- Membership
- Publications
- Meetings
- Competitions
- Community
- Programs
- Students
- High School Teachers
- Faculty and Departments
- Underrepresented Groups
- MAA Awards
- MAA Grants

- News
- About MAA

**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 5^{3} + 8^{3} + 11^{3} + 14^{3}.

Form a binomial having as its first term a literal quantity

Aand 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 ofAin 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:

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 ofAin the development of the above binomial. (Boyer, p. 239)

From 17^{4}, we are to subtract first 5^{4}, second 3^{4}\(\cdot\) 4, and third 54(5^{2} + 8^{2} + 11^{2} + 14^{2}) 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

Aof which the exponent is equal to the degree of the power proposed. (Boyer, p. 239)

Since the coefficient of *A*^{3} 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 5^{3} + 8^{3} + 11^{3} + 14^{3} = 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.

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

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

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

. . . . . .

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

______________________________________________________________________________

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

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

Janet Beery (University of Redlands), "Sums of Powers of Positive Integers - Solutions to Exercises 18-20," *Loci* (July 2010), DOI:10.4169/loci003284