1k+2k+3k+4k+...+nk = nk+1/(k+1) + nk/2 + ∑i=2k Ck+1i Bi nk+1-i
In which the Bi are those Bernoulli numbers (although Bernoulli mentiens them only in 1713 !).
This relation may also be written in a symbolic way, with "exponents" of B being indexes :
|∑i=1n (ik) = ["(n+B)k+1 - Bk+1"]/(k+1)|
The Bi may be calculated, from B0 = 1, by recurrence relations :
B2 - 2 B1 + 1 = B2 that is B1 = 1/2
B3 - 3 B2 + 3 B1 - 1 = B3 that is B2 = 1/6
B4 - 4 B3 + 6 B2 - 4 B1 + 1 = B4 that is B3 = 0
And generally from the symbolic notation (indexes as exponents of B) :
|" (B - 1)k = Bk "|
And the sum of powers :
11+21+31+41+...+n1 = n²/2 + n/2 = n(n+1)/2 (somme des entiers)
12+22+32+42+...+n2 = n³/3 + n²/2 + n/6 = n(n+1)(2n+1)/6 (somme des carrés)
13+23+33+43+...+n3 = n4/4 + n3/2 + n2/4 = n²(n+1)²/4 (somme des cubes)
14+24+34+44+...+n4 = n5/5 + n4/2 + n3/3 - n/30
15+25+35+45+...+n5 = n6/6 + n5/2 + 5n4/12 - n2/12
And the sum of 1000 first 10th powers in a blink ("intra semiquadrantem horae" wrote Bernoulli) :
∑i=1n i10 =(n11 + 11n10/2 + 55n9/6 - 330n7/30 + 462n5/42 - 165n3/30 + 11n x 5/66)/11
that is for n=1000 : 91 409 924 241 424 243 424 241 924 242 500.
Whoosh ! just for copying that, it takes more that a semiquadrantem horae (7mn30) !
We find also Bernoulli numbers in many realations :
For instance the Taylor series coeficients Bn in series X/(eX-1)
that is X/(eX-1) = ∑ Bn/n! Xn
Or in series for cotg(X)=1/X+∑1∞ (-1)n 22n B2n X2n-1/(2n)!
Finally the Riemann zeta function ζ(x) = ∑1/nx gives :
ζ(2n) = (-1)n-1 (2π)2n B2n/2(2n)!
Specifically the sum of inverse of squares ζ(2) = 1+1/4+1/9+1/16+...= π²/6
and the sum of inverse fourth powers ζ(4) = 1+1/16+1/81+1/256+...= π4/90