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

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.

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 $$v^7 = {n+5\choose 7}e + {n+4\choose 6}p + {n+3\choose 5}p^2 + {n+2\choose 4}p^3 + {n+1\choose 3}p^4$$

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