Question

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=a1

Answer 1

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