设数列{an},{bn},满足an=[lg(b1)+lg(b2)+...+lg(bn)]/n,证明{an}为等差数列的冲
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设数列{an},{bn},满足an=[lg(b1)+lg(b2)+...+lg(bn)]/n,证明{an}为等差数列的冲要条件
bn大于0,n属于自然数
证明{an}为等差数列的冲要条件是{bn}为等比数列
错了也没关系,
bn大于0,n属于自然数
证明{an}为等差数列的冲要条件是{bn}为等比数列
错了也没关系,
=====啊,等等
再问: ?怎么了?你会不?
再答: 马上
再问: 大哥~麻烦快点吧~急死我了~~~~~~~~~~~
再答: ①充分性,即:由“{bn}为等比数列”推出“{an}为等差数列” 设bn公比为q,∵b1>0,b2>0,∴q>0 则 lg(b1)+lg(b2)+lg(b3)+lg(b4)+...+lg(bn) =lg(b1)+lg(q·b1)+lg(q²b1)+lg(q³b1)+...+lg[q^(n-1)·b1] =lg(b1) + [lg(b1)+lg(q)] + [lgb1+lg(q²)]+ [lg(b1)+lg(q³)]...+[lg(b1)+lgq^(n-1)] =lg(b1) + [lg(b1)+lg(q)] + [lgb1+2lg(q)]+ [lg(b1)+3lgq(q)]...+[lg(b1)+(n-1)lg(q)] =nlg(b1)+[lg(q)+2lg(q)+3lg(q)+...+(n-1)lg(q)] =nlg(b1) + [lg(q)+(n-1)lg(q)](n-1)/2 =nlg(b1) + (1/2)n(n-1)lg(q) ∴an=[nlg(b1) + (1/2)n(n-1)lg(q)]/n = lg(b1) + (1/2)(n-1)lg(q) a(n+1)=lg(b1) + (1/2)nlg(q) a(n+1)-an=(1/2)nlg(q) - (1/2)(n-1)lg(q) =(1/2)lg(q) ∴数列{an}是以a1=lg(b1)为首项,(1/2)lg(q)为公差的等差数列。 ②必要性,即:由“{an}为等差数列”推出“{bn}为等比数列” 设an公差为d,an=a1+(n-1)d n·an=a1n+n(n-1)d 数列{lg(bn)}前n项和为Sn 构造等差数列{Cn},使C1=a1,公差为2d,前n项和为Tn 则 Tn=C1n+n(n-1)×2d/2 =a1n+n(n-1)d =n·an =lg(b1)+lg(b2)+...+lg(bn) =Sn 则T(n-1)=S(n-1) ∴Tn-T(n-1)=Sn-S(n-1) 即Cn=lg(bn) ∴数列{lg(bn)}是以a1为首项,2d为公差的等差数列 lg(bn)=a1+2(n-1)d lg[b(n+1)]-lg(bn)=2d lg[b(n+1)/bn]=2d b(n+1)/bn=10^(2d) ∴数列{bn}是以10^a1为首项,10^(2d)为公比的等比数列。
再问: ?怎么了?你会不?
再答: 马上
再问: 大哥~麻烦快点吧~急死我了~~~~~~~~~~~
再答: ①充分性,即:由“{bn}为等比数列”推出“{an}为等差数列” 设bn公比为q,∵b1>0,b2>0,∴q>0 则 lg(b1)+lg(b2)+lg(b3)+lg(b4)+...+lg(bn) =lg(b1)+lg(q·b1)+lg(q²b1)+lg(q³b1)+...+lg[q^(n-1)·b1] =lg(b1) + [lg(b1)+lg(q)] + [lgb1+lg(q²)]+ [lg(b1)+lg(q³)]...+[lg(b1)+lgq^(n-1)] =lg(b1) + [lg(b1)+lg(q)] + [lgb1+2lg(q)]+ [lg(b1)+3lgq(q)]...+[lg(b1)+(n-1)lg(q)] =nlg(b1)+[lg(q)+2lg(q)+3lg(q)+...+(n-1)lg(q)] =nlg(b1) + [lg(q)+(n-1)lg(q)](n-1)/2 =nlg(b1) + (1/2)n(n-1)lg(q) ∴an=[nlg(b1) + (1/2)n(n-1)lg(q)]/n = lg(b1) + (1/2)(n-1)lg(q) a(n+1)=lg(b1) + (1/2)nlg(q) a(n+1)-an=(1/2)nlg(q) - (1/2)(n-1)lg(q) =(1/2)lg(q) ∴数列{an}是以a1=lg(b1)为首项,(1/2)lg(q)为公差的等差数列。 ②必要性,即:由“{an}为等差数列”推出“{bn}为等比数列” 设an公差为d,an=a1+(n-1)d n·an=a1n+n(n-1)d 数列{lg(bn)}前n项和为Sn 构造等差数列{Cn},使C1=a1,公差为2d,前n项和为Tn 则 Tn=C1n+n(n-1)×2d/2 =a1n+n(n-1)d =n·an =lg(b1)+lg(b2)+...+lg(bn) =Sn 则T(n-1)=S(n-1) ∴Tn-T(n-1)=Sn-S(n-1) 即Cn=lg(bn) ∴数列{lg(bn)}是以a1为首项,2d为公差的等差数列 lg(bn)=a1+2(n-1)d lg[b(n+1)]-lg(bn)=2d lg[b(n+1)/bn]=2d b(n+1)/bn=10^(2d) ∴数列{bn}是以10^a1为首项,10^(2d)为公比的等比数列。
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