我又得到一個更快的解法了 !
a1=1 => 1
a2=1/3 => 1-(2/3)
a3=7/9 => 1-(2/3)+(2/3)^2
a4=13/27 => 1-(2/3)+(2/3)^2-(2/3)^3
..........
a101 = 1-(2/3)+(2/3)^2-(2/3)^3+ ......+(2/3)^100
= 1+(-2/3)+(-2/3)^2+(-2/3)^3+......+(-2/3)^100
= ( 用等比級數公式即可 )
= (3/5)[1+(2/3)^101]
= (3/5)[(3^101+2^101)/3^101]
= (3^101+2^101)/(5*3^100)

所以 答案依然是 A = 5 , B = 3 , C = 101

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孩子們, 別再問我為何每天都穿 KAPPA 了!

沒錯,

我也是得到這個解:
a_n = 1 - (2/3) A_{n-1}

解此式得
a_n = (3^n - (-1)^n (3-1)^n ) / (5* 3^{n-1})
當 n = 1 時 a_1 = 5/5 = 1.

所以當n = 101時,

a_101 = (3^101 + 2^101) / 5*3^100

因此 a = 5, b = 3, c = 101.