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.: 1^k+2^k+3^k+...).

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,...,a_n 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

List of some sum terms

k	Terms	Polynomial form
1	[1/2, 1/2]	$\sum _{i=1}^{n}i={\frac {n+n^{2}}{2}}$
2	[1/6, 1/2, 1/3]	$\sum _{i=1}^{n}i^{2}={\frac {n+3n^{2}+2n^{3}}{6}}$
3	[0, 1/4, 1/2, 1/4]	$\sum _{i=1}^{n}i^{3}={\frac {n^{2}+2n^{3}+n^{4}}{4}}$
4	[-1/30, 0, 1/3, 1/2, 1/5]	$\sum _{i=1}^{n}i^{4}={\frac {-n+10n^{3}+15n^{4}+6n^{5}}{30}}$
5	[0, -1/12, 0, 5/12, 1/2, 1/6]	$\sum _{i=1}^{n}i^{5}={\frac {-n^{2}+5n^{4}+6n^{5}+2n^{6}}{6}}$
6	[1/42, 0, -1/6, 0, 1/2, 1/2, 1/7]	$\sum _{i=1}^{n}i^{6}={\frac {n-7n^{3}+21n^{5}+21n^{6}+6n^{7}}{42}}$
7	[0, 1/12, 0, -7/24, 0, 7/12, 1/2, 1/8]	$\sum _{i=1}^{n}i^{7}={\frac {2n^{2}-7n^{4}+14n^{6}+12n^{7}+3n^{8}}{24}}$
8	[-1/30, 0, 2/9, 0, -7/15, 0, 2/3, 1/2, 1/9]	$\sum _{i=1}^{n}i^{8}={\frac {-3n+20n^{3}-42n^{5}+60n^{7}+45n^{8}+10n^{9}}{90}}$
9	[0, -3/20, 0, 1/2, 0, -7/10, 0, 3/4, 1/2, 1/10]	$\sum _{i=1}^{n}i^{9}={\frac {-3n^{2}+10n^{4}-14n^{6}+15n^{8}+10n^{9}+2n^{10}}{20}}$

Notes

Looking carefully at the table, checking terms, we can realize that:

$a_{k+1}\,$ is always 1/(k+1),

$a_{k}\,$ is always 1/2,

$a_{k-1}\,$ is always positive,

$a_{k-2}\,$ is always 0,

$a_{k-3}\,$ is always negative,

$a_{k-4}\,$ is always 0,

$a_{k-5}\,$ is always positive,

... and so on!

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:

k=1;  // k can take 1, 2, 3, ...
m=[];
m[k+1]=1;
for(i=2;i<=k+1;i++)
 m[i*(k+2)-1] = pow(i,k)+m[(i-1)*(k+2)-1];
  for(i=0;i<=k;i++)
   for(j=i*(k+2);j<=k+i*(k+2);j++)m[j]=pow(i+1,j-i*(k+2)+1);
    msolve(matrix(k+1,k+2,m))