By the principle of mathematical induction, prove n(n + 1) (n + 2) is

1.	By the principle of mathematical induction, prove n(n + 1) (n + 2) is divisible by 6, for all n ∈ N.
Answer» P(n): n(n + 1) (n + 2) is divisible by 6. P(1): 1 (2) (3) = 6 is divisible by 6 ∴ P(1) is true. Let us assume that P(k) is true for n = k That is, k (k + 1) (k + 2) = 6m for some m To prove P(k + 1) is true i.e. to prove (k + 1) (k + 2) (k + 3) is divisible by 6. P(k + 1) = (k + 1) (k + 2) (k + 3) = (k + 1)(k + 2)k + 3(k + 1)(k + 2) = 6m + 3(k + 1)(k + 2) In the second term either k + 1 or k + 2 will be even, whatever be the value of k. Hence second term is also divisible by 6. ∴ P (k + 1) is also true whenever P(k) is true. By Mathematical Induction P (n) is true for all values of n.

By the principle of mathematical induction, prove n(n + 1) (n + 2) is divisible by 6, for all n ∈ N.

Answer»

P(n): n(n + 1) (n + 2) is divisible by 6.

P(1): 1 (2) (3) = 6 is divisible by 6

∴ P(1) is true.

Let us assume that P(k) is true for n = k

That is, k (k + 1) (k + 2) = 6m for some m

To prove P(k + 1) is true i.e. to prove (k + 1) (k + 2) (k + 3) is divisible by 6.

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

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

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

In the second term either k + 1 or k + 2 will be even, whatever be the value of k.

Hence second term is also divisible by 6.

∴ P (k + 1) is also true whenever P(k) is true.

By Mathematical Induction P (n) is true for all values of n.