作业帮设数列{An}满足下列关系：a1=2a,An=2a-[a^2/(An-1)];Bn=1/(An-a),求证：（1）An≠a;(2)Bn是等差数列
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![设数列{An}满足下列关系：a1=2a,An=2a-[a^2/(An-1)];Bn=1/(An-a),求证：（1）An≠a;(2)Bn是等差数列](/uploads/image/z/13937262-6-2.jpg?t=%E8%AE%BE%E6%95%B0%E5%88%97%7BAn%7D%E6%BB%A1%E8%B6%B3%E4%B8%8B%E5%88%97%E5%85%B3%E7%B3%BB%EF%BC%9Aa1%3D2a%2CAn%3D2a-%5Ba%5E2%2F%28An-1%29%5D%3BBn%3D1%2F%28An-a%29%2C%E6%B1%82%E8%AF%81%EF%BC%9A%EF%BC%881%EF%BC%89An%E2%89%A0a%3B%282%29Bn%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97)
设数列{An}满足下列关系：a1=2a,An=2a-[a^2/(An-1)];Bn=1/(An-a),求证：（1）An≠a;(2)Bn是等差数列
设数列{An}满足下列关系：a1=2a,An=2a-[a^2/(An-1)];Bn=1/(An-a),求证：（1）An≠a;(2)Bn是等差数列
设数列{An}满足下列关系：a1=2a,An=2a-[a^2/(An-1)];Bn=1/(An-a),求证：（1）An≠a;(2)Bn是等差数列
1、反证法
假设A[n]=a,那么代入等式a=2*a-(a^2/A[n-1])
得出 A[n-1]=A[n]=a
因此可以推出 A[1]=A[2]=...=A[n]=a与题中A[1]=2a矛盾
2、
B[n]-B[n-1]=(A[n-1]-A[n])/(A[n]*A[n-1]-a*(A[n]+A[n-1])+a^2)
由A[n]*A[n-1]=2*a*A[n-1]-a^2带入上式
B[n]-B[n-1]=(A[n-1]-A[n])/(a*(A[n-1]-A[n]))=1/a=常数
因此 B[n]是等差数列
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