Question

数列{An}数等比数列,a1=2,{An+1}也是等比数列,则Sn是多少?

Answer 1

等比数列{an},公比为q,a1a3=a2^2

数列{an+1}也是等比数列,

即 (a1+1)(a3+1)=(a2+1)^2

得：a1+a3=2a2

(a1+a3)^2=4(a2)^2=4(a1a3)

(a1-a3)^2=0

a1=a3

即 {an}是常数列,an=a1=2

{an+1}也是常数列,每一项都是3

故 Sn=3n