设数列{an}的前n项和为Sn,an与2的算术平均值等于Sn与2的几何平均值,求an

Question

设数列{an}的前n项和为Sn,an与2的算术平均值等于Sn与2的几何平均值,求an

Answer 1

[a(n) + 2]/2 = [S(n)*2]^(1/2),S(n)>=0,a(n)>=-2.[a(n) + 2]^2 = 8S(n),[a(1)+2]^2 = 8S(1) = 8a(1),[a(1)]^2 + 4a(1) + 4 = 8a(1),a(1) = 2.[a(n+1)+2]^2 = 8S(n+1),8a(n+1)=8S(n+1)-8S(n)=[a(n+1)+2]^2-[a(n)+2]^2=[a(n+1)]^2 + 4a(n+1) + 4 - [a(n)+2]^2,[a(n+1)-2]^2 = [a(n)+2]^2,|a(n+1)-2| = a(n)+2,若a(n+1)>=2,则a(n+1)-2=a(n)+2,a(n+1)=a(n)+4,{a(n)}是首项为a(1)=2,公差为4的等差数列.a(n)=2 + 4(n-1),S(n)=2n+2n(n-1)=2n^2,满足题意.若-2