等差数列{an}的各项均为正数,a1=3,前n项和为Sn,{bn}等比,b1=1,且b2b2S2=64,b3S3=960求an,bn,1/S1+1/S2……+1/Sn

题目

等差数列{an}的各项均为正数,a1=3,前n项和为Sn,{bn}等比,b1=1,且b2b2S2=64,b3S3=960求an,bn,1/S1+1/S2……+1/Sn

答案

a1=3,所以S2=6+d,S3=9+3d
b1=1,b2=q,b3=q^2
所以 (6+d)q=64
(9+3d)q^2=960
相除
(9+3d)/(6+d)*q=15
q=15(6+d)/(9+3d)
代入(6+d)q=64
15(6+d)^2=64(9+3d)
因为是正数
所以d=2
q=15(6+d)/(9+3d)=8
an=2n+1
bn=8^(n-1)
Sn=2*n(n+1)/2+n=n^2+2n=n(n+2)
1/Sn=1/2*[1/n-1/(n+2)]
所以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)-1/(n+2)]
=(3n^2+13n)/(4n^2+12n+8)