已知等差数列$\{a_n\}$满足:$|a_1|+|a_2|+\cdots+|a_n|=|a_1+1|+|a_2+1|+\cdots+|a_n+1|=|a_1-1|+|a_2-1|+\cdots+|a_n-1|=98$ 则$n$的最大值为_____
分析:注意到$|a_k+1|+|a_k-1|\ge2|a_k|$当$a_k\ge1\mp a_k\le -1$时等号成立.故
$2*98=|a_1+1|+|a_2+1|+\cdots+|a_n+1|+|a_1-1|+|a_2-1|+\cdots+|a_n-1|\ge 2(|a_1|+|a_2|+\cdots+|a_n|)=2*98$
故$a_k\ge1\mp a_k\le -1$,不妨设公差$d\ge0$且$a_1,\cdots a_l\le-1,a_{k+1}\cdots+ a_n\ge1$则
$|a_1|+|a_2|+\cdots+|a_n|=-(a_1+a_2+\cdots+a_k)+(a_{k+1}+\cdots a_n)=98$且
$|a_1-1|+|a_2-1|+\cdots+|a_n-1|=k-(a_1+\cdots a_k)+(n-k)+a_{k+1}+\cdots +a_n=98$故$n=2k$
$98=|a_1|+|a_2|+\cdots+|a_n|=-(a_1+a_2+\cdots+a_k)+(a_{k+1}+\cdots a_n)=k^2d$
注意到$d=a_{k+1}-a_k\ge2$
故$k^2=\dfrac{98}{d}\le49,n=2k\le14$