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.求和:1/S1+1/S2+……+1/Sn

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.求和:1/S1+1/S2+……+1/Sn
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.
求和:1/S1+1/S2+……+1/Sn

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.求和:1/S1+1/S2+……+1/Sn
设 公比为q, 公差为d.
b2*S2=q*(3+3+d)=64
b3*S3=q^2*(9+3d)=960
q(6+d) = 64
q^2(3+d) = 320
(6+d)^2 = (3+d) * 64*64/320 = (3+d)*64/5
5d^2 + 60d + 180 = 192 + 64d
5d^2 - 4d - 12 = 0
(5d + 6)(d -2) = 0
d =2
(d = -6/5 导致 an 不能始终为正数,所以舍去）
q = 8
an = 2n + 1
bn = 8^(n-1)
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Sn = (a1 + an)*n/2 = (3 + 2n +1)*n/2 = n(n+2)
1/Sn = 1/[n(n+2)] = (1/2)*[1/n - 1/(n+2)]
（1/S1）+（1/S2）+.+（1/Sn）
= (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)]
注意上面 1/3 - 1/3 , 1/4 -1/4 …… 1/(n-1) - 1/(n-1), 1/n - 1/n 相消.最后中括弧内残留
1 + 1/2 - 1/(n+1) - 1/(n+2) =
因此
（1/S1）+（1/S2）+.+（1/Sn）
= [3/2 - 1/(n+1) - 1/(n+2)] /2
祝你学习愉快

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