在数列an中,a1=1,且对任意实数n∈N*,都有,an+1=an+2^n,(1)求证:数列an/2^n是等差数列;(2)设数列an的前n项和为sn,求证:对任意的n∈N*,都有s(n+1)-4an=1

问题描述:

在数列an中,a1=1,且对任意实数n∈N*,都有,an+1=an+2^n,
(1)求证:数列an/2^n是等差数列;
(2)设数列an的前n项和为sn,求证:对任意的n∈N*,都有s(n+1)-4an=1