在各项为正的等比数列{an}中,首项a1=1/2,数列{bn}=log1/2an(1/2为log的底),且b1+b2+b3=6.(1)求数列{an}的通项公式.(2)求证:a1b1+a2b2+.+anbn
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![在各项为正的等比数列{an}中,首项a1=1/2,数列{bn}=log1/2an(1/2为log的底),且b1+b2+b3=6.(1)求数列{an}的通项公式.(2)求证:a1b1+a2b2+.+anbn](/uploads/image/z/2131882-34-2.jpg?t=%E5%9C%A8%E5%90%84%E9%A1%B9%E4%B8%BA%E6%AD%A3%E7%9A%84%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%AD%2C%E9%A6%96%E9%A1%B9a1%3D1%2F2%2C%E6%95%B0%E5%88%97%7Bbn%7D%3Dlog1%2F2an%281%2F2%E4%B8%BAlog%E7%9A%84%E5%BA%95%EF%BC%89%2C%E4%B8%94b1%2Bb2%2Bb3%3D6.%EF%BC%881%EF%BC%89%E6%B1%82%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F.%EF%BC%882%EF%BC%89%E6%B1%82%E8%AF%81%EF%BC%9Aa1b1%2Ba2b2%2B.%2Banbn)
在各项为正的等比数列{an}中,首项a1=1/2,数列{bn}=log1/2an(1/2为log的底),且b1+b2+b3=6.(1)求数列{an}的通项公式.(2)求证:a1b1+a2b2+.+anbn
在各项为正的等比数列{an}中,首项a1=1/2,数列{bn}=log1/2an(1/2为log的底),且b1+b2+b3=6.
(1)求数列{an}的通项公式.(2)求证:a1b1+a2b2+.+anbn
在各项为正的等比数列{an}中,首项a1=1/2,数列{bn}=log1/2an(1/2为log的底),且b1+b2+b3=6.(1)求数列{an}的通项公式.(2)求证:a1b1+a2b2+.+anbn
(1)b1+b2+b3=log1/2 (a1a2a3)=6 ,所以a1a2a3=(1/2)^6
又an是等比数列,所以a1a3=(a2)² 故(a2)³=(1/2)^6 得 a2=(1/2)²=1/4
所以公比q=a2/a1=1/2
故an=a1q^(n-1)=(1/2)^n
(2)bn=log1/2an=log1/2 【(1/2)^n】=n
所以anbn=n*(1/2)^n
a1b1+a2b2+.+anbn=1(1/2)+2(1/2)^2+3(1/3)^3+n(1/2)^n ①
1/2(a1b1+a2b2+.+anbn)=1(1/2)^2+2(1/2)^3+3(1/3)^4+n(1/2)^(n+1) ②
①-②得
1/2(a1b1+a2b2+.+anbn)=1/2+(1/2)^2+(1/2)^3+(1/3)^4+(1/2)^n-n(1/2)^(n+1)
=1/2 ×(1-(1/2)^n)/(1-1/2)-n(1/2)^(n+1)
=1-(1/2)^n-n(1/2)^(n+1)<1
所以a1b1+a2b2+.+anbn
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
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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。
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1. b1+b2+b3=log1/2(a1a2a3)=6
a1a2a3=1/64
a2=1/4 a1=1/2 q=1/2
an=(1/2)^n
2. bn=log1/2an=n
an*bn=n/2^n
Sn=a1b1+a2b2+......+anbn
Sn=1/2+2/2^2+3/2^3+……+n/2^n
Sn/...
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1. b1+b2+b3=log1/2(a1a2a3)=6
a1a2a3=1/64
a2=1/4 a1=1/2 q=1/2
an=(1/2)^n
2. bn=log1/2an=n
an*bn=n/2^n
Sn=a1b1+a2b2+......+anbn
Sn=1/2+2/2^2+3/2^3+……+n/2^n
Sn/2= 1/2^2+2/2^3+3/2^4+……+(n-1)/2^n+n/2^(n+1) 相减
Sn/2=1/2+1/2^2+1/2^3+……+1/2^n-n/2^(n-1)
=1/2(1-1/2^n)/(1-1/2)-n/2^(n-1)
Sn=2-1/2^n-2n/2^(n-1)
Sn=<2
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