已知数列an满足a1=1/5,且当n>=2,n属于正整数时,有(a(n-1))/(an)=(2a(n-1)+1)/(1-
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已知数列an满足a1=1/5,且当n>=2,n属于正整数时,有(a(n-1))/(an)=(2a(n-1)+1)/(1-2an)
(1)求数列an的通项an
(2)试问a1a2是否是数列an中的项?如果是,是第几项,如果不是,说明理由
(1)求数列an的通项an
(2)试问a1a2是否是数列an中的项?如果是,是第几项,如果不是,说明理由
![已知数列an满足a1=1/5,且当n>=2,n属于正整数时,有(a(n-1))/(an)=(2a(n-1)+1)/(1-](/uploads/image/z/7548955-43-5.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97an%E6%BB%A1%E8%B6%B3a1%EF%BC%9D1%2F5%2C%E4%B8%94%E5%BD%93n%3E%3D2%2Cn%E5%B1%9E%E4%BA%8E%E6%AD%A3%E6%95%B4%E6%95%B0%E6%97%B6%2C%E6%9C%89%28a%28n-1%29%29%2F%28an%29%3D%282a%28n-1%29%2B1%EF%BC%89%2F%281-)
(a(n-1))/(an)=(2a(n-1)+1)/(1-2an)
=>a(n-1)*(1-2an)=an*(2a(n-1)+1)
=>a(n-1)-2ana(n-1)=2ana(n-1)+an
=>a(n-1)-an=4ana(n-1)
=>1/an-1/a(n-1)=4
设bn=1/an
=>bn-b(n-1)=4,且b1=1/a1=5
=>bn=5+4(n-1)=4n+1
=>an=1/bn=1/(4n+1)
=>a1a2=1/5*1/9=1/45=1/(4*11+1
=>n=11
=>第十一项
=>a(n-1)*(1-2an)=an*(2a(n-1)+1)
=>a(n-1)-2ana(n-1)=2ana(n-1)+an
=>a(n-1)-an=4ana(n-1)
=>1/an-1/a(n-1)=4
设bn=1/an
=>bn-b(n-1)=4,且b1=1/a1=5
=>bn=5+4(n-1)=4n+1
=>an=1/bn=1/(4n+1)
=>a1a2=1/5*1/9=1/45=1/(4*11+1
=>n=11
=>第十一项
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