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)

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