题目

1,x1,x2...Xn,成等比数列,x1 x2..xn>0,x1*x2*...xn=?
x1,x2...Xn,2成等比数列,x1 x2..xn>0,x1*x2*...xn=?

提问时间：2020-12-18

答案

设公比为q
则a1=1
a(n+2)=a1*q^(n+1)=2 即q^(n+1)=2 q=2^[1/(n+1)]
所以x1*x2*...xn=(a1*q)*(a1*q^2)*...*[a1*q^(n-1)]
=q^[1+2+...+(n-1)]
=q^[n(n-1)/2]
=2^[n(n-1)/2(n+1)]
=2^[(n²-n)/(2n+2)]