^n {i^4 } + \sum_{p = 1}^n {\sum_{i = 1}^p {i^3 } }$$

or

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

$$(1^4 + 2^4 + 3^4 + \cdots + n^4 ) + \sum\limits_{p = 1}^n {(1^3 + 2^3 + 3^3 + \cdots + p^3 )};$$

or

$$\displaystyle{1^4 + 2^4 + 3^4 + \cdots + n^4 =}$$ $$(n + 1)(1^3 + 2^3 + 3^3 + \cdots + n^3 ) - \sum_{p = 1}^n {(1^3 + 2^3 + 3^3 + \cdots + p^3 )}.$$

If we now apply the formula for the sum of the cubes, the equation becomes $$1^4 + 2^4 + 3^4 + \cdots + n^4 = (n + 1)\left( {{{n(n + 1)} \over 2}} \right)^2 - \sum_{p = 1}^n {\left( {{{p(p + 1)} \over 2}} \right)^2 }$$

$$= (n + 1)\left( {{{n^4 } \over 4} + {{n^3 } \over 2} + {{n^2 } \over 4}} \right) - \sum_{p = 1}^n {\left( {{{p^4 } \over 4} + {{p^3 } \over 2} + {{p^2 } \over 4}} \right)}$$

$$= {{n^5 } \over 4} + {{3n^4 } \over 4} + {{3n^3 } \over 4} + {{n^2 } \over 4} - {1 \over 4}\sum_{p = 1}^n {p^4 } - {1 \over 2}\sum_{p = 1}^n {p^3 } - {1 \over 4}\sum_{p = 1}^n {p^2 }$$

$$= {{n^5 } \over 4} + {{3n^4 } \over 4} + {{3n^3 } \over 4} + {{n^2 } \over 4} - {1 \over 4}\left( {1^4 + 2^4 + \cdots + n^4 } \right) - {1 \over 2}\left( {1^3 + 2^3 + \cdots + n^3 } \right)$$

$$\displaystyle{{-\frac{1}{4}}\left(1^2 + 2^2 + \cdots + n^2 \right).}$$

Collecting sums of fourth powers and applying the formulas for sums of cubes and squares, we have

$${5 \over 4}\left( {1^4 + 2^4 + \cdots + n^4 } \right) = {{n^5 } \over 4} + {{3n^4 } \over 4} + {{3n^3 } \over 4} + {{n^2 } \over 4} - {1 \over 2}\left( {{{n^4 } \over 4} + {{n^3 } \over 2} + {{n^2 } \over 4}} \right)$$

$$\displaystyle{{-\frac{1}{4}}\left({\frac{n(n+1)(2n+1)}{6}} \right),}$$

or $$1^4 + 2^4 + \cdots + n^4 = {{n^5 } \over 5} + {{n^4 } \over 2} + {{n^3 } \over 3} - {n \over {30}}$$

for all positive integers n.  Al-Haytham actually used n = 4 in his work, then stated the general result in words (Katz, pp. 256-257).  Translating al-Haytham’s words into our symbols, we have (Katz, p. 258) $$1^4 + 2^4 + 3^4 + \cdots + n^4 = \left( {{n \over 5} + {1 \over 5}} \right)n\left( {n + {1 \over 2}} \right)\left( {(n + 1)n - {1 \over 3}} \right)$$

for all positive integers n.

The righthand side of this equation is a multiple of $${{n(n + 1)(2n + 1)} \over 6},$$

the sum of the first n positive integer squares, hence also of $${{n(n + 1)} \over 2},$$

the sum of the first n positive integers.

Exercise 9:  Draw a rectangular figure like Figure 6 that illustrates the identity $$(4 + 1)\sum\limits_{i = 1}^4 {i^2 } = \sum_{i = 1}^4 {i^3 } + \sum_{p = 1}^4 {\sum_{i = 1}^p {i^2 } }.$$

Your rectangle will have height 5 and width 30.  Your figure need not be drawn to scale!

Exercise 10:  Draw a rectangular figure like Figure 6 or 7 that illustrates the identity $$(n + 1)\sum_{i = 1}^n {i = } \sum_{i = 1}^n {i^2 } + \sum_{p = 1}^n {\sum_{i = 1}^p i }.$$

Provide the details of the computation that shows that letting k = 1 in the equation $$(n + 1)\sum_{i = 1}^n {i^k = } \sum_{i = 1}^n {i^{k + 1} } + \sum_{p = 1}^n {\sum_{i = 1}^p {i^k } }$$

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

Exercise 11:  Draw a rectangular figure like Figure 7 that illustrates the identity $$(n + 1)\sum_{i = 1}^n {i^2 = } \sum_{i = 1}^n {i^3 } + \sum_{p = 1}^n {\sum_{i = 1}^p {i^2 }}.$$

Provide the details of the computation that shows that letting k = 2 in the equation $$(n + 1)\sum_{i = 1}^n {i^k = } \sum_{i = 1}^n {i^{k + 1} } + \sum_{p = 1}^n {\sum_{i = 1}^p {i^k }}$$

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

Exercise 12:  Show how to write al-Haytham’s formula $$\left( {{n \over 5} + {1 \over 5}} \right)n\left( {n + {1 \over 2}} \right)\left( {(n + 1)n - {1 \over 3}} \right)$$

for the sum of the first n fourth powers as a rational multiple of $${{n(n + 1)(2n + 1)} \over 6}.$$

Solutions to these exercises can be found by clicking 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.