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=

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=

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……①b3S3=(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d=2,q=8∴an=a1+(n-1)d=2n+1bn=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=(1/2)*{(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)]}=3/4-(2n+3)/[2(n+1)(n+2)]======以下答案可供参考======供参考答案1:(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)=64q(6+d)=64b3S3=q^2 (a1+a2+a3)=q^2 (3a2)=960联立：q=8,d=2则：an=2n+1bn=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)]

这个答案应该是对的