已知等比数列1,x1,x2,…,x2n,2,求x1·x2·x3·…·x2n.

问题描述:

已知等比数列1,x1,x2,…,x2n,2,求x1·x2·x3·…·x2n.

x1·x2·x3·…·x2n
=(1*2)^n=2^n

an=q^n-1
2=q^(2n+1)
x1·x2·x3·…·x2n
=q·q2·q3·…·q2n
=q^(1+2+..+2n)
=q^(1+2n)*n
2^n

x1·x2·x3·…·x2n=(x1·x2n)·(x2·x2n-1)·…·(xn·xn+1)
等比数列中 x1·x2n=x2·x2n-1=…=xn·xn+1=1·2=2
所以 x1·x2·x3·…·x2n=2^n

已知等比数列1,x1,x2,…,x2n,2
所以1*2=x1*x2n=x2*x2n-1=...=xn*xn+1=2
所以x1·x2·x3·…·x2n=2^n

∵1,x1,x2,…,x2n,2成等比数列,公比q
∴2=1·q2n+1
x1x2x3…x2n=q·q2·q3…q2n=q1+2+3+…+2n
=q^(1+2n)*n
2^n