等差数列｛an｝的各项均为正数,a1=3,前n项和为Sn,｛bn｝为等比数列,b1=1,且b2S2=b4,b3S3=960求an与bn；求1/S1+1/S2+.+1/Sn.
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(1)设an=a1+(n-1)d,d为公差,d≠0,
bn=b1q^(n-1),q≠0,1,则：
b2S2=b1q(a1+a2)=64
q(6+d)=64
b3S3=q^2 (a1+a2+a3)=q^2 (3a2)=960
联立：q=8,d=2
则：an=2n+1
bn=8^(n-1)
(2)Sn=n(n+2)
1/Sn=1/2 * [1/n - 1/(n+2)]=
(1/S1)+(1/S2)+…(1/Sn)=1/2 *
[1/1-1/3+1/2-1/4+...+1/(n-1)-1/(n+1)+1/n-1/(n+2)]
=[1+1/2-1(n+1)-1/(n+2)]
=3/4 - 1/[2(n+1)] - 1/[2(n+2)]