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解答


a1 = 10^1 + 3*4^1 + 5 = 27 = 9*3
a2 = 10^2 + 3*4^2 + 5 = 153 = 9*17
a3 = 10^3 + 3*4^3 + 5 = 1197 = 9*133


當n = 1時
a1 = 10^1 + 3*4^1 + 5 = 27 = 9*3
n = 1時成立

令n = k時成立
即ak = 10^k + 3*4^k + 5 = 9t (t為正整數)
則當n = k+1時
a(k+1)
= 10^(k+1) + 3*4^(k+1) + 5
= 10*10^k + 4*3*4^k + 5
= (10^k + 3*4^k + 5) + 9*10^k + 3*3*4^k
= 9t + 9(10^k) + 9(4^k)
= 9(t + 10^k + 4^k)
n = k+1時亦成立

由數學歸納法得證
對於所有的自然數n
an恆被9整除

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