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# Sums of Powers of Positive Integers

## Solutions to Exercises 9-12

Exercise 9.

Figure 19.  Each side of the equation $$(4 + 1)\sum_{i = 1}^4 {i^2 } = \sum_{i = 1}^4 {i^3 } + \sum_{p = 1}^4 {\sum_{i = 1}^p {i^2 } }$$ is the area of the rectangle.

Exercise 10.   Let k = 1 in Figure 7.  Then each side of the equation $$(n + 1)\sum_{i = 1}^n i = \sum_{i = 1}^n {i^2 } + \sum_{p = 1}^n {\sum_{i = 1}^p i }$$

is the area of the rectangle.  Letting k = 1 in the equation $$(n + 1)\sum_{i = 1}^n {i^k} = \sum_{i = 1}^n {i^{k + 1}} + \sum_{p = 1}^n {\sum_{i = 1}^p {i^k }}$$

gives the equation $$(n + 1)\sum_{i = 1}^n i = \sum_{i = 1}^n {i^2 } + \sum_{p = 1}^n {\sum_{i = 1}^p i }$$ or

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

$$\displaystyle{1^2 + 2^2 + 3^2 + \cdots + n^2 =}$$ $$(n + 1)(1 + 2 + 3 + \cdots + n) - \sum_{p = 1}^n {(1 + 2 + 3 + \cdots + p)}$$ or \eqalign{1^2 + 2^2 + 3^2 + \cdots + n^2 & = (n + 1)\left( {{{n(n + 1)} \over 2}} \right) - \sum_{p = 1}^n {{{p(p + 1)} \over 2}}\cr &= (n + 1)\left( {{{n(n + 1)} \over 2}} \right) - {1 \over 2}\sum_{p = 1}^n {p^2 } - {1 \over 2}\sum_{p = 1}^n p \cr &= (n + 1)\left( {{{n(n + 1)} \over 2}} \right) - {1 \over 2}\left( {1^2 + 2^2 + 3^2 + \cdots + n^2 } \right)}

$$\displaystyle{- {1 \over 2}\left( {1 + 2 + 3 + \cdots + n} \right)}$$

or \eqalign{{3 \over 2}\left( {1^2 + 2^2 + 3^2 + \cdots + n^2 } \right) &= (n + 1)\left( {{{n(n + 1)} \over 2}} \right) - {1 \over 2}{{n(n + 1)} \over 2}\cr &= \left( {n + {1 \over 2}} \right)\left( {{{n(n + 1)} \over 2}} \right)} or $$1^2 + 2^2 + 3^2 + \cdots + n^2 = {{n(n + 1)(2n + 1)} \over 6}.$$

Exercise 11.  Let k = 2 in Figure 7.  Then each side of the equation $$(n + 1)\sum_{i = 1}^n i^2 = \sum_{i = 1}^n {i^3 } + \sum_{p = 1}^n {\sum_{i = 1}^p i^2}$$

is the area of the rectangle.  Letting k = 2 in the equation $$(n + 1)\sum_{i = 1}^n {i^k} = \sum_{i = 1}^n {i^{k + 1}} + \sum_{p = 1}^n {\sum_{i = 1}^p {i^k }}$$

gives the equation $$(n + 1)\sum_{i = 1}^n i^2 = \sum_{i = 1}^n {i^3} + \sum_{p = 1}^n {\sum_{i = 1}^p i^2 }$$ or

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

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

If we now apply the formula for the sum of the squares, the equation becomes

$$1^3 + 2^3 + 3^3 + \cdots + n^3 = (n + 1)\left( {{{n(n + 1)(2n + 1)} \over 6}} \right) - \sum_{p = 1}^n {{{p(p + 1)(2p + 1)} \over 6}}$$ $$= (n + 1)\left( {{{n(n + 1)(2n + 1)} \over 6}} \right) - {1 \over 3}\sum_{p = 1}^n {p^3 } - {1 \over 2}\sum_{p = 1}^n {p^2 } - {1 \over 6}\sum_{p = 1}^n p$$

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

$$\displaystyle{- {1 \over 2}\left( {1^2 + 2^2 + 3^2 + \cdots + n^2 } \right) - {1 \over 6}\left( {1 + 2 + 3 + \cdots + n} \right)}.$$

Collecting sums of cubes and applying the formula for the sum of the squares again, we have

$$\displaystyle{{4 \over 3}\left( 1^3 + 2^3 + 3^3 + \cdots + n^3 \right) =}$$ $$(n + 1)\left( {n(n + 1)(2n + 1) \over 6} \right) - {1 \over 2}{{n(n + 1)(2n + 1)} \over 6} - {1 \over 6}{{n(n + 1)} \over 2}$$ or \eqalign{1^3 + 2^3 + 3^3 + \cdots + n^3 &= {3 \over 4}\left( {{{n(n + 1)} \over {12}}\left( {2(n + 1)(2n + 1) - (2n + 1) - 1} \right)} \right)\cr &= {1 \over 4}\left( {{{n(n + 1)} \over 4}\left( {(2n + 1)^2 - 1} \right)} \right)\cr &= {1 \over 4}\left( {{{n(n + 1)} \over 4}\left( {4n^2 + 4n} \right)} \right)} or $$1^3 + 2^3 + 3^3 + \cdots + n^3 = \left( n(n + 1) \over 2 \right)^2.$$

Exercise 12.

\eqalign{\left({n \over 5} + {1 \over 5} \right)n\left( n + {1 \over 2} \right)\left( (n + 1)n - {1 \over 3} \right) &={1 \over 5}(n + 1)n\left({1 \over 2} \right)(2n + 1)\left((n + 1)n - {1 \over 3} \right)\cr &={n(n + 1)(2n + 1) \over 6}\left({3 \over 5}(n + 1)n - {1 \over 5} \right).}

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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.