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)],记bn=(1-an)/(1+an),证明bn是等比,并由此求数列an的通项
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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)],记bn=(1-an)/(1+an),证明bn是等比,并由此求数列an的通项](/uploads/image/z/7138368-0-8.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%E8%AE%B0bn%3D%281-an%29%2F%281%2Ban%29%2C%E8%AF%81%E6%98%8Ebn%E6%98%AF%E7%AD%89%E6%AF%94%2C%E5%B9%B6%E7%94%B1%E6%AD%A4%E6%B1%82%E6%95%B0%E5%88%97an%E7%9A%84%E9%80%9A%E9%A1%B9)
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)],记bn=(1-an)/(1+an),证明bn是等比,并由此求数列an的通项
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)],记bn=(1-an)/(1+an),证明bn是等比,并由此求数列an的通项
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)],记bn=(1-an)/(1+an),证明bn是等比,并由此求数列an的通项
按题意,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)],由题目条件可知(a1+an)/[(1+a1)(1+an)]=(a2+an-1)/[(1+a2)(1+an-1)],因此b1*bn=b2*bn-1,bn/bn-1=b2/b1=(1/9)/(1/3)=1/3,因此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),而bn的通项公式为bn=(1/3)^n,因此an的通项为an=[1-(1/3)^n]/[1+(1/3)^n].