对于函数f(x),若存在x 0 ∈R,使f(x 0 )=x 0 成立,则称x 0 为f(x)的不动点.如果函数 f(x)= x 2 +a bx-c (b,c∈ N * ) 有且仅有两个不动点0和2,且 f(-2)<- 1 2 . (1)求实数b,c的值; (2)已知各项不为零的数列{a n }的前n项之和为S n ,并且 4 S n ?f( 1 a n )=1 ,求数列{a n }的通项公式; (3)求证: (1- 1 a n ) a n+1 < 1 e <(1- 1 a n ) a n .
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