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:
∑
i
=
1
n
i
k
=
1
+
2
k
+
3
k
+
⋯
+
n
k
{\displaystyle \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.
∑
i
=
1
n
i
k
=
a
1
n
+
a
2
n
2
+
⋯
+
a
k
n
k
+
a
k
+
1
n
k
+
1
{\displaystyle \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
∑
i
=
1
n
i
k
=
∑
j
=
1
k
+
1
a
j
n
j
{\displaystyle \sum _{i=1}^{n}i^{k}=\sum _{j=1}^{k+1}a_{j}n^{j}}
a 1 , a 2 ,...,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.
The simplest form of this form is the normal arithmetic progression, k = 1.
∑
i
=
1
n
i
=
1
+
2
+
3
+
⋯
+
n
=
a
1
n
+
a
2
n
2
{\displaystyle \sum _{i=1}^{n}i=1+2+3+\dots +n=a_{1}n+a_{2}n^{2}}
This is well known, where a 1 = a 2 = 0.5
To prove this, assume the solution for n = 1, 2.
For n = 1,
∑
i
=
1
1
i
=
1
=
a
1
(
1
)
+
a
2
(
1
)
2
=
a
1
+
a
2
{\displaystyle \sum _{i=1}^{1}i=1=a_{1}(1)+a_{2}(1)^{2}=a_{1}+a_{2}}
or
a
1
+
a
2
=
1
{\displaystyle a_{1}+a_{2}=1\,}
With n = 2
∑
i
=
1
2
i
=
1
+
2
=
a
1
(
2
)
+
a
2
(
2
)
2
=
2
a
1
+
4
a
2
{\displaystyle \sum _{i=1}^{2}i=1+2=a_{1}(2)+a_{2}(2)^{2}=2a_{1}+4a_{2}}
or
2
a
1
+
4
a
2
=
3
{\displaystyle 2a_{1}+4a_{2}=3\,}
Solving both equations results:a 1 = a 2 = 0.5
Again, we can solve the following power series for k = 2
∑
i
=
1
n
i
2
=
1
2
+
2
2
+
3
2
+
⋯
+
n
2
=
a
1
n
+
a
2
n
2
+
a
3
n
3
{\displaystyle \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,
∑
i
=
1
1
i
2
=
1
=
a
1
(
1
)
+
a
2
(
1
)
2
+
a
3
(
1
)
2
=
a
1
+
a
2
+
a
3
{\displaystyle \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
{\displaystyle a_{1}+a_{2}+a_{3}=1\,}
With n = 2
∑
i
=
1
2
i
=
1
+
2
2
=
a
1
(
2
)
+
a
2
(
2
)
2
+
a
3
(
2
)
3
{\displaystyle \sum _{i=1}^{2}i=1+2^{2}=a_{1}(2)+a_{2}(2)^{2}+a_{3}(2)^{3}}
or
2
a
1
+
4
a
2
+
8
a
3
=
5
{\displaystyle 2a_{1}+4a_{2}+8a_{3}=5\,}
With n = 3
∑
i
=
1
3
i
=
1
+
2
2
+
3
2
=
a
1
(
3
)
+
a
2
(
3
)
2
+
a
3
(
3
)
3
{\displaystyle \sum _{i=1}^{3}i=1+2^{2}+3^{2}=a_{1}(3)+a_{2}(3)^{2}+a_{3}(3)^{3}}
or
3
a
1
+
9
a
2
+
27
a
3
=
14
{\displaystyle 3a_{1}+9a_{2}+27a_{3}=14\,}
Solving the 3 equations results:a 1 = 1/6,a 2 = 1/2,a 3 = 1/3
List of some sum terms
edit
k
Terms
Polynomial form
1
[1/2, 1/2]
∑
i
=
1
n
i
=
n
+
n
2
2
{\displaystyle \sum _{i=1}^{n}i={\frac {n+n^{2}}{2}}}
2
[1/6, 1/2, 1/3]
∑
i
=
1
n
i
2
=
n
+
3
n
2
+
2
n
3
6
{\displaystyle \sum _{i=1}^{n}i^{2}={\frac {n+3n^{2}+2n^{3}}{6}}}
3
[0, 1/4, 1/2, 1/4]
∑
i
=
1
n
i
3
=
n
2
+
2
n
3
+
n
4
4
{\displaystyle \sum _{i=1}^{n}i^{3}={\frac {n^{2}+2n^{3}+n^{4}}{4}}}
4
[-1/30, 0, 1/3, 1/2, 1/5]
∑
i
=
1
n
i
4
=
−
n
+
10
n
3
+
15
n
4
+
6
n
5
30
{\displaystyle \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]
∑
i
=
1
n
i
5
=
−
n
2
+
5
n
4
+
6
n
5
+
2
n
6
6
{\displaystyle \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]
∑
i
=
1
n
i
6
=
n
−
7
n
3
+
21
n
5
+
21
n
6
+
6
n
7
42
{\displaystyle \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]
∑
i
=
1
n
i
7
=
2
n
2
−
7
n
4
+
14
n
6
+
12
n
7
+
3
n
8
24
{\displaystyle \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]
∑
i
=
1
n
i
8
=
−
3
n
+
20
n
3
−
42
n
5
+
60
n
7
+
45
n
8
+
10
n
9
90
{\displaystyle \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]
∑
i
=
1
n
i
9
=
−
3
n
2
+
10
n
4
−
14
n
6
+
15
n
8
+
10
n
9
+
2
n
10
20
{\displaystyle \sum _{i=1}^{n}i^{9}={\frac {-3n^{2}+10n^{4}-14n^{6}+15n^{8}+10n^{9}+2n^{10}}{20}}}
Looking carefully at the table, checking terms, we can realize that:
a
k
+
1
{\displaystyle a_{k+1}\,}
a
k
{\displaystyle a_{k}\,}
a
k
−
1
{\displaystyle a_{k-1}\,}
a
k
−
2
{\displaystyle a_{k-2}\,}
a
k
−
3
{\displaystyle a_{k-3}\,}
a
k
−
4
{\displaystyle a_{k-4}\,}
a
k
−
5
{\displaystyle a_{k-5}\,}
... 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 ))
