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stangerzv

New Math Formula: Sums of Power for Arithmetic Series

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Posted (edited)

This is part of my works. I have formulated a new formulation for sums of power and it works for any numbers (i.e. real & complex arithmetic progression). The generalize equation can generate any power p (it works fine also with complex p).

Let the p-th power of an arithmetic series as follows
[math]\sum_{i=1}^n x_i^p = x_1^p + x_2^p + x_3^p + \cdots + x_n^p[/math]

The general equation for the sum is given as follows

[math]\sum^{n}_{i=1} {x_i}^p=\sum^{u}_{m=0}\phi_m s^{2m}\frac{[\sum^{n}_{i=1} x_i] ^{p-2m}}{n^{p-(2m+1)}}[/math]
where: [math]p-(2m+1)\ge 1[/math] if p is even and  [math]p-(2m+1)\le 1[/math] if p is odd, [math]\phi_m[/math] is a coefficient,[math]\sum^{n}_{i=1} {x_i}[/math] sum of n-th term, [math]u=\frac {p-1}{2}[/math] for odd p and [math]u=\frac {p}{2}[/math] for even p and s is the difference between terms (i.e [math]s=x_{i+1}-x_{i}[/math]).

Below are the equations for p=2-7

[math]\\\sum_{i=0}^{n}x_{i}^{2}=\frac{\left [ \sum_{i=0}^{n}x_{i} \right ]^2}{n}+\frac{n(n^2-1)s^2}{12}\\\\\\\sum_{i=0}^{n}x_i^3=\frac{\left [ \sum_{i=0}^{n}x_i \right ]^3}{n^2}+\frac{(n^2-1)s^2\left [ \sum_{i=0}^{n} x_i\right ]}{4}[/math]

The value s is the common difference of successive terms in arithmetic progression and [math]\sum_{i=0}^{n}x_i[/math] is the sum of arithmetic terms. The beauty of this equation is that when you set n=2, it describes the Fermat's Last Theorem in a polynomial forms and if you set p to be negative, you can get new form of Riemmann's Zeta Function.


Here, you can see how the coefficients are repetitive:
[math]\\\sum_{i=0}^{n}x_i^4=\frac{\left [ \sum_{i=0}^{n}x_i \right ]^4}{n^3}+\frac{(n^2-1)s^2\left [ \sum_{i=0}^{n}x_i \right ]^2}{2n}+\frac{n(3n^2-7)(n^2-1)s^4}{240}\\\\\\\sum_{i=0}^{n}x_i^5=\frac{\left [ \sum_{i=0}^{n}x_i \right ]^5}{n^4}+\frac{5(n^2-1)s^2\left [ \sum_{i=0}^{n} x_i\right ]^3}{6n^2}+\frac{(3n^2-7)(n^2-1)s^4\left [ \sum_{i=0}^{n}x_i \right ]}{48}[/math]

[math]\\\sum_{i=0}^{n}x_i^6=\frac{\left [ \sum_{i=0}^{n}x_i \right ]^6}{n^5}+\frac{5(n^2-1)s^2\left [ \sum_{i=0}^{n}x_i \right ]^4}{4n^3}+\frac{(3n^2-7)(n^2-1)s^4\left [ \sum_{i=0}^{n}x_i \right ]^2}{16n}+\frac{n(3n^4-18n^2+31)
(n^2-1)s^6}{1344}\\\\\\\sum_{i=0}^{n}x_i^7=\frac{\left [ \sum_{i=0}^{n}x_i \right ]^7}{n^6}+\frac{7(n^2-1)s^2\left [ \sum_{i=0}^{n}x_i \right ]^5}{4n^4}+\frac{7(3n^2-7)(n^2-1)s^4\left [ \sum_{i=0}^{n}x_i \right ]^3}{48n^2}+\frac{(3n^4-18n^2+31)
(n^2-1)s^6\left [ \sum_{i=0}^{n}x_i \right ]}{192}[/math]

Perhaps, by looking at this new formulation, someone could work out an alternative shorter proof for Fermat's Last Theorem.

Edited by stangerzv

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