Question

设数列[an}的前n项和为Sn,已知a1=1且满足3Sn的平方=an(3Sn-1),1求{1/Sn}为等差数列 1.求{

Answer 1

1.n>1,

3Sn^2=An(3Sn -1)

An=Sn-S(n-1)下标

3Sn^2=(3Sn -1)[Sn-S(n-1)]=3Sn^2-3SnS(n-1)-Sn+S(n-1)

3SnS(n-1)=S(n-1)-Sn

[S(n-1)-Sn]/[SnS(n-1)]=3

1/Sn - 1/S(n-1) = 3

1/Sn=1+3(n-1)=3n-2

当n=1时也满足.

所以{1/Sn}为等差数列

2.bn=Sn/3n+1=1/[(3n-2)(3n+1)]=(1/3)[1/(3n-2) - 1/(3n+1)]

3Tn=(1-1/4)+(1/4-1/7)+(1/7-1/10)+.+[1/(3n-2) - 1/(3n+1)]

=1- 1/(3n+1)=3n/(3n+1)

Tn=n/(3n+1)