User:Email4mobile/sum of power arithmetic progression

I'm trying to call this "sum of power arithmetic progression" because it is a generalized arithmetic progression for powers. (i.e.: 1k+2k+3k+...).

One can generalize this in the following form:


 * $$\sum_{i=1}^n i^k = 1 + 2^k + 3^k +\dots + n^k$$

Since we are discussing a sequence of natural numbers, then we expect that the total sum is on the form of a polynomial.


 * $$\sum_{i=1}^n i^k = a_1 n + a_2 n^2 +\dots + a_k n^k + a_{k+1} n^{k+1}$$

or simply


 * $$\sum_{i=1}^n i^k = \sum_{j=1}^{k+1} a_j n^j $$

Using the previous relation, one can find the values of the terms: a1, a2,...,an by verifying a minimum of k+1 different values for the formula (example n=1, 2, 3, ..., k+1) and then putting them into k+1 linear equations to be solved.

Examples
The simplest form of this form is the normal arithmetic progression, k = 1.


 * $$\sum_{i=1}^n i = 1 + 2 + 3 +\dots + n = a_1 n + a_2 n^2$$

This is well known, where a1 = a2 = 0.5

To prove this, assume the solution for n= 1, 2.

For n = 1,


 * $$\sum_{i=1}^1 i = 1 = a_1(1) + a_2(1)^2 = a_1 + a_2$$

or
 * $$a_1 + a_2= 1\,$$

With n = 2
 * $$\sum_{i=1}^2 i = 1 + 2 = a_1(2) + a_2(2)^2 = 2a_1 + 4a_2$$

or


 * $$2a_1 + 4a_2 = 3\,$$

Solving both equations results:a1 = a2 = 0.5

Again, we can solve the following power series for k = 2


 * $$\sum_{i=1}^n i^2 = 1^2 + 2^2 + 3^2 +\dots + n^2 = a_1 n + a_2 n^2+ a_3 n^3$$

For n = 1,


 * $$\sum_{i=1}^1 i^2 = 1 = a_1(1) + a_2(1)^2 + a_3(1)^2= a_1 + a_2 + a_3$$

or


 * $$a_1 + a_2 + a_3= 1\,$$ ...(1)

With n = 2


 * $$\sum_{i=1}^2 i = 1 + 2^2 = a_1(2) + a_2(2)^2 + a_3(2)^3$$

or


 * $$2a_1 + 4a_2 + 8a_3= 5\,$$ ...(2)

With n = 3


 * $$\sum_{i=1}^3 i = 1 + 2^2 +3^2 = a_1(3) + a_2(3)^2 + a_3(3)^3$$

or


 * $$3a_1 + 9a_2 + 27a_3 = 14\,$$ ...(3)

Solving the 3 equations results:a1 = 1/6,a2 = 1/2,a3 = 1/3

Script
Note: I generated the previous table values using my online Math wizard (JScript based). The script to be used in the command line is: