Sums of Powers of Positive Integers - Solutions to Exercises 21-22

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

Exercise 21. According to Bernoulli’s formula,

\eqalign{\int n^4 & = {1 \over {4 + 1}}n^{4 + 1} + {1 \over 2}n^4 + {4 \over 2}An^{4 - 1} + {{4 \cdot 3 \cdot 2} \over {2 \cdot 3 \cdot 4}}Bn^{4 - 3} \cr &= {1 \over 5}n^5 + {1 \over 2}n^4 + {4 \over 2} \cdot {1 \over 6}n^3 + \left( { - {1 \over {30}}} \right)n = {1 \over 5}n^5 + {1 \over 2}n^4 + {1 \over 3}n^3 - {1 \over {30}}n}

and

\eqalign{\int n^5 &= {1 \over {5 + 1}}n^{5 + 1} + {1 \over 2}n^5 + {5 \over 2}An^{5 - 1} + {{5 \cdot 4 \cdot 3} \over {2 \cdot 3 \cdot 4}}Bn^{5 - 3} \cr &= {1 \over 6}n^6 + {1 \over 2}n^5 + {5 \over 2} \cdot {1 \over 6}n^4 + {5 \over 2}\left( { - {1 \over {30}}} \right)n^2 = {1 \over 6}n^6 + {1 \over 2}n^5 + {5 \over {12}}n^4 - {1 \over {12}}n^2.}

Students may have noticed already that, since A = 1/6, the coefficient of nc – 1 is c/12. According to Bernoulli’s formula,

$${\int n^{11}} = {1\over{11+1}}n^{11+1}+ {1\over 2} n^{11}+ {{11}\over 2} A n^{11 - 1}+ {{11.\,10.\,9} \over {2.3.4}}Bn^{11 - 3}+ {{11.\,10.\,9.\,8.\,7} \over {2.3.4.5.6}}Cn^{11 - 5}$$

$$+ {{11.\,10.\,9.\,8.\,7.\,6.\,5} \over {2.3.4.5.6.7.8}}Dn^{11 - 7}+ {{11.\,10.\,9.\,8.\,7.\,6.\,5.4.3} \over {2.3.4.5.6.7.8.9.10}}En^{11 - 9}$$

$$= {1 \over {12}}n^{12} + {1 \over 2}n^{11} + {{11} \over 2} \cdot {1 \over 6}n^{10} + {{11.\,10.\,9} \over {2.3.4}}\left( { - {1 \over {30}}} \right)n^8 + {{11.\,10.\,9.\,8.\,7} \over {2.3.4.5.6}} \cdot {1 \over {42}}n^6$$

$$+ {{11.\,10.\,9}\over{2.3.4}}\left( {-{1\over{30}}}\right) n^4 + {{11}\over 2} \cdot {5\over {66}}n^2$$

$$= {1 \over {12}}n^{12} + {1 \over 2}n^{11} + {{11} \over {12}}n^{10} - {{11} \over 8}n^8 + {{11} \over 6}n^6 - {{11} \over 8}n^4 + {5 \over {12}}n^2.$$

Exercise 22. According to Bernoulli’s formula,

$${\int n^{12}={1 \over {12 + 1}}n^{12 + 1} + {1 \over 2}n^{12} + {{12} \over 2}An^{12 - 1} + {{12.\,11.\,10} \over {2.3.4}}Bn^{12 - 3} + {{12.\,11.\,10.\,9.\,8} \over {2.3.4.5.6}}Cn^{12 - 5}}$$

$$+ {{12.\,11.\,10.\,9.\,8.\,7.\,6} \over {2.3.4.5.6.7.8}}Dn^{12 - 7}+ {{12.\,11.\,10.\,9.\,8.\,7.\,6.\,5.\,4} \over {2.3.4.5.6.7.8.9.10}}En^{12 - 9}$$

$$+ {{12.\,11.\,10.\,9.\,8.\,7.\,6.\,5.\,4.\,3.\,2} \over {2.3.4.5.6.7.8.9.10.11.12}}Fn^{12 -11}$$

$$= {1 \over {13}}n^{13} + {1 \over 2}n^{12} + {{12} \over 2} \cdot {1 \over 6}n^{11} + {{12.\,11.\,10} \over {2.3.4}}\left( { - {1 \over {30}}} \right)n^9 + {{12.\,11.\,10.\,9.\,8} \over {2.3.4.5.6}} \cdot {1 \over {42}}n^7$$

$$+ {{12.\,11.\,10.\,9} \over {2.3.4.5}}\left( { - {1 \over {30}}} \right)n^5 + {{12.11} \over {2.3}} \cdot {5 \over {66}}n^3 + Fn$$

$$= {1 \over {13}}n^{13} + {1 \over 2}n^{12} + n^{11} - {{11} \over 6}n^9 + {{22} \over 7}n^7 - {{33} \over {10}}n^5 + {5 \over 3}n^3 + Fn,$$

where $${1 \over {13}} + {1 \over 2} + 1 - {{11} \over 6} + {{22} \over 7} - {{33} \over {10}} + {5 \over 3} + F = 1,$$

or

$$F = 1 - {{210 + 1365 + 2730 - 5005 + 8580 - 9009 + 4550} \over {2730}} = {{2730 - 3421} \over {2730}} = - {{691} \over {2730}},$$

whence

$$\int n^{12} = {1 \over {13}}n^{13} + {1 \over 2}n^{12} + n^{11} - {{11} \over 6}n^9 + {{22} \over 7}n^7 - {{33} \over {10}}n^5 + {5 \over 3}n^3 - {{691} \over {2730}}n.$$

Janet Beery (University of Redlands), "Sums of Powers of Positive Integers - Solutions to Exercises 21-22," Loci (July 2010), DOI:10.4169/loci003284