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Sums of Powers of Positive Integers - Solutions to Exercises 13-14

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

Exercise 13. The constant difference table for sums of fifth powers is in Figure 20. Blue entries are obtained from knowing that the next highest entry in the fifth powers column is 0; that is, from knowing that 05 = 0. Red entries are obtained from knowing the constant difference is 120. Blue entries may be obtained in this way, too.

Difference table for sums of 5th powers

Figure 20. Constant difference table for sums of fifth powers

Substituting the values from the first row of the table in Figure 20 into Harriot’s formula for v6, we obtain the following:

$$\eqalign{720 v^6 &= 120nnnnnn + 1080nnnnn + 3000nnnn + 1800nnn - 3120nn - 2880n \cr & \quad \quad \quad \quad \quad \quad \;\;\; - 720nnnnn - 3600nnnn - 3600nnn + 3600nn + 4320n \cr & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \;\quad \;\;\; + 900nnnn + 1800nnn - 900nn - 1800n \cr & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \;\;\; + 0nnn\;\;\; - 0nn\;\;\; - 0n \cr & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \;\quad \quad \quad \quad \quad \quad \; + 360nn - 360n \cr & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \;\quad \;\quad \quad \quad \quad \quad \quad \quad \quad \quad + 720n - 720 \cr & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \;\;\; + 720\cr & = 120nnnnnn + 360nnnnn + 300nnnn + 0nnn - 60nn + 0n + 0\cr &= 120nnnnnn + 360nnnnn + 300nnnn - 60nn,}$$ or $$v^6 = {1 \over {720}}\left( {120nnnnnn + 360nnnnn + 300nnnn - 60nn} \right)$$ $$= {1 \over 6}nnnnnn + {1 \over 2}nnnnn + {5 \over {12}}nnnn - {1 \over {12}}nn,$$

or, in modern notation,

$$\sum_{k = 1}^n {k^5 } = {1 \over 6}n^6 + {1 \over 2}n^5 + {5 \over {12}}n^4 - {1 \over {12}}n^2.$$

Exercise 14.

We have

$$\eqalign{v^5 = & {{(n + 3)(n + 2)(n + 1)n(n - 1)} \over {5 \cdot 4 \cdot 3 \cdot 2}}e + {{(n + 2)(n + 1)n(n - 1)} \over {4 \cdot 3 \cdot 2}}p\cr & + {{(n + 1)n(n - 1)} \over {3 \cdot 2}} p^2 + {{n(n - 1)} \over {2}} p^3 + (n - 1) p^4 + p^5,}$$

$$\eqalign{v^6 & = {{(n + 4)(n + 3)(n + 2)(n + 1)n(n - 1)} \over {6 \cdot 5 \cdot 4 \cdot 3 \cdot 2}}e + {{(n + 3)(n + 2)(n + 1)n(n - 1)} \over {5 \cdot 4 \cdot 3 \cdot 2}}p\cr &+ {{(n + 2)(n + 1)n(n - 1)} \over {4 \cdot 3 \cdot 2}} p^2 + {{(n + 1)n(n - 1)} \over {3 \cdot 2}} p^3 + {{n(n - 1)} \over 2} p^4 + (n - 1) p^5 + {p^6},}$$ and

$$\eqalign{v^7 & = {{(n + 5)(n + 4)(n + 3)(n + 2)(n + 1)n(n - 1)} \over {7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2}}e\cr & + {{(n + 4)(n + 3)(n + 2)(n + 1)n(n - 1)} \over {6 \cdot 5 \cdot 4 \cdot 3 \cdot 2}}p + {{(n + 3)(n + 2)(n + 1)n(n - 1)} \over {5 \cdot 4 \cdot 3 \cdot 2}} p^2\cr & + {{(n + 2)(n + 1)n(n - 1)} \over {4 \cdot 3 \cdot 2}} p^3 + {{(n + 1)n(n - 1)} \over {3 \cdot 2}} p^4 + {{n(n - 1)} \over 2} p^5 + (n - 1) p^6 + p^7}$$ or $$v^7 = {n+5\choose 7}e + {n+4\choose 6}p + {n+3\choose 5}p^2 + {n+2\choose 4}p^3 + {n+1\choose 3}p^4$$

\(\displaystyle{+ {n \choose 2}p^5 + (n - 1) p^6 + p^7.}\)

Dummy View - NOT TO BE DELETED