tbff8
幼苗
共回答了18个问题采纳率:83.3% 举报
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
显然bn是单调减小的,令bn=0,则有6-(n+1)/2=0,n=11
而b10=1/2,b11=0,b12=-1/2
1年前
6