数列an,bn满足an+1=an²/an+bn,bn+1=bn²/an+bn,a1=3,b1=1,(I)令C=an-bn,求Cn.(II)bn前n项和为Sn,证Sn<3/2.(详)
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![数列an,bn满足an+1=an²/an+bn,bn+1=bn²/an+bn,a1=3,b1=1,(I)令C=an-bn,求Cn.(II)bn前n项和为Sn,证Sn<3/2.(详)](/uploads/image/z/5303846-38-6.jpg?t=%E6%95%B0%E5%88%97an%2Cbn%E6%BB%A1%E8%B6%B3an%2B1%3Dan%26%23178%3B%2Fan%2Bbn%2Cbn%2B1%3Dbn%26%23178%3B%2Fan%2Bbn%2Ca1%3D3%2Cb1%3D1%2C%28I%29%E4%BB%A4C%3Dan-bn%2C%E6%B1%82Cn.%EF%BC%88II%29bn%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E8%AF%81Sn%EF%BC%9C3%2F2.%EF%BC%88%E8%AF%A6%EF%BC%89)
数列an,bn满足an+1=an²/an+bn,bn+1=bn²/an+bn,a1=3,b1=1,(I)令C=an-bn,求Cn.(II)bn前n项和为Sn,证Sn<3/2.(详)
数列an,bn满足an+1=an²/an+bn,bn+1=bn²/an+bn,a1=3,b1=1,(I)令C=an-bn,求Cn.
(II)bn前n项和为Sn,证Sn<3/2.(详)
数列an,bn满足an+1=an²/an+bn,bn+1=bn²/an+bn,a1=3,b1=1,(I)令C=an-bn,求Cn.(II)bn前n项和为Sn,证Sn<3/2.(详)
由An,Bn的递推式,及Cn的定义式,可知
C(n+1)=A(n+1)-B(n+1)=An^2/(An+Bn)-Bn^2/(An+Bn)=(An^2-Bn^2)/(An+Bn)=An-Bn=Cn.
即Cn为常数列.又C1=A1-B1=2,所以Cn=2.
由An,Bn的递推式,两式相除(已知An,Bn任意一项不为0),得
A(n+1)/B(N+1)=(An/Bn)^2.
由数学归纳法,可证A(n+1)/B(n+1)=(A1/B1)^(2^n)=3^(2^n).
代入Bn的递推式,得
B(n+1)=Bn/(3^(2^(n-1))+1)=…=B1/[(3^2+1)*(3*4+1)*(3^8+1)…*(3^(2^(n-1))+1)].
显然Bn为正数列,所以Sn是递增的.
下面证明Bn1.
注意到,(3^2+1)*(3*4+1)*(3^8+1)…*(3^(2^(n-2))+1)>(3^2)*3*3*…*3=3^(n-1).
所以就有Bn