a1+1,a3+1,a7+1成等比数列,则
(a3+1)^2=(a1+1)(a7+1)
(a1+2d+1)^2=(a1+1)(a1+6d+1)
d=2代入,整理,得
4a1=12
a1=3
an=a1+(n-1)d=3+2(n-1)=2n+1
Sn=(a1+an)n/2=(3+2n+1)n/2=(n+2)n
1/Sn=1/[n(n+2)]=(1/2)[1/n -1/(n+2)]
Tn=1/S1+1/S2+...+1/Sn
=(1/2)[1/1-1/3+1/2-1/4+...+1/n -1/(n+2)]
=(1/2)[1+1/2+...+1/n-(1/3+1/4+...+1/(n+2)]
=(1/2)[1+1/2 -1/(n+1)-1/(n+2)]
=3/4 -1/[2(n+1)] -1/[2(n+2)]

截图吧，我从来不手写拍照的。