设f(x)=x3,等差数列｛an｝中a3=7,a1+a2+a3=12,记Sn=f（三次根号（an+1））,

题目

设f(x)=x3,等差数列｛an｝中a3=7,a1+a2+a3=12,记Sn=f（三次根号（an+1））,
令bn=anSn,数列｛1/bn｝的前n项和为Tn,求{an}的通项公式和sn

答案

a1+2d=7
3a1+3d=12
解得a1=1 d=3
an=1+3(n-1)=3n-2
Sn=a(n+1)=3(n+1)-2=3n+1
估计你还想求Tn
1/bn=1/(3n-2)/(3n+1)=1/(3n-2)-1/(3n+1)
Tn=1-1/4+1/4-1/7+1/7-1/10+...+1/(3n-2)-1/(3n+1)
=1-1/(3n+1)
=3n/(3n+1)