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a[n+1] = 3a[n] + 1. 也就是說,數列的每個數等於前面那個數的三倍加上1. 這個可以解出來通項 a[n+1] + 1/2 = 3a[n] + 3/2. 所以,a[n] + 1/2是等比數列 a[n] = aq^n - 1/2 a[1]=0,1/2 = aq a[2]=1,3/2 = aq^2 q = 3,a = 1/6 所以通項a[n] =(1/6)* 3^n - 1/2.
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