Prove that the product of three consecutive positive integers is divisible by 6.

Explanation

Let `P(n)=n(n+1)(n+2)` , where n is a positive integer
Step I For `n=1`,
`P(1)=1(1+1)(1+2)=1.2.3=6`, which is divisible by 6.
Therefore , the result is true for `n=1`.
Step II Let us assume that the result in true for `n=k`, where k is a positive integer
Then , `P(k)=k(k+1)(k+2)` is divisible by 6.
`rArr P(k)=6r`, where r is an integer .

Step III For `n=k+1`, where k is a positive integer .
`P(k+1)=(k+1)(k+1+1)(k+2+1)`
`=(k+1)(k+2)(k+3)`
Now , `P(k+1)-P(k)=(k+1)(k+2)(k+3)-k(k+1)(k+2)`
`=(k+1)(k+2)(k+3-k)`
`=3(k+1)(k+2)`
`rArr P(k+1)=P(k)+3(k+1)(k+2)`
But we know that , `P(k)` is divisible by 6. Also , `39k+1)(k+2)` is divisible by 6 for all positive integer. This shows that the result is true for `n=k+1`. Hence , by the principle of mathematical induction , the result is true for all positive integer.