Question

一个等比数列前n项和为S,前n项倒数和为T,则其前n项之积为?

Answer 1

S=A1*(1-q^n)/(1-q)

T=(1/A1)*(1-1/q^n)/(1-1/q)=(1/A1)*(1/q^n)(q^n-1)/(1/q)(q-1)

两式相除

S/T=(A1)^2*q^(n-1)

n项的积＝A1*A2*A3*...An

=(A1)^n*q^(1+2+3+...+n-1)

=(A1)^n*q^[n(n-1)/2]

=[(A1)^2*q^(n-1)]^(n/2)

=(S/T)^(n/2)