回復 1# Exponential 的帖子
a_n = (1/2)a_(n-1) + (1/2)b_(n-1)
b_n = (1/2)a_(n-1) + (1/2)c_(n-1)
c_n = (1/2)b_(n-1) + (1/2)c_(n-1)
a_n - c_n = (1/2)a_(n-1) - (1/2)c_(n-1)
a_1 = c_1 = 0
a_n = c_n
b_n = a_(n-1)
a_n = (1/2)a_(n-1) + (1/2)a_(n-2)
由特徵方程可得 a_n = (1/3) - (1/3)(-1/2)^(n-1)
