已知数列an满足a1=1,且an=2an-1+2^n 1、求证an/2^n是等差数列2、求an的通项公式

n = an/2^(n-1)b = a/2^(n-2)bn - b = an/2^(n-1) - a/2^(n-2)= (an - 2a )/2^(n-1)把 已知条件 a = 2an+2^n 即 an = 2a + 2^(n-1) 代入上式bn - b = 2^(n-1)/2^(n-1)= 1因此 bn 是等差数列