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Nothing has afforded me so convincing a proof of the unity of the Deity as these purely mental conceptions of numerical and mathematical science which have been by slow degrees vouchsafed to man, and are still granted in these latter times by the Differential Calculus, now superseded by the Higher Algebra, all of which must have existed in that sublimely omniscient Mind from eternity.

Martha Somerville (ed.), Personal Recollections of Mary Somerville, Boston, 1874.

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# Sums of Powers of Positive Integers

## Solutions to Exercises 18-20

Exercise 18.  We repeat Carl Boyer’s translation of Pascal’s instructions for finding the sum of powers of an arithmetic progression, so that we may more easily apply them to our arithmetic progression.

Given any numbers whatever in arithmetic progression, each being raised to the same (integral) power, to find the sum of these powers.  (Boyer, p. 239)

Since our arithmetic progression is 5, 8, 11, 14, in what follows our first or smallest term is 5, our last term is 14, our difference is 3, and our number of terms is 4.  Our positive integer power is 3 and we are to find a formula for the sum 53 + 83 + 113 + 143.

Form a binomial having as its first term a literal quantity A and for second term the difference of the given progression.  Raise this binomial to a power of which the exponent is one more than the power proposed, noting the coefficients of the successive powers of A in the resulting development.  (Boyer, p. 239)

Our instructions are to form the binomial A + 3 and then to expand (A + 3)4.  Using the Binomial Theorem, we have

$$\displaystyle{\left( {A + 3} \right)^4 =}$$

$$A^4 + 4A^3 \cdot 3 + {4\choose 2}A^2 \cdot 3^2 + {4\choose 3}A \cdot 3^3 + 3^4 = A^4 + 12A^3 + 54A^2 + 108A + 81.$$

We are to remember the coefficients 12, 54, and 108.

Raise to the same power as the binomial the number which in the progression follows immediately after the last given term.  From the result obtained subtract the following:

1st.  The first term of the progression – that is, the smallest of the given terms, itself raised to this same power (one greater than the degree proposed).

2nd.  The difference of the progression raised to this same power, then multiplied by the number of given terms.

3rd.  The sums of similar powers of degree lower than the degree proposed, multiplied, respectively, by the coefficients of the same powers of A in the development of the above binomial.  (Boyer, p. 239)

From 174, we are to subtract first 54, second 34$$\cdot$$ 4, and third 54(52 + 82 + 112 + 142) and 108(5 + 8 + 11 + 14), to obtain the expression

$$17^4 - \left( {5^4 + 3^4\cdot 4 + 54\left( {5^2 + 8^2 + 11^2 + 14^2 } \right) + 108\left( {5 + 8 + 11 + 14} \right)} \right).$$

The remainder found is a multiple of the sum sought, and contains this sum as many times as unity is contained in the coefficient of the power of A of which the exponent is equal to the degree of the power proposed.  (Boyer, p. 239)

Since the coefficient of A3 is 12, the result (“remainder”) of the previous step is 12 times the “sum sought”; that is,

$$17^4 - \left( {5^4 + 3^4 \cdot 4 + 54\left( {5^2 + 8^2 + 11^2 + 14^2 } \right) + 108\left( {5 + 8 + 11 + 14} \right)} \right)$$ $$=12\left( {5^3 + 8^3 + 11^3 + 14^3 } \right).$$

Indeed, each side of the equation equals 56544, and 53  + 83  + 113  + 143  = 56544/12 = 4712.

Exercise 19.  Substituting m = 5, we have \eqalign{6\sum_{k = 1}^n k^5 &=(n + 1)^6 - \left( 1 + n + 15\sum_{k = 1}^n k^4 + 20\sum_{k = 1}^n k^3 + 15\sum_{k = 1}^n k^2 + 6\sum_{k = 1}^n k \right) \cr &=n^6 + 6n^5 + 15n^4 + 20n^3 + 15n^2 + 6n + 1 - 1 - n\cr & \quad - 15\left({n^5 \over 5} + {n^4 \over 2} + {n^3 \over 3} - {n \over 30} \right) - 20\left({n^4 \over 4} + {n^3 \over 2} + {n^2 \over 4} \right)\cr & \quad - 15\left( {n^3 \over 3} + {n^2 \over 2} + {n \over 6} \right) - 6\left({n^2 \over 2} + {n \over 2} \right)\cr &=n^6 + 3n^5 + {5 \over 2}n^4 - {1 \over 2}n^2.}

Therefore, on dividing by m + 1 = 6, we get $$\sum_{k=1}^n k^5 = {1\over 6}n^6+{1\over 2}n^5 +{5\over 12}n^4-{1\over 12}n^2.$$

Exercise 20.  From the Binomial Theorem or direct computation, $$(x + 1)^6 - x^6 = 6x^5 + 15x^4 + 20x^3 + 15x^2 + 6x + 1.$$

Substituting x = 1, 2, 3, …, n gives the list of n equations shown below.  Summing these n equations gives the equations shown below the line, which can be solved for the sum of the fifth powers.

$$\quad 2^6 - 1^6 = 6 \cdot 1^5 + 15 \cdot 1^4 + 20 \cdot 1^3 + 15 \cdot 1^2 + 6 \cdot 1 + 1$$

$$\quad 3^6 - 2^6 = 6 \cdot 2^5 + 15 \cdot 2^4 + 20 \cdot 2^3 + 15 \cdot 2^2 + 6 \cdot 2 + 1$$

$$\quad 4^6 - 3^6 = 6 \cdot 3^5 + 15 \cdot 3^4 + 20 \cdot 3^3 + 15 \cdot 3^2 + 6 \cdot 3 + 1$$

. . .                                                           . . .

$$(n + 1)^6 - n^6 = 6 \cdot n^5 + 15 \cdot n^4 + 20 \cdot n^3 + 15 \cdot n^2 + 6 \cdot n + 1$$

______________________________________________________________________________

$$(n + 1)^6 - 1^6 = 6\sum_{k = 1}^n {k^5 } + 15\sum_{k = 1}^n {k^4 } + 20\sum_{k = 1}^n {k^3 } + 15\sum_{k = 1}^n {k^2 } + 6\sum_{k = 1}^n k + \sum_{k = 1}^n 1$$

$${\rm{or}}\quad (n + 1)^6 - (n + 1) = 6\sum_{k = 1}^n {k^5 } + 15\sum_{k = 1}^n {k^4 } + 20\sum_{k = 1}^n {k^3 } + 15\sum_{k = 1}^n {k^2 } + 6\sum_{k = 1}^n k$$

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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.