Abd Shukor's Formula

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The Abd Shukor's Formula for Sums of Power for an arbitrary arithmetic progression, named after Abd Shukor.. The formula is given as follows

The sum of the p-th power of the first nth term of an arithmetic progression is given as follows

$\sum _{i=1}^{n}x_{i}^{p}=x_{1}^{p}+x_{2}^{p}+x_{3}^{p}+\cdots +x_{n}^{p}$

The sum can be calculated by using the closed form general equation given below

$\sum _{i=1}^{n}{x_{i}}^{p}=\sum _{m=0}^{u}\phi _{m}s^{2m}{\frac {[\sum _{i=1}^{n}x_{i}]^{p-2m}}{n^{p-(2m+1)}}}$

where: $p-(2m+1)\geq 1$ if p is even and $p-(2m+1)\leq 1$ if p is odd, $\phi _{m}$ is a coefficient,$\sum _{i=1}^{n}{x_{i}}$ is an arithmetic sum of the nth term, $u={\frac {p-1}{2}}$ for odd p and $u={\frac {p}{2}}$ for even p and s is the difference between consecutive terms (i.e $s=x_{i+1}-x_{i}$).

The coefficient $\phi _{m}$ is given as follows

$\phi _{m}={\frac {1}{2^{2m}}}{\frac {1}{(2m+1)}}{\binom {p}{2m}}{\frac {1}{2^{2m}}}{\frac {1}{(2m+1)}}{\binom {p}{2m}}[n^{2m}-2\sum _{t=1}^{m}[(2t+1)(2^{2t-1}-1){\binom {m}{t}}B_{t}n^{2(m-t)}{\frac {\prod _{j=0}^{t-1}(1+2(m-j))}{\prod _{j=0}^{t}(1+2(t-j))}}]$

By expanding the equation for the first 10th power equations yields

$\sum _{i=1}^{n}x_{i}=\phi _{0}[\sum _{i=1}^{n}x_{i}]$

$\sum _{i=1}^{n}{x_{i}}^{2}=\phi _{0}{\frac {[\sum _{i=1}^{n}x_{i}]^{2}}{n}}+\phi _{1}ns^{2}$

$\sum _{i=1}^{n}{x_{i}}^{3}=\phi _{0}{\frac {[\sum _{i=1}^{n}x_{i}]^{3}}{n^{2}}}+\phi _{1}[\sum _{i=1}^{n}x_{i}]s^{2}$

$\sum _{i=1}^{n}{x_{i}}^{4}=\phi _{0}{\frac {[\sum _{i=1}^{n}x_{i}]^{4}}{n^{3}}}+\phi _{1}{\frac {[\sum _{i=1}^{n}x_{i}]^{2}s^{2}}{n}}+\phi _{2}ns^{4}$

$\sum _{i=1}^{n}{x_{i}}^{5}=\phi _{0}{\frac {[\sum _{i=1}^{n}x_{i}]^{5}}{n^{4}}}+\phi _{1}{\frac {[\sum _{i=1}^{n}x_{i}]^{3}s^{2}}{n^{2}}}+\phi _{2}[\sum _{i=1}^{n}x_{i}s^{4}$

$\sum _{i=1}^{n}{x_{i}}^{6}=\phi _{0}{\frac {[\sum _{i=1}^{n}x_{i}]^{6}}{n^{5}}}+\phi _{1}{\frac {[\sum _{i=1}^{n}x_{i}]^{4}s^{2}}{n^{3}}}+\phi _{2}{\frac {[\sum _{i=1}^{n}x_{i}]^{2}s^{4}}{n}}+\phi _{3}ns^{6}$

$\sum _{i=1}^{n}{x_{i}}^{7}=\phi _{0}{\frac {[\sum _{i=1}^{n}x_{i}]^{7}}{n^{6}}}+\phi _{1}{\frac {[\sum _{i=1}^{n}x_{i}]^{5}s^{2}}{n^{4}}}+\phi _{2}{\frac {[\sum _{i=1}^{n}x_{i}]^{3}s^{4}}{n^{2}}}+\phi _{3}[\sum _{i=1}^{n}x_{i}s^{6}$

$\sum _{i=1}^{n}{x_{i}}^{8}=\phi _{0}{\frac {[\sum _{i=1}^{n}x_{i}]^{8}}{n^{7}}}+\phi _{1}{\frac {[\sum _{i=1}^{n}x_{i}]^{6}s^{2}}{n^{5}}}+\phi _{2}{\frac {[\sum _{i=1}^{n}x_{i}]^{4}s^{4}}{n^{3}}}+\phi _{3}{\frac {[\sum _{i=1}^{n}x_{i}]^{2}s^{6}}{n}}+\phi _{4}ns^{8}$

$\sum _{i=1}^{n}{x_{i}}^{9}=\phi _{0}{\frac {[\sum _{i=1}^{n}x_{i}]^{9}}{n^{8}}}+\phi _{1}{\frac {[\sum _{i=1}^{n}x_{i}]^{7}s^{2}}{n^{6}}}+\phi _{2}{\frac {[\sum _{i=1}^{n}x_{i}]^{5}s^{4}}{n^{4}}}+\phi _{3}{\frac {[\sum _{i=1}^{n}x_{i}]^{3}s^{6}}{n^{2}}}+\phi _{4}[\sum _{i=1}^{n}x_{i}s^{8}$

$\sum _{i=1}^{n}{x_{i}}^{10}=\phi _{0}{\frac {[\sum _{i=1}^{n}x_{i}]^{10}}{n^{9}}}+\phi _{1}{\frac {[\sum _{i=1}^{n}x_{i}]^{8}s^{2}}{n^{7}}}+\phi _{2}{\frac {[\sum _{i=1}^{n}x_{i}]^{6}s^{4}}{n^{5}}}+\phi _{3}{\frac {[\sum _{i=1}^{n}x_{i}]^{4}s^{6}}{n^{3}}}+\phi _{4}{\frac {[\sum _{i=1}^{n}x_{i}]^{2}s^{8}}{n}}+\phi _{5}ns^{10}$

Solving the first 10th power yields

$\sum _{i=1}^{n}{x_{i}}=\sum _{i=1}^{n}{x_{i}}$

$\sum _{i=1}^{n}{x_{i}}^{2}={\frac {[\sum _{i=1}^{n}x_{i}]^{2}}{n}}+{\frac {n(n^{2}-1)}{12}}s^{2}$

$\sum _{i=1}^{n}{x_{i}}^{3}={\frac {[\sum _{i=1}^{n}x_{i}]^{3}}{n^{2}}}+{\frac {(n^{2}-1)s^{2}}{4}}[\sum _{i=1}^{n}x_{i}]$

$\sum _{i=1}^{n}{x_{i}}^{4}={\frac {[\sum _{i=1}^{n}x_{i}]^{4}}{n^{3}}}+(n^{2}-1)s^{2}{\frac {[\sum _{i=1}^{n}x_{i}]^{2}}{2n}}+{\frac {n(3n^{2}-7)(n^{2}-1)s^{4}}{240}}$

$\sum _{i=1}^{n}{x_{i}}^{5}={\frac {[\sum _{i=1}^{n}x_{i}]^{5}}{n^{4}}}+5(n^{2}-1)s^{2}{\frac {[\sum _{i=1}^{n}x_{i}]^{3}}{6n^{2}}}+{\frac {(3n^{2}-7)(n^{2}-1)s^{4}}{48}}[\sum _{i=1}^{n}x_{i}]$

$\sum _{i=1}^{n}{x_{i}}^{6}={\frac {[\sum _{i=1}^{n}x_{i}]^{6}}{n^{5}}}+5(n^{2}-1)s^{2}{\frac {[\sum _{i=1}^{n}x_{i}]^{4}}{4n^{3}}}+(3n^{2}-7)(n^{2}-1)s^{4}{\frac {[\sum _{i=1}^{n}x_{i}]^{2}}{16n}}+{\frac {n(3n^{4}-18n^{2}+31)(n^{2}-1)s^{6}}{1344}}$

$\sum _{i=1}^{n}{x_{i}}^{7}={\frac {[\sum _{i=1}^{n}x_{i}]^{7}}{n^{6}}}+7(n^{2}-1)s^{2}{\frac {[\sum _{i=1}^{n}x_{i}]^{5}}{4n^{4}}}+7(3n^{2}-7)(n^{2}-1)s^{4}{\frac {[\sum _{i=1}^{n}x_{i}]^{3}}{48n^{2}}}+(3n^{4}-18n^{2}+31)(n^{2}-1)s^{6}{\frac {[\sum _{i=1}^{n}x_{i}]}{192}}$

$\sum _{i=1}^{n}{x_{i}}^{8}={\frac {[\sum _{i=1}^{n}x_{i}]^{8}}{n^{7}}}+7(n^{2}-1)s^{2}{\frac {[\sum _{i=1}^{n}x_{i}]^{6}}{3n^{5}}}+7(3n^{2}-7)(n^{2}-1)s^{4}{\frac {[\sum _{i=1}^{n}x_{i}]^{4}}{24n^{3}}}+(3n^{4}-18n^{2}+31)(n^{2}-1)s^{6}{\frac {[\sum _{i=1}^{n}x_{i}]^{2}}{48n}}+{\frac {n(5n^{6}-55n^{4}+239n^{2}-381)(n^{2}-1)s^{8}}{11520}}$

$\sum _{i=1}^{n}{x_{i}}^{9}={\frac {[\sum _{i=1}^{n}x_{i}]^{9}}{n^{8}}}+3(n^{2}-1)s^{2}{\frac {[\sum _{i=1}^{n}x_{i}]^{7}}{n^{6}}}+21(3n^{2}-7)(n^{2}-1)s^{4}{\frac {[\sum _{i=1}^{n}x_{i}]^{5}}{40n^{4}}}+(3n^{4}-18n^{2}+31)(n^{2}-1)s^{6}{\frac {[\sum _{i=1}^{n}x_{i}]^{3}}{16n^{2}}}+(5n^{6}-55n^{4}+239n^{2}-381)(n^{2}-1)s^{8}{\frac {[\sum _{i=1}^{n}x_{i}]}{1280}}$

$\sum _{i=1}^{n}{x_{i}}^{10}={\frac {[\sum _{i=1}^{n}x_{i}]^{10}}{n^{9}}}+15(n^{2}-1)s^{2}{\frac {[\sum _{i=1}^{n}x_{i}]^{8}}{4n^{7}}}+7(3n^{2}-7)(n^{2}-1)s^{4}{\frac {[\sum _{i=1}^{n}x_{i}]^{6}}{8n^{5}}}+5(3n^{4}-18n^{2}+31)(n^{2}-1)s^{6}{\frac {[\sum _{i=1}^{n}x_{i}]^{4}}{32n^{3}}}+(5n^{6}-55n^{4}+239n^{2}-381)(n^{2}-1)s^{8}{\frac {[\sum _{i=1}^{n}x_{i}]^{2}}{256n}}+{\frac {n(3n^{10}-55n^{8}+462n^{6}-2046n^{4}+4191n^{2}-2555)s^{10}}{33792}}$

Derivation of Sums of Power for Integers using Generalized Equation for Sums of Power for an arbitrary arithmetic progression.

Setting $x_{1}=1$ and $s=1$, the general equation could be used to derive the Faulhaber's Sum of Power for integers

Power sums for Integers generalize equation is given as follows:

$\sum _{i=1}^{n}{x_{i}}^{p}=\sum _{i=1}^{n}i^{p}=\sum _{j=0}^{m}[\phi _{m}{\frac {[\sum _{i=1}^{n}i]^{p-2j}}{n^{p-(2j+1)}}}]$

Since $\sum _{i=1}^{n}i={\frac {n(n+1)}{2}}$, the equation above becomes

$\sum _{i=1}^{n}{x_{i}}^{p}=\sum _{i=1}^{n}i^{p}=\sum _{j=0}^{m}[\phi _{j}{\frac {[{\frac {n(n+1)}{2}}]^{p-2j}}{n^{p-(2j+1)}}}=n\sum _{j=0}^{m}[\phi _{j}[{\frac {(n+1)}{2}}]^{p-2j}]$

where $\phi _{0}=1$, $m={\frac {p-1}{2}}$ for odd p and ${\frac {p}{2}}$ for even p

For p=2

$\sum _{i=1}^{n}i^{2}=n\sum _{j=0}^{1}[\phi _{j}[{\frac {(n+1)}{2}}]^{2-2j}]=n[\phi _{0}({\frac {n+1}{2}})^{2}+\phi _{1}]=n[({\frac {n+1}{2}})^{2}+({\frac {n^{2}-1}{12}})]={\frac {n^{2}(n+1)^{2}}{4n}}+{\frac {n(n^{2}-1)}{12}}={\frac {n(n+1)}{12}}(3(n+1)+(n-1))={\frac {n(n+1)}{12}}(4n+2)={\frac {n(n+1)(2n+1)}{6}}={\frac {1}{6}}(2n^{3}+3n^{2}+n)$

Abd Shukor's Sums of Power Formula

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