Question

在等比数列an中,a1=2,若数列an+1也是等比数列,求an

Answer 1

an=2

a(n+1)*a(n-1)=an^2

[a(n+1)+1][a(n-1)+1]=(an+1)^2

乘开 即有 an^2+a(n+1)+a(n-1)+1=an^2+2an+1

a(n+1)+a(n-1)=2an=2[a(n+1)+a(n-1)]^0.5

[a(n+1)-a(n-1)]^2=0

a(n+1)=a(n-1)

所以,该数列为常数数列

an=2