已知数列an的前n项和为Sn,Sn=2-【（2/n）+1】an（n≥1）（1）求证 数列an/n是等比数列
2个回答
(1)
a1=S1=2-[(2/1)+1]a1
整理,得4a1=2 a1=1/2
n≥2时,
Sn=2-[(2/n)+1]an
Sn-1=2-[2/(n-1) +1]a(n-1)
Sn-Sn-1=an=2-[(2/n)+1]an-2+[2/(n-1) +1]a(n-1)
整理,得
[2(n+1)/n]an=[(n+1)/(n-1)]a(n-1)
2an/n=a(n-1)/(n-1)
(an/n)/[a(n-1)/(n-1)]=1/2,为定值.
a1/1=(1/2)/1=1/2
数列{an/n}是以1/2为首项,1/2为公比的等比数列.
an/n=(1/2)×(1/2)^(n-1)=1/2ⁿ
(2)
an=n/2ⁿ
2ⁿ×an=2ⁿ×n/2ⁿ=n
Tn=1+2+...+n=n(n+1)/2
1/Tn=2/[n(n+1)]=2[1/n-1/(n+1)]
An=1/T1+1/T2+1/T3+...+1/Tn
=2[1-1/2+1/2-1/3+1/3-1/4+...+1/n-1/(n+1)]
=2[1-1/(n+1)]
=2n/(n+1)
An/[2/(nan)]
=[2n/(n+1)]/[2/(n²/2ⁿ)]
=n³/[(n+1)×2ⁿ]
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