Sums of Powers of Positive Integers - Blaise Pascal (1623-1662), France

The French mathematician and philosopher, Blaise Pascal, is best known in mathematics for his analysis of Pascal’s triangle (which he called the “arithmetical triangle”), for his pioneering work in what is now known as probability theory, and for inventing a calculating machine. Although Pascal corresponded with Fermat, the two mathematicians probably did not share with one another their work on sums of powers of integers, as Pascal’s approach to describing formulas for these sums was quite different from Fermat’s.

In his famous Traite du Triangle Arithmetique or Treatise on the Arithmetical Triangle, written in 1654 and published in 1665, Pascal described in words a general formula for the sum of powers of the first n terms of an arithmetic progression (Pascal, p. 39 of “X. Potestatum numericarum summa”), of which the sum of powers of the first n positive integers is a special case. What follows is Carl Boyer’s translation of this passage from Latin to English, alternating with our translation of Pascal’s words to modern symbols and to our specific situation of the sum of powers of the first n positive integers.

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 1, 2, 3, 4, … , n, in what follows our first or smallest term is 1, our last term is n, our difference is 1, and our number of terms is n. If our positive integer power is m, then we are to find a formula for the sum $$1^m + 2^m + 3^m + \cdots + n^m \quad {\rm or}\quad \sum_{k = 1}^n{k^m}. $$

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 + 1 and then to expand (A + 1)^m+1. Using what we now call the Binomial Theorem, we have

\(\displaystyle{\left( {A + 1} \right)^{m + 1} =}\)

$$A^{m + 1} + (m + 1)A^m + {m+1\choose 2}A^{m - 1} + {m+1\choose 3}A^{m - 2} + \cdots + (m + 1)A + 1.$$

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:

1^st. 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).

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

3^rd. 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 (n + 1)^m+1, we are to subtract first 1^m+1, second 1^m+1n = n, and third

$${m+1\choose 2}\sum_{k = 1}^n k^{m - 1} +{m+1\choose 3}\sum_{k = 1}^n k^{m - 2} + \cdots + (m + 1)\sum_{k = 1}^n k,$$

to obtain the expression

$$(n + 1)^{m + 1} - \left(1 + n + {m+1\choose 2}\sum_{k = 1}^n k^{m - 1} +{m+1\choose 3}\sum_{k = 1}^n k^{m - 2}+\cdots + (m + 1)\sum_{k = 1}^n k\right).$$

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 A^m is m + 1, the result (“remainder”) of the previous step is m + 1 times the “sum sought”; that is, $$(n + 1)^{m + 1} - \left(1 + n + {m+1\choose 2}\sum_{k = 1}^n k^{m - 1}+{m+1\choose 3}\sum_{k = 1}^n k^{m - 2}+\cdots + (m + 1)\sum_{k = 1}^n k\right)$$

\( = (m+1)\,\left( 1^m + 2^m + 3^m + \cdots + n^m\right). \)

Dividing both sides of this equation by m + 1 results in a formula for the sum \(1^m + 2^m + 3^m + \cdots + n^m\) in terms of the sums of all smaller powers of the same integers. Thus if we want to use this formula to write a polynomial expression for the sum of the fifth powers, then we must first have computed polynomial expressions for the sums of the integers, the squares, the cubes, and the fourth powers.

Just before stating the general formula or algorithm given above, Pascal illustrated both how and why it worked by using it to compute 5³ + 8³ + 11³ + 14³ (Pascal, pp. 36-38). Note that his arithmetic progression was 5, 8, 11, 14, with first term 5, last term 14, difference 3, and number of terms n = 4, and his exponent was m = 3. His final result was

17⁴

– 5 – 8 – 11 – 14 multiplied by 108

– 5² – 8² – 11² – 14² multiplied by 54

– 81 – 81 – 81 – 81

– 5⁴

equals 5³ + 8³ + 11³ + 14³ multiplied by 12 (Pascal, p. 38),

or, in more modern notation, 17⁴ – 108(5 + 8 + 11 + 14)

– 54(5² + 8² + 11² + 14²)

– 81(4)

– 5⁴

= 12(5³ + 8³ + 11³ + 14³).

The formula in the second-to-last paragraph often is written as $$(n + 1)^{m + 1} - (n + 1) = {m+1\choose 2}\sum_{k = 1}^n k^{m - 1}+{m+1\choose 3}\sum_{k = 1}^n k^{m - 2}+\cdots + (m + 1)\sum_{k = 1}^n k.$$

A.W.F. Edwards outlined Pascal’s justification for the case m = 3 as follows (Edwards, pp. 82-83). Again, in his actual justification, Pascal used the numerical example above with just four terms, but with an arithmetic progression more complicated than 1, 2, 3, 4. From the Binomial Theorem or direct computation, $$(x + 1)^4 - x^4 = 4x^3 + 6x^2 + 4x + 1.$$

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 cubes:

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

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

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

$$\quad \quad \vdots \quad \quad \quad \vdots \quad \quad \quad$$

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

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

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

This argument can be generalized from m = 3 to any positive integer exponent m greater than or equal to 2.

Exercise 18: Apply Pascal’s written instructions for computing the sum of the powers of the first n terms in an arithmetic progression to the power 3 and the arithmetic progression 5, 8, 11, 14. That is, show how to apply Pascal’s instructions to compute 5³ + 8³ + 11³ + 14³.

Exercise 19: Use Pascal’s formula

$$\left( {n + 1} \right)^{m + 1} - \left( {1 + n + {m+1\choose 2}\sum_{k = 1}^n {k^{m - 1} } + {m+1\choose 3}\sum_{k = 1}^n {k^{m - 2} } + \cdots + (m + 1)\sum_{k = 1}^n k } \right)$$

\( = (m+1)\,\left( 1^m + 2^m + 3^m + \cdots + n^m\right) \)

with m = 5 to derive a formula for the sum of the first n fifth powers, $$\sum_{k = 1}^n {k^5 }.$$

Exercise 20: Give A.W.F. Edwards’ justification for Pascal’s case m = 5. Begin by expanding $$(x + 1)^6 - x^6.$$

Solutions to these exercises can be found by clicking here.