n(n+1) (n+5) ia a multiple of 3.

Let P(n) = n(n+1) (n+3)
P(1) is a multiple of 3
Let P(k) be a multiple of 3
(i.e) P(k) = K(k+1) (k+5) = 3m…………..(A)
To prove P(k+1) is a multiple of 3 using result…….(A)
P(k+1) = {(k+1)(k+2)}(k+6)
          = {k(k+1)+2(k+1)}(k+5+1)}
          = k(k+1)(k+5) + 2(k+1) (k+5) + k(k+1) +2(k+1)
          = 3m +
          = 3m (1+ )
which is divisible by 3
hence the result.
P(K+1) is true.
By the Principle of mathematical induction, P(n) is true for all values of n where n N
Hence proved.