Prove by using the principle of mathematical induction n(n + 1)(n + 2) is divisible by 6 for all n N.

Asked by Topperlearning User | 4th Jun, 2014, 01:23: PM

Expert Answer:

Let P(n): n(n + 1)(n + 2) is divisible by 6.
P(1): 1(1 + 1)(1 + 2) = 6 which is divisible by 6. Thus P(n) is true for n = 1.
Let P(k) be true for some natural number k.
i.e.  P(k): k(k + 1)(k + 2) is divisible by 6.
Now we prove that P(k + 1) is true whenever P(k) is true.
Now, (k + 1)(k + 2)(k + 3) = k(k + 1)(k + 2) + 3(k + 1)(k + 2)
Since, we have assumed that k(k + 1)(k + 2) is divisible by 6, also (k + 1)(k + 2) is divisible by 6 as either of (k + 1) and ( k + 2) has to be even number.
P(k + 1) is true.
Thus P(k + 1) is true whenever P(k) is true.
By principle mathematical induction n(n + 1)(n + 2) is divisible by 6 for all n  N. 

Answered by  | 4th Jun, 2014, 03:23: PM