在1与100之间插入n正数,使n+2个数成等比数列,则插入的n个数的积是多少啊

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a1=1a(n+2)=100
a(n+2)=a1·q^(n+1)
q^(n+1)=100
a2·a3·a4·……·an·a(n+1)
=(a1·q)·(a1·q^2)·(a1·q^3)·……·[a1·q^(n-1)]·(a1·q^n)
=q^[1+2+3+……+(n-1)+n]
=q^[n(n+1)/2]
=[q^(n+1)]^(n/2)
=100^(n/2)
=10^n======以下答案可供参考======
供参考答案1:
数列为1 10 100