Question

已知数列{an}为等比数列,前三项为a,1/2a+1/2,1/3a+1/3,且Sn=a1+a2+a3+…+an,则Tn=

Answer 1

等比数列

[1/2(a+1)]^2=a*[1/3(a+1)]

1/4*(a+1)^2=a/3*(a+1)

等比数列的项不等于0

a+1不等于0

所以(a+1)/4=a/3

a=3

所以a1=3,a2=2,q=2/3

所以bn=an^2,则b1=9,

bn/b(n-1)=[an/a(n-1)]^2=q^2=4/9

所以Tn=9*[1-(4/9)^n]/(1-4/9)=81/5-(81/5)*(4/9)^n