Question -
Answer -
Let P (n): n (n + 1) (n + 5) is a multiple of 3
Let us check for n = 1,
P (1): 1 (1 + 1) (1 + 5)
: 2 × 6
: 12
P (n) is true for n = 1. Where, P (n) is a multiple of 3
Now, let us check for P (n) is true for n = k, and have toprove that P (k + 1) is true.
P (k): k (k + 1) (k + 5) is a multiple of 3
: k(k + 1) (k + 5) = 3λ … (i)
We have to prove,
(k + 1)[(k + 1) + 1][(k + 1) + 5] is a multiple of 3
(k + 1)[(k + 1) + 1][(k + 1) + 5] = 3μ
So,
= (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 + 3k2 + 15k + 12
= 3(λ + k2 + 5k + 4)
= 3μ
P (n) is true for n = k + 1
Hence, P (n) is true for all n ∈ N.