彰師大105年度離散數學問題8?

(a)
等式左右皆乘以 n+1 , 等式恆成立:
(n+1) a_n = ( n+1 - 1 ) a_n-1
(n+1) a_n = n * a_n-1

2*a1 = a0
3*a2 = 2*a1
4*a3 = 3*a2
5*a4 = 4*a3
....................
n*a_n-1 = (n-1)a_n-2
(n+1)a_n = n*a_n-1

以上 n 個式子相加得 :
(n+1)a_n = a0 = 1
a_n = 1/(n+1) ..... Ans

(b)
a_n - 5*a_n-1 + 6*a_n-2 = 0
特徵方程式為 :
x² - 5x + 6 = 0
x = 2 , 3
故可設 a_n = p(2^n) + q(3^n)

代入初始值得 :
a0 = p + q = 1
a1 = 2p + 3q = 0
解得 p = 3 , q = - 2 , 因此:
a_n = 3(2^n) - 2(3^n) ..... Ans