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The most painful thing about mathematics is how far away you are from being able to use it after you have learned it.

In J. R. Newman (ed.) The World of Mathematics, New York: Simon and Schuster, 1956.

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# Sums of Powers of Positive Integers

## Solutions to Exercises 15-17

Exercise 15.  Beginning with our simplification of Fermat’s formula, we have

$${{n(n + 1)(2n + 1)(3n^2 + 3n - 1)} \over {30}} = {{n(n + 1)(2n + 1)} \over 6} \cdot {{(3n^2 + 3n - 1)} \over 5}.$$

Beginning with Fermat’s formula, we have

\eqalign{ & {1 \over 5}\left( {(4n + 2)\,\left( {{{n(n + 1)} \over 2}} \right)^2 - {{n(n + 1)(2n + 1)} \over 6}} \right) \cr & = {2 \over 5}(2n + 1)\,\left( {{{n(n + 1)} \over 2}} \right)^2 - {1 \over 5} \cdot {{n(n + 1)(2n + 1)} \over 6}\cr & = {{n(n + 1)(2n + 1)} \over 6}\left( {6 \cdot {2 \over 5} \cdot {{n(n + 1)} \over 4} - {1 \over 5}} \right)\cr & = {{n(n + 1)(2n + 1)} \over 6} \cdot {{(3n^2 + 3n - 1)} \over 5}. \cr}

Exercise 16.  If Tn6  is the nth sixth order triangular number, then, by definition, Tn6 is the sum of the first n triangulopyramidal numbers; that is, $$T_n^6 = \sum_{k = 1}^n {TP_k } \quad {\rm or} \quad T_n^6 = \sum_{k = 1}^n {T_k^5 }.$$

Perhaps Fermat would have called Tn6 a “pyramidopyramidal number” and written, “The last side multiplied by the triangulopyramidal of the next greater side makes six times the pyramidopyramidal.”  This translates to $$nTP_{n + 1} = 6T_n^6 \quad {\rm or} \quad n \cdot {{(n + 1)(n + 2)(n + 3)(n + 4)(n + 5)} \over {2 \cdot 3 \cdot 4 \cdot 5}} = 6T_n^6 \quad {\rm or}$$

$$T_n^6 = {{n(n + 1)(n + 2)(n + 3)(n + 4)(n + 5)} \over {2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}}.$$

Exercise 17.  Using our formulas for TPn and Tn (from Exercise 16), the equation $$T_n^6 = \sum_{k = 1}^n {TP_k }$$

from Exercise 16 becomes $${{n(n + 1)(n + 2)(n + 3)(n + 4)(n + 5)} \over {2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}} = \sum_{k = 1}^n {{{k(k + 1)(k + 2)(k + 3)(k + 4)} \over {2 \cdot 3 \cdot 4 \cdot 5}}}\quad {\rm or}$$

$${{n(n + 1)(n + 2)(n + 3)(n + 4)(n + 5)} \over {2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}} = \sum_{k = 1}^n {{{k^5 + 10k^4 + 35k^3 + 50k^2 + 24k} \over {120}}}.$$

Solving for the sum of the fifth powers, we obtain

$$\displaystyle{{1 \over {120}}\sum_{k = 1}^n {k^5 } =}$$

$${{n(n + 1)(n + 2)(n + 3)(n + 4)(n + 5)} \over {720}} - {1 \over {12}}\sum_{k = 1}^n {k^4 } - {7 \over {24}}\sum_{k = 1}^n {k^3 } - {5 \over {12}}\sum_{k = 1}^n {k^2 } - {1 \over 5}\sum_{k = 1}^n k.$$

If we now substitute our formulas for the sums of the first n positive integers and of their squares, cubes, and fourth powers, multiply both sides of the equation by 120, and simplify, we will obtain a formula for the sum of the first n fifth powers as follows:

\eqalign{\sum_{k = 1}^n {k^5 } & = 120 \cdot {{n^6 + 15n^5 + 85n^4 + 225n^3 + 274n^2 + 120n} \over {720}}\cr & \quad - {{120} \over {12}}\left( {{{n^5 } \over 5} + {{n^4 } \over 2} + {{n^3 } \over 3} - {n \over {30}}} \right) - {{7 \cdot 120} \over {24}}\left( {{{n^4 } \over 4} + {{n^3 } \over 2} + {{n^2 } \over 4}} \right)\cr & \quad - {{5 \cdot 120} \over {12}}\left( {{{n^3 } \over 3} + {{n^2 } \over 2} + {n \over 6}} \right) - {{120} \over 5}\left( {{{n^2 } \over 2} + {n \over 2}} \right) \cr &= {1 \over 6}n^6 + {1 \over 2}n^5 + {5 \over {12}}n^4 + 0n^3 - {1 \over {12}}n^2 + 0n\cr & = {1 \over 6}n^6 + {1 \over 2}n^5 + {5 \over {12}}n^4 - {1 \over {12}}n^2.}

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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.