数列an中 a1=0 a2=2 a（n+2）-6a（n+1）+5an=2^n 求通项an

1、设b[n]=a[n]+⅓×2^n,那么
b[1] = a[1] + ⅓×2 = ⅔
b[2] = a[2] + ⅓×2^2 = 10/3
且有 a[n] = b[n] - ⅓×2^n,代入 a[n+2] - 6a[n+1] + 5a[n] = 2^n 并整理,有
b[n+2] - 6b[n+1] + 5b[n] = 0
2、数列 b[n] 的通项公式可以由特征方程给出
b[n] = 5^n × 2/15
或者这样得到：
由b[n+2] - 6b[n+1] + 5b[n] = 0有 b[n+2] - b[n+1] = 5(b[n+1] - b[n]),所以
c[n] = b[n+1] - b[n] 是比例为5的等比数列,且 c[1] = b[2] - b[1] = 8/3,
所以c[n] = 8/3 × 5^(n-1),从而
b[n] = c[n-1] + b[n-1] = c[n-1] + c[n-2] + b[n-2] = ...= c[n-1] + c[n-2] + c[n-3] + ...+c[1] + b[1]
=5^n × 2/15
3、由b[n]的通项公式,可以得到a[n]的通项公式为
a[n] = 5^n × 2/15 - ⅓×2^n