问题 :已知数列 {an},a1=1 ,an =2an-1(n≥ 2 ,n∈N ) ,求an.易知 {an}是等比数列 ,公比为 2 ,所以an =2 n-1.若将上述问题改为 :【例 1】　已知数列 {an},a1=1 ,an =2an-1+ 1 (n≥ 2 ,n∈N ) .求an.我们可以通过化归思想 ,利用待定系数法转化为等比数列问题 .事实上 ,我们希望 Question: Known series {an}, a1=1, an = 2an-1(n≥2, n∈N), find an. Easy to know {an} is a geometric sequence with a common ratio of 2, so an = 2 N-1. If the above problem is changed to: [Example 1] Known sequence {an}, a1=1, an = 2an-1 + 1 (n≥2, n∈N) . Seeking an. We can pass Return to the idea, use the undetermined coefficient method to convert to the problem of geometric progression. In fact, we hope that