题目

数列｛an｝的前n项和,且Sn=n∧²an,则an/an+1=

提问时间：2021-02-01

答案

n≥1时,Sn=n²an=n²[Sn-S(n-1)]
(n²-1)Sn=n²S(n-1)
Sn/S(n-1)=n²/(n²-1)=n²/[(n+1)(n-1)]
S(n-1)/S(n-2)=(n-1)²/[n(n-2)]
…………
S2/S1=2²/(3×1)
连乘
Sn/S1=(2²×3²×...×n²)/[(3×1)(4×2)...(n+1)(n-1)]
=(2²×3²×...×n²)/[3×4×...×(n+1)×1×2×...×(n-1)]
=(2²×3²×...×n²)/[1×2×3²×4²×...(n-1)²×n(n+1)]
=4n²/[2n(n+1)]
=2n/(n+1)
Sn=[2n/(n+1)]S1
an=Sn-S(n-1)=[2n/(n+1) -2(n-1)/n]S1
a(n+1)=S(n+1)-Sn=[2(n+1)/(n+2) -2n/(n+1)]S1
an/a(n+1)
=[2n/(n+1)-2(n-1)/n]/[2(n+1)/(n+2)-2n/(n+1)]
=[n(n+2)n-(n-1)(n+1)(n+2)]/[(n+1)n(n+1)-n(n+2)n]
=(n+2)/n
=1 +2/n