数列an的每一项都为正数,a1=1/2,a2=4/5,且对满足m+n=p+q的正整数m,n,p,q都有(am+an)/[(1+am)(1+an)]=(ap+aq)/[(1+ap)(1+aq)],求数列an的通项别说什么b1,bn的,先从题目入手的有没有,
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![数列an的每一项都为正数,a1=1/2,a2=4/5,且对满足m+n=p+q的正整数m,n,p,q都有(am+an)/[(1+am)(1+an)]=(ap+aq)/[(1+ap)(1+aq)],求数列an的通项别说什么b1,bn的,先从题目入手的有没有,](/uploads/image/z/15102274-58-4.jpg?t=%E6%95%B0%E5%88%97an%E7%9A%84%E6%AF%8F%E4%B8%80%E9%A1%B9%E9%83%BD%E4%B8%BA%E6%AD%A3%E6%95%B0%2Ca1%3D1%2F2%2Ca2%3D4%2F5%2C%E4%B8%94%E5%AF%B9%E6%BB%A1%E8%B6%B3m%2Bn%3Dp%2Bq%E7%9A%84%E6%AD%A3%E6%95%B4%E6%95%B0m%2Cn%2Cp%2Cq%E9%83%BD%E6%9C%89%28am%2Ban%29%2F%5B%281%2Bam%29%281%2Ban%29%5D%3D%28ap%2Baq%29%2F%5B%281%2Bap%29%281%2Baq%29%5D%2C%E6%B1%82%E6%95%B0%E5%88%97an%E7%9A%84%E9%80%9A%E9%A1%B9%E5%88%AB%E8%AF%B4%E4%BB%80%E4%B9%88b1%2Cbn%E7%9A%84%EF%BC%8C%E5%85%88%E4%BB%8E%E9%A2%98%E7%9B%AE%E5%85%A5%E6%89%8B%E7%9A%84%E6%9C%89%E6%B2%A1%E6%9C%89%EF%BC%8C)
数列an的每一项都为正数,a1=1/2,a2=4/5,且对满足m+n=p+q的正整数m,n,p,q都有(am+an)/[(1+am)(1+an)]=(ap+aq)/[(1+ap)(1+aq)],求数列an的通项别说什么b1,bn的,先从题目入手的有没有,
数列an的每一项都为正数,a1=1/2,a2=4/5,且对满足m+n=p+q的正整数m,n,p,q都有(am+an)/[(1+am)(1+an)]=(ap+aq)/[(1+ap)(1+aq)],求数列an的通项
别说什么b1,bn的,先从题目入手的有没有,
数列an的每一项都为正数,a1=1/2,a2=4/5,且对满足m+n=p+q的正整数m,n,p,q都有(am+an)/[(1+am)(1+an)]=(ap+aq)/[(1+ap)(1+aq)],求数列an的通项别说什么b1,bn的,先从题目入手的有没有,
由题意,b1*bn=(1-a1)(1-an)/[(1+a1)(1+an)]=1-2(a1+an)/[(1+a1)(1+an)],b2*bn-1=(1-a2)(1-an-1)/[(1+a2)(1+an-1)]=1-2(a2+an-1)/[(1+a2)(1+an-1)]
因为(am+an)/[(1+am)(1+an)]=(ap+aq)/[(1+ap)(1+aq)]
所以(a1+an)/...
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由题意,b1*bn=(1-a1)(1-an)/[(1+a1)(1+an)]=1-2(a1+an)/[(1+a1)(1+an)],b2*bn-1=(1-a2)(1-an-1)/[(1+a2)(1+an-1)]=1-2(a2+an-1)/[(1+a2)(1+an-1)]
因为(am+an)/[(1+am)(1+an)]=(ap+aq)/[(1+ap)(1+aq)]
所以(a1+an)/[(1+a1)(1+an)]=(a2+an-1)/[(1+a2)(1+an-1)]
所以b1*bn=b2*bn-1,bn/bn-1=b2/b1=(1/9)/(1/3)=1/3
所以bn是首项为1/3,公比为1/3的等比数列
由bn=(1-an)/(1+an)可知an=(1-bn)/(1+bn)
又因bn的通项公式为bn=(1/3)^n,
所以an的通项为an=[1-(1/3)^n]/[1+(1/3)^n]
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