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:
Since we are discussing a sequence of natural numbers, then we expect that the total sum is on the form of a polynomial.
or simply
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.
The simplest form of this form is the normal arithmetic progression, k = 1.
This is well known, where a1 = a2 = 0.5
To prove this, assume the solution for n= 1, 2.
For n = 1,
or
With n = 2
or
Solving both equations results:a1 = a2 = 0.5
Again, we can solve the following power series for k = 2
For n = 1,
or
- ...(1)
With n = 2
or
- ...(2)
With n = 3
or
- ...(3)
Solving the 3 equations results:a1 = 1/6,a2 = 1/2,a3 = 1/3
List of some sum terms
[edit]
k
|
Terms
|
Polynomial form
|
1
|
[1/2, 1/2]
|
|
2
|
[1/6, 1/2, 1/3]
|
|
3
|
[0, 1/4, 1/2, 1/4]
|
|
4
|
[-1/30, 0, 1/3, 1/2, 1/5]
|
|
5
|
[0, -1/12, 0, 5/12, 1/2, 1/6]
|
|
6
|
[1/42, 0, -1/6, 0, 1/2, 1/2, 1/7]
|
|
7
|
[0, 1/12, 0, -7/24, 0, 7/12, 1/2, 1/8]
|
|
8
|
[-1/30, 0, 2/9, 0, -7/15, 0, 2/3, 1/2, 1/9]
|
|
9
|
[0, -3/20, 0, 1/2, 0, -7/10, 0, 3/4, 1/2, 1/10]
|
|
Looking carefully at the table, checking terms, we can realize that:
is always 1/(k+1),
is always 1/2,
is always positive,
is always 0,
is always negative,
is always 0,
is always positive,
... and so on!
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))