Download 07 New Formula for the Sum of Powers

07 New Formula for the Sum of Powers
7.1 Expression with binomial coefficients of a factorial
Formula 7.1.1 ( S. Ruiz )
When
n
is a natural number, the following expression holds for arbitrary
n
n-r
n ! = Σ(-1)
r=0
 r
n
x.
(x + r)n
Proof
n
2
and the differences sequence are as follows.
1 4 9 16 25 
3 5 7 9 
n 2:
d 1:
d 2:
2
2 
2
Each difference is calculated as follows.
d1: 3 = 22 -12 , 5 = 32 -22
, 7 = 42 -32 , 
d2: 2 = 32 -222 +12 , 2 = 42 -232 +22 , 
If we focus on
d 2 , we find that this can be expressed as follows for arbitrary x .
2! = 2Ｃ 2(x +2)2- 2Ｃ 1(x +1)2+ 2Ｃ 0 x 2
Next,
n
3
and the differences sequence are as follows.
n :
d 1:
1 8 27 64 125 21 
7 19 37 61 91 
d 2:
12 18 24 30 
3
d 3:
6
6
6

These differences are also calculated as follows.
d1: 7 = 23 -13 , 19 = 33 -23 ,
37 = 43 -33 , 61 = 53 -43 , 
d2: 12 = 33 -223 +13 , 18 = 43 -233 +23 , 24 = 53 -243 +33 , 
d3: 6 = 43 -333 +323 -13 , 6 = 53 -343 +333 -23 , 
If we focus on
d 3 , we find that this can be also expressed as follows for arbitrary x .
3! = 3Ｃ 3(x +3)3- 3Ｃ 2(x +2)3+ 3Ｃ 1(x +1)3- 3Ｃ 0 x 3
Next,
n 4 and the differences sequence of are as follows.
n 4:
d 1:
d 2:
d 3:
d 4:
1 16 81 256 625 1296 2401 
15 65 175 369 671 1105 
50 110 194 302 434
60
84 108 132 
24
24
24 
-1-

These differences are also calculated as follows.
d 1: 15 = 24 - 14 , 65 = 34 - 24 , 175 = 44 - 34 , 369 = 54 - 44 , 671 = 64 - 54 , 
d 2: 50 = 34 - 224 + 14 , 110 = 44 - 234 + 24 , 175 = 54 - 244 + 34 , 
d 3: 60 = 44 - 334 + 324 - 14 , 84 = 54 - 344 + 334 - 24 , 
d 4: 24 = 54 - 444 + 634 - 424 + 14 , 24 = 64 - 454 + 644 - 434 + 24 , 
And
d4
can be also expressed as follows for arbitrary
x.
4! = 4Ｃ 4(x +4)4- 4Ｃ 3(x +3)4+ 4Ｃ 2(x +2)4- 4Ｃ 1(x +1)4+ 4Ｃ 0 x 4
Thus, in general, we obtain the following.
n
n-r
n ! = Σ(-1)
r=0
Example: When
 r  (x + r)
n
n
x =-0.3 , x =
-2-
7.2 Expression with binomial coefficients of a power (Part1)
Formula 7.2.1
When
m,n
are natural numbers, the following expression holds.
m
n m = Σ mBr nＣ r
r=0
Where,
r
r-s
m
mB r = Σ(-1) rＣ s s
r =0, 1, 2, ,m
s=0
Proof
n
2
and the difference sequence are as follows.
0 1 4 9 16 25 
1 3 5 7 9 
2 2 2 2 
n 2:
d 1:
d 2:
On the other hand, 2 times of Pascal's triangle is as follows.
2
2 2
2 4 2
2 6 6 2
2 8 12 8 2

Comparing both, we obtain the followings.
d2(n) = 2 nＣ 0
n -1


d1(n) = 1 + 2Σ rＣ 0 = 1 + 2 nＣ 1
r=0
n -1
rＣ 0 = nＣ 1
Σ

r=0
Then
n -1
n -1
n -1
r=0
r=0
n 2 = 0 + Σ1 + 2 rＣ 1 = Σ1 + 2Σ rＣ 1
r=0
n -1
Here,
1 = nＣ 1
Σ
r=0
n -1
,
rＣ 1
Σ
r=0
= nＣ 2 0Ｃ 1 = 0 .
Then
n 2 = 1 nＣ 1 +2 nＣ 2
(2.n)
Coefficients (blue and magenta) of the right hand side are corresponding with the first terms of the difference
sequence of
n2 .
And the first terms of the difference sequences are given as follows.
d1: 1 = 12 - 02 , 3 = 22 - 12 , 
d2: 2 = 22 -212 + 02 , 
That is, the r-th coefficient 2Br of the right side of (2.n) can be expressed as follows.
r
2B r
= Σ(-1)r-s rＣ s s 2
s=0
-3-
Next,
n
3
and the difference sequence are as follows.
n 3:
d 1:
0
1
8
1
d 2:
7
27 64 125 216 
19 37 61 91 
12 18 24 30 
6
d 3:
6
6
6

6
On the other hand, 6 times of Pascal's triangle is as follows.
6 nＣ 0: 6 6 6 6 
Comparing both, we obtain the followings.
d3(n) = 6 nＣ 0
n -1
n -1
rＣ 0 = nＣ 1
Σ

r=0


d2(n) = 6 + 6ΣrＣ 0 = 6 + 6 nＣ 1
r=0
n -1
d1(n) = 1 + Σ6 + 6 rＣ 1
r=0
n -1
n -1
r=0
r=0
= 1 + 6Σ1 + 6Σ rＣ 1
n -1
Here,
1 = nＣ 1
Σ
r=0
n -1
,
rＣ 1
Σ
r=0
= nＣ 2 0Ｃ 1 = 0 .
Then,
d1(n) = 1 + 6 nＣ 1 + 6 nＣ 2
Then
n -1
n 3 = 0 + Σ1 + 6 rＣ 1 + 6 rＣ 2
r=0
n -1
n -1
n -1
r=0
r=0
r=0
= Σ1 + 6Σ rＣ 1 + 6Σ rＣ 2
Furthermore,
n -1
1
Σ
r=0
= nＣ 1 ,
n -1
rＣ 1
Σ
r=0
= nＣ 2 ,
n -1
rＣ 2
Σ
r=0
= nＣ 3
rＣ s = 0 r < s 
Using these,
3
n = 1 nＣ 1 + 6 nＣ 2 + 6 nＣ 3
(3.n)
Coefficients (blue and magenta) of the right hand side are corresponding with the first terms of the difference
sequence of
n3 .
And the first terms of the difference sequences are given as follows.
d1: 1 = 13 - 03 , 7 = 23 - 13 , 19 = 33 - 23 , 
d2: 6 = 23 - 213 + 03 , 12 = 33 - 223 + 13 , 
d3: 6 = 33 - 323 + 313 - 03 , 
That is, the r-th coefficient 3Br of the right side of (3.n) can be expressed as follows.
r
r-s
3
3B r = Σ(-1) rＣ s s
s=0
Hereafter by induction, we obtain the desired expression.
-4-
Example:
34
0
1
0-s
4
=0 ,
1Ｃ s s
4B 0 = Σ(-1)
1-s
4
=1
1Ｃ s s
4B1 = Σ(-1)
s=0
2
4B 2
= Σ(-1)2-s2Ｃ s s 4 = 14 ,
4B 4
= Σ(-1)4-s4Ｃ s s 4 = 24
s=0
3
4B 3
= Σ(-1)3-s3Ｃ s s 4 = 36
s=0
s=0
4
s=0
From these,
4
n = 0 nＣ 0 +1 nＣ 1 +14 nＣ 2 +36 nＣ 3 +24 nＣ 4
Substituting
n =3
for this,
34 = 03Ｃ 0 +13Ｃ 1 +143Ｃ 2 +363Ｃ 3 +243Ｃ 4 = 81
Formula 7.2.1 is extensible to the real number for
n
using general binomial coefficients.
Formula7.2.2
When
m is a natural number and x
m
x m = Σ mBr
r=0
is a positive number, the following expression holds.
 r
x
Where,
r
mB r
Example:
= Σ(-1)r-s rＣ s s m
r =0, 1, 2, ,m
s=0
0.53
0
3B 0
1
= Σ(-1)0-s1Ｃ s s 3 = 0 ,
s=0
2
3B 2
= Σ(-1)2-s2Ｃ s s 3 = 6 ,
3B 1
= Σ(-1)1-s1Ｃ s s 3 = 1
3B 3
= Σ(-1)3-s3Ｃ s s 3 = 6
s=0
3
s=0
s=0
Then,
0.53 = 0
 0   1   2   3  = 0.125
0.5
0.5
+1
0.5
+6
0.5
+6
Note
The first few steps of the triangle of these coefficients are as follows.
1B 1
1
2B 1 2B 2
1
3B 1 3B 2 3B 3
4B 1 4B 2 4B 3 4B 4
1
=
1
5B 1 5B 2 5B 3 5B 4 5B 5
6B 1 6 B 2 6 B 3 6 B 4 6 B 5 6 B 6
7B 1 7B 2 7B 3 7B 4 7B 5 7B 6 7B 7
2
1
1
62
6
14
30
6
36
150
540
24
240
1560
120
1800
720
1 126 1806 8400 16800 15120 5040


-5-
According to On-Line Encyclopedia of Integer Sequences , this coefficient mBr is already known. ( A019538 )
And this coefficient has the following relation to Stirling numbers of the 2nd kind.
mB r
= r! S2(m,r)
-6-
7.3 Expression with binomial coefficients of a power (Part2)
Formula 7.3.1
When
m ,n
are natural numbers, the following expression holds.
m -1
n m = Σ mDr+1 n+rＣ m
r=0
Where, mDr
r =1, 2, ,m
are Eulerian Numbers which are given by
r-1
mDr
= Σ(-1)s m +1Ｃ s(r -s)m
m =1, 2, 3, 
s=0
Proof
From Formula 7.2.1 ,
2
n = nＣ 1 +2 nＣ 2
= nＣ 1 + nＣ 2 + nＣ 2
= n +1Ｃ 2 + nＣ 2
n 3 = nＣ 1 +6 nＣ 2 +6 nＣ 3
= nＣ 1 + nＣ 2+5 nＣ 2 +6 nＣ 3
= n +1Ｃ 2 +5 nＣ 2 +6 nＣ 3
= n +1Ｃ 2 +5nＣ 2 + nＣ 3+ nＣ 3
= n +1Ｃ 2 +5 n +1Ｃ 3 + nＣ 3
= n +1Ｃ 2 + n +1Ｃ 3+4 n +1Ｃ 3 + nＣ 3
= n +2Ｃ 3 +4 n +1Ｃ 3 + nＣ 3
n 4 = nＣ 1 +14 nＣ 2 +36 nＣ 3 +24 nＣ 4
= nＣ 1 + nＣ 2+13 nＣ 2 +36 nＣ 3 +24 nＣ 4
= n +1Ｃ 2 +13 nＣ 2 +36 nＣ 3 +24 nＣ 4
= n +1Ｃ 2 +13nＣ 2 + nＣ 3+23 nＣ 3 +24 nＣ 4
= n +1Ｃ 2 +13 n +1Ｃ 3 +23 nＣ 3 +24 nＣ 4
= n +1Ｃ 2 + n +1Ｃ 3+12 n +1Ｃ 3 +23nＣ 3 + nＣ 4+ nＣ 4
= n +2Ｃ 3 +12 n +1Ｃ 3 +23 n +1Ｃ 4 + nＣ 4
= n +2Ｃ 3 +12n +1Ｃ 3 + n +1Ｃ 4+11 n +1Ｃ 4 + nＣ 4
= n +2Ｃ 3 +12 n +2Ｃ 4 +11 n +1Ｃ 4 + nＣ 4
= n +2Ｃ 3 + n +2Ｃ 4+11 n +2Ｃ 4 +11 n +1Ｃ 4 + nＣ 4
= n +3Ｃ 4 +11 n +2Ｃ 4 +11 n +1Ｃ 4 + nＣ 4
n 5 = nＣ 1 +30 nＣ 2 +150 nＣ 3 +240 nＣ 4 +120 nＣ 5
= nＣ 1 + nＣ 2+29 nＣ 2 +150 nＣ 3 +240 nＣ 4 +120 nＣ 5
= n +1Ｃ 2 +29 nＣ 2 +150 nＣ 3 +240 nＣ 4 +120 nＣ 5
= n +1Ｃ 2 +29nＣ 2 + nＣ 3+121 nＣ 3 +240 nＣ 4 +120 nＣ 5
= n +1Ｃ 2 +29 n +1Ｃ 3 +121 nＣ 3 +240 nＣ 4 +120 nＣ 5
= n +1Ｃ 2 + n +1Ｃ 3+28 n +1Ｃ 3 +121nＣ 3 + nＣ 4+119 nＣ 4 +120 nＣ 5
-7-
= n +2Ｃ 3 +28 n +1Ｃ 3 +121 n +1Ｃ 4 +119 nＣ 4 +120 nＣ 5
= n +2Ｃ 3 +28n +1Ｃ 3 + n +1Ｃ 4+93 n +1Ｃ 4 +119nＣ 4 + nＣ 5+ nＣ 5
= n +2Ｃ 3 +28 n +2Ｃ 4 +93 n +1Ｃ 4 +119 n +1Ｃ 5 + nＣ 5
= n +2Ｃ 3 + n +2Ｃ 4+27 n +2Ｃ 4 +93n +1Ｃ 4 + n +1Ｃ 5+26 n +1Ｃ 5 + nＣ 5
= n +3Ｃ 4 +27 n +2Ｃ 4 +93 n +2Ｃ 5 +26 n +1Ｃ 5 + nＣ 5
= n +3Ｃ 4 +27n +2Ｃ 4 + n +2Ｃ 5+66 n +2Ｃ 5 +26 n +1Ｃ 5 + nＣ 5
= n +3Ｃ 4 +27 n +3Ｃ 5 +66 n +2Ｃ 5 +26 n +1Ｃ 5 + nＣ 5
= n +3Ｃ 4 + n +3Ｃ 5+26 n +3Ｃ 5 +66 n +2Ｃ 5 +26 n +1Ｃ 5 + nＣ 5
= n +4Ｃ 5 +26 n +3Ｃ 5 +66 n +2Ｃ 5 +26 n +1Ｃ 5 + nＣ 5

These coefficients are "Coefficients that make powers constant (Eulerian Numbers) " which are mentioned in
" 01 Zeta Generating Functions ". Therefore, by induction, we obtain the desired expression.
Example:
73, 34
73 = 7Ｃ 3 +48Ｃ 3 + 9Ｃ 3
= 35+456+84
= 343
34 = 3Ｃ 4+114Ｃ 4 +115Ｃ 4 + 6Ｃ 4 = 0+111+115+15 = 81
Formula 7.3.1 is also extensible to the real number for
n
.
Formula 7.3.2
When
m is a natural number and x
m -1
x m = Σ mDr+1
r=0
Where, mDr
is a positive number, the following expression holds.
m
x +r
r =1, 2, ,m
are Eulerian Numbers which are given by
r-1
mDr
m =1, 2, 3, 
= Σ(-1)s m +1Ｃ s(r -s)m
s=0
0.73 , 0.94
0.7
1.7
2.7
+4
+
0.73 =
3
3
3
0.9
1.9
2.9
3.9
0.94 =
+11
+11
+
4
4
4
4
Example:
= 0.343
     
        = 0.6561
Note: Eulerian Triangle
The first 7 steps of Eulerian Triangle are as follows.
1D1
1
2D1 2D2
1
3D1 3D2 3D3
4D1 4D2 4D3 4D4
5D1 5D2 5D3 5D4 5D5
6D1 6D2 6D3 6D4 6D5 6D6
7D1 7D2 7D3 7D4 7D5 7D6 7D7
1
=
1
4
1
1 11 11 1
1 26
66 26 1
1 57 302 302 57 1
1 120 1191 2416 1191 120 1
-8-
7.4 New formula for the sum of powers
Formula 7.4.1
m,n
When
are natural numbers, the following expression holds.
n
m
k =1
r=1
Σk m = Σ mDr n+rＣ m+1
r =1,2, ,m
Where, mDr
are Eulerian Numbers which are given by
r-1
mDr
m =1, 2, 3, 
= Σ(-1)s m +1Ｃ s(r -s)m
s=0
Proof
From Formula 7.3.1 ,
2
n = n +1Ｃ 2 + nＣ 2
Then,
n
k
Σ
k =1
2
n
n
k =1
k =1
= Σ k+1Ｃ 2 +Σ kＣ 2
Here,
n
n
n -1
Ｃ 2 = n +2Ｃ 3 , ΣkＣ 2 = Σk +1Ｃ 2 = n +1Ｃ 3
Σ
k =1
k =1
k =1
k +1
Substituting these for the above,
n
k
Σ
k =1
2
= n +2Ｃ 3 + n +1Ｃ 3
This is what the all subscripts of
n2
are increased one. .
Next, from Formula 7.3.1 ,
3
n = n +2Ｃ 3 +4 n +1Ｃ 3 + nＣ 3
Then,
n
k
Σ
k =1
3
n
n
n
k =1
k =1
k =1
= Σ k+2Ｃ 3 +4Σ k+1Ｃ 3 +Σ kＣ 3
Here,
n
n
n
Ｃ 3 = n +3Ｃ 4 , Σk+1Ｃ 3 = n +2Ｃ 4 , ΣkＣ 3 = n +1Ｃ 4
Σ
k =1
k =1
k =1
k +2
Substituting these for the above,
n
k
Σ
k =1
3
= n +3Ｃ 4 +4 n +2Ｃ 4 + n +1Ｃ 4
This is also what the all subscripts of
n3
are increased one.
When the 4th degree, we obtain the following in a similar way.
n
k
Σ
k =1
4
= n +4Ｃ 5 +11 n +3Ｃ 5 +11 n +2Ｃ 5+ n +1Ｃ 5
Hereafter by induction, we obtain the desired expression.
100
Example:
k
Σ
k =1
5
Ｃ 6 +26102Ｃ 6 +66103Ｃ 6 +26104Ｃ 6 + 105Ｃ 6 = 171708332500
101
21006 +61005 +51004 -1002
12
Formula 7.4.1 is also extensible to the real number for
n
-9-
.
= 171708332500
Formula 7.4.2
m is a natural number and x
When
x
m
k m = Σ mDr
Σ
k =0
r=1
Where, mDr

is a positive number, the following expression holds.

x +r
m +1
r =1,2, ,m
are Eulerian Numbers which are given by
r-1
mDr
= Σ(-1)s m +1Ｃ s(r -s)m
0.9
Example:
m =1, 2, 3, 
s=0
k
Σ
k =0
4
 5   5   5   5  = 0.659148
1.9
2.9
+11
3.9
+11
4.9
+
60.95 +150.94 +100.93 -0.91
30
= 0.659148
Note: Calculation method one by one for Eulerian Numbers
Eulerian Number can be calculated also by the algorithm one by one as follows.
2
3
4
5
6
1 1
2, 2
1
4
1
3, 2 2, 3
1 11 11 1
4, 2 3, 3 2, 4
1 26 66 26 1
5, 2 4, 3 3,4 2, 5
1
57 302 302 57 1
Calculating formula
4 = 12 + 12
11 = 13 + 42
26 = 14 +112 ,
66 = 113 + 113
57 = 15 +262 , 302 = 264 + 663


- 10 -
7.5 Formula for the sum of powers of real numbers
In this section, we calculate the following the sum of powers. Where,
m,n
are natural numbers and
a,b
are real numbers.
Sm ,n = (1a +b)m +(2a +b)m +(3a +b)m +  +(na +b)m
Formula 7.5.1
m,n
When
are natural numbers and
n
a,b
are real numbers, the following expression holds.
m
(a k +b)m = b m nＣ 1 + ΣmＣ r a r b
Σ
k =1
r=1
Where, rDs
s =1, 2, ,r
Σ
r Ds n+sＣ r+1
s=1
are Eulerian Numbers which are given by
s-1
r Ds
r
m-r
= Σ(-1)t r +1Ｃ t(s -t)r
r =1 , 2 ,  ,m
t=0
Proof
m
m
r=0
r=0
(a k + b)m = ΣmＣ r(a k)r b m-r = ΣmＣ r k r a r b m-r
Then
n
n
m
(a k +b)m = ΣΣmＣ r k r a r b
Σ
k =1
k=1 r=0
m-r
m
n
= ΣmＣ r a r b m-rΣk r
r=0
k =1
n
m
n
k=1
r=1
k=1
= mＣ 0 a 0 b mΣk 0 + ΣmＣ r a r b m-rΣk r
i.e.
n
(a k +b)m = b m nＣ 1
Σ
k =1
m
n
r=1
k =1
+ ΣmＣ r a r b m-rΣk r


n
k
Σ
k=1
0

= nＣ 1
From Formula 7.4.1 in the previous section,
n
r
Σk r = Σ r Ds n+sＣ r+1
k =1
s=1
s-1
r Ds
= Σ(-1)t r +1Ｃ t(s -t)r
r =1 , 2 , 3 ,  ,m
t=0
Substituting these for the above,
n
m
(a k +b)m = b m nＣ 1 + ΣmＣ r a r b
Σ
k =1
r=1
m-r
r
Σ
r Ds n+sＣ r+1
s=1
Although this seems a complicated formula, if this is written down about
The coefficients of
a,b
are the
m -th step of Pascal's Triangle
n
n
1
(a k +b)
Σ
k =1
2
= b nＣ 1 + a n +1Ｃ 2
= b 2nＣ 1 + 2abn +1Ｃ 2
+ a 2n +1Ｃ 3 + n +2Ｃ 3
n
(a k +b)
Σ
k =1
3
= b 3nＣ 1 + 3ab 2n +1Ｃ 2
+ 3a 2bn +1Ｃ 3 + n +2Ｃ 3
+ a 3n +1Ｃ 4 +4 n +2Ｃ 4 + n +3Ｃ 4
- 11 -
, it is as follows.
and the coefficients of n+sＣ r+1 are the upper
m steps of the Eulerian Triangle.
(a k +b)
Σ
k =1
m =1, ,5
n
(a k +b)
Σ
k =1
4
= b 4nＣ 1 + 4ab 3n +1Ｃ 2
+ 6a 2b 2n +1Ｃ 3 + n +2Ｃ 3
+ 4a 3bn +1Ｃ 4 +4 n +2Ｃ 4 + n +3Ｃ 4
+ a 4n +1Ｃ 5 +11 n +2Ｃ 5 +11 n +3Ｃ 5+ n +4Ｃ 5
n
(a k +b)
Σ
k =1
5
= b 5nＣ 1 + 5ab 4n +1Ｃ 2
+ 10a 2b 3n +1Ｃ 3 + n +2Ｃ 3
+ 10a 3b 2n +1Ｃ 4 +4 n +2Ｃ 4 + n +3Ｃ 4
+ 5a 4bn +1Ｃ 5 +11 n +2Ｃ 5 +11 n +3Ｃ 5+ n +4Ｃ 5
+ a 5n +1Ｃ 6 +26n +2Ｃ 6 +66n +3Ｃ 6 +26n +4Ｃ 6 + n +5Ｃ 6
The strong point of this formula is that the calculations are essentially performed by integers. And if there are
non-integer calculations, the result becomes easier. We show two examples below.
Example 1
50
( k - e)
Σ
k =1
4
= e 450Ｃ 1 - 4 e 351Ｃ 2
+ 6 2e 251Ｃ 3 + 52Ｃ 3
- 4 3e51Ｃ 4 +4 52Ｃ 4 + 53Ｃ 4
+  451Ｃ 5 +11 52Ｃ 5 +11 53Ｃ 5+ 54Ｃ 5
= 50e 4 - 41275 e 3 + 6(20825 + 22100) 2e 2
- 4(249900 + 4270725+292825) 3e
+ (2349060 + 112598960 + 112869685+3162510) 4
i.e.
50
( k - e)
Σ
k =1
4
= 50e 4 - 5100 e 3 + 257550 2e 2 - 6502500 3e + 65666665 4
Example 2
100

Σ
k =1
3
3
2 k + 3 =  3
2
Ｃ 1 + 3 2  3
100
2
+ 3 2
Ｃ2
101
3101Ｃ 3 + 102Ｃ 3
3
+  2 101Ｃ 4 +4 102Ｃ 4 + 103Ｃ 4
= 1003 3 + 50509 2 + (166650+171700)6 3
+ (4082925+44249575+4421275)2 2
i.e.
100

Σ
k =1
3
2 k + 3 = 51050450 2 + 2030400 3
2011.12.24
2015.11.14 added Sec.5
Kano. Kono
Alien's Mathematics
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