Let P(n): n(n + 1) (n + 5) is multiple by 3 for all n∈N
Let P(n) is true for n=1
P(1): 1(1 + 1) (1 + 5)
= 2 × 6
= 12
Since, it is multiple of 3
So, P(n) is true for n = 1
Now, Let P(n) is true for n = k
P(k): k(k + 1) (k + 5)
= k(k + 1) (k + 5) is a multiple of 3
Then, k(k + 1) (k + 5) = 3λ - - - - - (1)
We have to show,
= (k + 1)[(k + 1) + 1][(k + 1) + 5] is a multiple of 3
= (k + 1)[(k + 1) + 1][(k + 1) + 5] = 3μ
Now,
= (k + 1)[(k + 1) + 1][(k + 1) + 5]
= (k + 1)(k + 2)[(k + 1) + 5]
= [k(k + 1) + 2(k + 1)][(k + 5) + 1]
= k(k + 1)(k + 5) + k(k + 1) + 2(k + 1)(k + 5) + 2(k + 1)
= 3λ + k2 + k + 2(k2 + 6k + 5) + 2k + 2
= 3λ + k2 + k + 2k2 + 12k + 10 + 2k + 2
= 3λ + 3k 2 + 15k + 12
= 3(λ + k 2 + 5k + 4)
= 3μ
Therefore, P(n) is true for n = k + 1
Hence, P(n) is true for all n∈N