Question

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=64，b3S3=960

1.求数列｛an｝与｛bn｝的通项公式
（2）求1/S1+1/S2+...+1/Sn
详细过程，谢了

Answer 1

1.
设等差数列{an}公差为d,等比数列{bn}公比为q,则:
b2S2=(b1*q)(a1+a1+d)=q(6+d)=64……①
b3S3=(b1*q^2)(a1+a1+d+a1+2d)=(q^2)(9+3d)=960即(q^2)(3+d)=320……②
联立①②得q=8或q=40/3,则d=2或d=-6/5(舍去),故镇历d=2,q=8
∴an=a1+(n-1)d=2n+1
bn=b1*q^(n-1)=8^(n-1)=2^(3n-3)
2.
Sn=(a1+an)*n/2=n(n+2)
1/Sn=1/[n(n+2)]=(1/2)*[1/n-1/(n+2)]
∴原式=(1/拆旅并2)*{(1-1/3)+(1/2-1/4)+(1/3-1/5)+…+[1/(n-2)-1/n]+[1/(n-1)-1/(n+1)]+[1/n-1/(n+2)]}
=(1/2)*[1+1/2-1/(n+1)-1/(n+2)]
=(1/2)*{3/2-(2n+3)/[(n+1)(n+2)]}
=3/旅迹4-(2n+3)/[2(n+1)(n+2)]

Answer 2

解：
(1)
设an=a1+(n-1)d,d为公差，d≠谈胡厅0，
bn=b1q^(n-1),q≠0,1，则：做或
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/含隐n-1/(n+2)]
=[1+1/2-1(n+1)-1/(n+2)]
=3/4 - 1/[2(n+1)] - 1/[2(n+2)]