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.

2006-11-17 21:43:23 補充：
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 number3k(k+1) = 6T所以 3k(k+1) 可以被 6 整除。

when n=1
1^3+5*1=6
hence it is divisible by 6
suppose n=k is true,k^3+5k is divisible by 6
for n=k+1
(k+1)^3+5(k+1)
K^3+3k^2+3k+1+5k+5
=k^3+5k + 3k^2+3k+6
since 3k^2+3k=3k(k+1) is divisivle by 6 ,since k(k+1)must be even

hence n=k+1 is true
se,n^3+5n is divisible by 6 ,for any positive n