F.4 A.maths

Let P(n) be n^3+5n is divisible by 6 for all natural numbers n .

When n = 1,

1^3+5(1)
=1+5
=6
which is divisible by 6
So P(0) is true.

Assume P(k) is true, i.e.k^3+5(k) is divisible, i.e.
k^3+5(k) = 3m where m is a natural number and k is any natural number.

When n = k+1,
(k+1)^3+5(k+1)
= (k^3+3k²+3k+1) + 5k + 5
= (k^3 + 5k) + 3k² + 3k + 1
= 6m + 3k² + 3k + 6 .............. by assumsion of P(k)
= 6m + 3(k)(k+1) + 6 .... (*)

As for k(k+1) either k or k+1 is even number,
so k(k+1) is divisible by 2,
so k(k+1) = 2T, where T = k(k+1)/2

(* continued)
= 6m + 6T + 6
= 6(m+T+1)
which is divisible by 6
So P(n+1) is true.

By Mathematical Induction, n^3+5n is divisible by 6 for all natural numbers n.

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小小補充，這題題目最難的地方應該是要從 3k(k+1) 這部份抽出 6 的因子。這個是需要像上面我寫一般去把 k(k+1) 先抽出 2 因子再做：

k(k+1) = 2T where T is a natural number
3k(k+1) = 6T
所以 3k(k+1) 可以被 6 整除。

Let p(n) be n^3 + 5n is divisible by 6.
p(1):
1^3 + 5*1 = 6 is divisible by 6
assume p(k) is true
We assume k^3 + 5*k = 6m where m is constant.
For n=k+1,
(k+1)^3 + 5(k+1)
= k^3 + 3*k^2 + 3k + 1 + 5k + 5
=[ k^3 + 5k] + 3*k^2 + 3k + 6
=6m + 3(k)(k+1) + 6
=6m + 6[(k)(k+1)/2] + 6
=6 * [m+1 + k*(k+1)/2 ]
=6 * [m+1 + (1+2+3+....+k) ] is divisible by 6
p(k+1) is true.
By MI, p(n) is true.