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已知数列an是首项a1=32,公比q=1/2的等比数列,数列bn满足bn=1/n(log2a1+log2a2+…+log2an)
试求bn的前n项和最大值
人气:361 ℃ 时间:2020-06-05 02:37:52
解答
an=a1*q^(n-1)=32*(1/2)^(n-1)=(1/2)^(-5)*(1/2)^(n-1)=(1/2)^(n-6)=2^(6-n),bn=1/n(log2a1+log2a2+…+log2an)=1/n(log2(2^5)+log2(2^4)+…+log2(2^(6-n))=1/n(6-1+(6-2)+……+(6-n))=1/n[6n-n(n+1)/2]=6-(n+1)/2...
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