1,x1,x2...Xn,成等比数列,x1 x2..xn>0,x1*x2*...xn=?
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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=?
x1,x2...Xn,2成等比数列,x1 x2..xn>0,x1*x2*...xn=?
![1,x1,x2...Xn,成等比数列,x1 x2..xn>0,x1*x2*...xn=?](/uploads/image/z/18239506-34-6.jpg?t=1%2Cx1%2Cx2...Xn%2C%E6%88%90%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%2Cx1+x2..xn%3E0%2Cx1%2Ax2%2A...xn%3D%3F)
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)]
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)]
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