2Sn^2=a1^3+a2^3··+an^3,所以2S(n-1)^2=a1^3+a2^3··+a(n-1)^3,(n>=2),所以2Sn^2-2S(n-1)^2=an^3,2(Sn+S(n-1))(Sn-S(n-1))=an^3,所以2(Sn+S(n-1))=an^2,所以2(S(n-1)+S(n-2))=a(n-1)^2,(n>=3)两式相减得2an+2a(n-1)=an^2-a(n-1)^2,an-a(n-1)=2(n>=3)因为2Sn^2=a1^3+a2^3··+an^3,所以a1=2,a2=4,所以a2-a1=2,所以an为等差数列,an=2n,所以cn=3^n+(-1)^(n-1)y-2^n,c(n+1)-cn=2*3^n-2^n+(-1)^... |