第2个回答 2012-05-05
1.
设等比数列{an}公比为q
b1+b2+b3=log(1/2)(a1)+log(1/2)(a2)+log(1/2)(a3)
=log(1/2)(a1a2a3)
=log(1/2)(a2³)
=3log(1/2)(a2)=6
log(1/2)(a2)=2
a2=1/4
q=a2/a1=(1/4)/(1/2)=1/2
an=a1q^(n-1)=(1/2)(1/2)^(n-1)=1/2ⁿ
bn=log(1/2)(1/2ⁿ)=nlog(1/2)(1/2)=n
数列{an}通项公式为an=1/2ⁿ,数列{bn}通项公式为bn=n。
2.
anbn=n/2ⁿ
令a1b1+a2b2+...+anbn=Sn
Sn=a1b1+a2b2+...+anbn=1/2 +2/2²+3/2³+...+n/2ⁿ
Sn/2=1/2²+2/2³+...+(n-1)/2ⁿ+n/2^(n+1)
Sn-Sn/2=Sn/2=1/2+1/2²+1/2³+...+1/2ⁿ -n/2^(n+1)
=(1/2)(1-1/2ⁿ)/(1-1/2) -n/2^(n+1)
=1 -1/2ⁿ-(n/2)/2ⁿ
Sn=2-2/2ⁿ-n/2ⁿ<2-0-0=2
Sn<2
a1b1+a2b2+...+anbn<2,不等式成立。