1)假设公差为t
则a3=a2+t
a4=a2+2t
==>3a2+3t=9==>a2+t=3==>a3=3
又a2＋1,a3＋3,a4＋8为等比数列
==》(3-t+1)(3+t+8)=36
==>t=1
==>an=n
==>bn=3*2^(n-1)
2)Sn=[(n+1)n]/2
==>1/S1＋1/S2＋...1/Sn=2/(1*2)+2/(2*3)+...2/[(n+1)n]
又1/[(n+1)n]=1/n-1/(n+1)
==>1/S1＋1/S2＋...1/Sn=2/(1*2)+2/(2*3)+...2/[(n+1)n]
=2[1-1/2+1/2-1/3+1/3-1/4.+n/1-1/(n+1)]
=2[1-1/(n+1)]
=2n/(n+1)