數學歸納法(急!!)

其實只要能證明b是對的，就能證明a是對。因為一個數能被6整除，是一定能被3整除的。(因3是6的因數)

設P(n)為命題，對於所有n

n(n + 1)(n + 2)能被6整除

當n = 1

(1)(1 + 1)(1 + 2) = 6 = 6 X 1

所以P(1)正確

假設P(k)正確

k(k + 1)(k + 2) = 6N，N為一整數

當n = k + 1

(k + 1)(k + 2)(k + 3)

= k(k + 1)(k + 2) + 3(k + 1)(k + 2)

= 6N + 3(k + 1)(k + 2)

對於自然數k，(k + 1)(k + 2)必有一個為雙數

即(k + 1)(k + 2)可寫成2M，M為一自然數

所以6N + 3(k + 1)(k + 2) = 6N + 3(2M) = 6(N + M)

能被6整除

所以，根據數學歸納法，P(n)對於所有自然數n皆正確。

a.) prove n(n+1)(n+2) is divisible by 3 for all n are natural numbers, we have
n(n+1)(n+2) = 3n
1) for n=0, we have
L.H.S.= 0 (0+1)(0+2)
= 0
The statement is true for n=0

2) for n=1, we have
L.H.S. = 1 (1+1)(1+2)
= 2(3)
The statement is true for n=1

3) Assume the statement is also true when n=k for any natural number k, we have
k(k+1)(k+2) = 3k

4) when n=k+1, we have
L.H.S. = (k+1)(k+1+1)(k+1+2)
= (k+1)(k+2)(k+3)
= k(k+1)(k+2) + 3(k+1)(k+2)
= 3k + 3(k+1)(k+2)
= 3 [k + (k+1)(k+2)]
The statement is also true when n=k+1

By mathematical induction, we proof that n(n+1)(n+2) is divisible by 3 for any natural number n.



b.) prove n(n+1)(n+2) is divisible by 6 for all n are natural numbers, we have
n(n+1)(n+2) = 6n
1) for n=0, we have
L.H.S.= 0 (0+1)(0+2)
= 0
The statement is true for n=0

2) for n=1, we have
L.H.S. = 1 (1+1)(1+2)
= 6
The statement is true for n=1

3) Assume the statement is also true when n=k for any natural number k, we have
k(k+1)(k+2) = 6k

4) when n=k+1, we have
L.H.S. = (k+1)(k+1+1)(k+1+2)
= (k+1)(k+2)(k+3)
= k(k+1)(k+2) + 3(k+1)(k+2)
= 6k + 6(3)
= 6 (k+3)
The statement is also true when n=k+1

By mathematical induction, we proof that n(n+1)(n+2) is divisible by 6 for any natural number n.