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Sums of Powers of Positive Integers - Solutions to Exercises 15-17

Author(s): 
Janet Beery (University of Redlands)

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 Tn6 (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:

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

Janet Beery (University of Redlands), "Sums of Powers of Positive Integers - Solutions to Exercises 15-17," Loci (July 2010), DOI:10.4169/loci003284

Dummy View - NOT TO BE DELETED

Sums of Powers of Positive Integers