已知An=2A(n-1)+2^n-1(n≥2）a1=5 a2=13 a3=33 a4=81.{（An+α）/2^n}为等差数列,求a

An=2A(n-1)+2^n-1 等式两边同时除以2^n可得 An/2^n=A(n-1)/2^(n-1)-1/2^n+1 ……①{（An+α）/2^n}为等差数列,所以（An+α）/2^n=（A（n-1）+α）/2^（n-1）+d （d为参数）整理得An/2^n=A（n-1）+α/2^n+d ……②对...