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Aan=n^2+nBan=n^2-nCan=3n-1Dan=4n-2

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
A an=n^2+n
B an=n^2-n
C an=3n-1
D an=4n-2

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