Question

设数列{an}(an>0)的前n项和是Sn,且an与2的算数平均值等于Sn与2的几何平均值,则{an}的通项为

Answer 1

(an+2)/2=根号2Sn

所以(an+2)^2=8Sn

(an-1+2)^2=8Sn-1

2式相减

(an+2)^2-(an-1+2)^2=8an

an^2-an-1^2-4an-4an-1=0

(an+an-1)(an-an-1)-4(an+an-1)=0

(an+an-1)(an-an-1-4)=0

又有an＞0

所以an-an-1-4=0

an=an-1+4

所以an 是一个以4为公差的等差数列

(a1+2)^2=8a1

(a1-2)^2=0

a1=2

所以an=2+(n-1)*4=4n-2