已知等比数列{an}满足an>0,n=1,2,…,且a5·a2n-5=22n(n≥3),则当n≥1时,log2a1+log2a3+…+log2a2n-1=A.n(2n-1)B.(n+1)2C.n2D.(n-1)2
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C解析考点:数列与函数的综合.分析:由题意,等比数列{an}a>0,n=1,2,…,且a5?a2n-5=22n(n≥3),又当n>1时,log2a1+log2a3+log2a5+…+log2a2n-1=log2a1a3a5…a2n-1.由等比数列的性质m+n=s+t,aman=asat.求出a1a3a5…a2n-1的值,即可求出正确