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?

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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?
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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?

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?
规律如下
An=6*（n+1）+n
当An=2001
n=285

由规律可知：6*(n+1)+n=2001
解得：n=285

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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；

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