题干

对于数列A:a1,a2,…,an,若满足ai∈{0,1}(i=1,2,3,…,n),则称数列A为“0﹣1数列”.若存在一个正整数k(2≤k≤n﹣1),若数列{an}中存在连续的k项和该数列中另一个连续的k项恰好按次序对应相等,则称数列{an}是“k阶可重复数列”,例如数列A:0,1,1,0,1,1,0.因为a1,a2,a3,a4与a4,a5,a6,a7按次序对应相等,所以数列{an}是“4阶可重复数列”.


(Ⅰ)分别判断下列数列A:1,1,0,1,0,1,0,1,1,1.是否是“5阶可重复数列”?如果是,请写出重复的这5项;

(Ⅱ)若项数为m的数列A一定是“3阶可重复数列”,则m的最小值是多少?说明理由;

(III)假设数列A不是“5阶可重复数列”,若在其最后一项am后再添加一项0或1,均可使新数列是“5阶可重复数列”,且a4=1,求数列{an}的最后一项am的值.

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答案(点此获取答案解析)

解:(Ⅰ)是“5阶可重复数列”,10101.

(Ⅱ)因为数列{an}的每一项只可以是0或1,所以连续3项共有23=8种不同的情形.

若m=11,则数列{an}中有9组连续3项,则这其中至少有两组按次序对应相等,即项数为11的数列{an}一定是“3阶可重复数列”;若m=10,数列0,0,1,0,1,1,1,0,0,0不是“3阶可重复数列”;则3≤m<10时,均存在不是“3阶可重复数列”的数列

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