已知正项数列an的前n项和为Sn对任意n都有a1^3+a2^3+...+an^3=Sn^2求数列an的通向公式

(A1)^3=(S1)^2=(A1)^2
An>0 A1=1
(A1)^3+(A2)^3+……+(A(n-1))^3=(S(n-1))^2
与原式相减
(An)^3=(Sn)^2-(S(n-1))^2=(Sn-S(n-1))(Sn+S(n-1))=An(Sn+S(n-1))
(An)^2=Sn+S(n-1)
由此推出(A(n+1))^2=S(n+1)+Sn
两式相减
(A(n+1))^2-(An)^2=A(n+1)+An
(A(n+1)+An)(A(n+1)-An)=A(n+1)+An
A(n+1)+An≠0
A(n+1)-An=1
{An}是以1为首项，1为公差的等差数列
An=n