sumes de potencies

1^{0}+...+n^{0} = n
1^{1}+...+n^{1} = (1/2)n(n+1)
1^{2}+...+n^{2} = (1/6)n(n+1)(2n+1)
1^{3}+...+n^{3} = (1/4)n^{2}(n+1)^{2}
1^{4}+...+n^{4} = (1/30)n(n+1)(2n+1)(3n^{2}+3n+(-1))
1^{5}+...+n^{5} = (1/12)n^{2}(n+1)^{2}(2n^{2}+2n+(-1))
1^{6}+...+n^{6} = (1/42)n(n+1)(2n+1)(3n^{4}+6n^{3}+(-3)n+1)
1^{7}+...+n^{7} = (1/24)n^{2}(n+1)^{2}(3n^{4}+6n^{3}+(-1)n^{2}+(-4)n+2)
1^{8}+...+n^{8} = (1/90)n(n+1)(2n+1)(n^{2}+n+(-1))(5n^{4}+10n^{3}+(-5)n+4)+...
                            ...+(1/90)n(n+1)(2n+1)
1^{9}+...+n^{9} = (1/20)n^{2}(n+1)^{2}(n^{2}+n+(-1))(2n^{4}+4n^{3}+(-n^{2})+(-3)n+3)


he arribat amb polinomis de sumes de potencies fins a n^{9} y metode del binomi para sumes de potencies fins a n^{5}


método del binomi para sumes de potencies:


(n+1)^{n}+(-1)=[ n // 1 ]·∑ ( k^{n+(-1)} )+...+[ n // n+(-1) ]·∑ ( k )+n


teorema:
si (1/oo^{n}) = 0 ==> lim[n-->oo]( (1/n^{m+1})·( 1^{m}+...+n^{m} ) ) = ( 1/(m+1) )