Question

有一列数a1，a2，a3，a4。。。an-1，an其中a1=6*2+1，a2=6*3+2，。。。当an=2001时，n等于多少？

有一列数a1，a2，a3，a4。。。an-1，an其中a1=6*2+1，a2=6*3+2，。。。当an=2001时，n等于多少？

Answer 1

由a1=6*2+1，a2=6*3+2，。。。可知an=6*(n+1)+n;
当an=2001时，6*(n+1)+n=2001；求出n=285；

Answer 2

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Answer 3

规律如下
An=6*（n+1）+n

当An=2001
n=285

Answer 4

由规律可知：6*(n+1)+n=2001
解得：n=285