题目

已知数列{an} 的是一个各项为正数的等比数列,Sn为它的前n项和,Sn’=1/a1+1/a2+…+1/an,Pn=a1a2a3…an,求

提问时间：2021-03-03

答案

设数列{An}首项为a,公比为q
Sn=a(1-q^n)/(1-q)
数列{1/An}前n项和为Sn',首项为1/a,公比1/q
Sn'=1/a×(1-(1/q)^n)/(1-(1/q))=q(1-(1/q^n))/a(q-1)
Pn=A1A2A3……An
=a^n×q^(0+1+2+……+n-1)
=a^n×q^(n(n-1)/2)
Sn/Sn'=[a(1-q^n)/(1-q)]÷[q(1-(1/q^n))/a(q-1)]
化简得
Sn/Sn'=a^2×q^(n-1)
(Sn/Sn')^(n/2)=[a^2×q^(n-1)]^(n/2)
=a^n×q^(n(n-1)/2)
=Pn