Question 19:
Prove the following by using the principle of mathematical induction for all n ∈ N: n (n + 1) (n + 5) is a multiple of 3.
Answer
Let the given statement be P(n), i.e.,
P(n): n (n + 1) (n + 5), which is a multiple of 3.
It can be noted that P(n) is true for n = 1 since 1 (1 + 1) (1 + 5) = 12, which is a multiple of 3.
Let P(k) be true for some positive integer k,
i.e., k (k + 1) (k + 5) is a multiple of 3.
∴k (k + 1) (k + 5) = 3m, where m ∈ N … (1)
We shall now prove that P(k + 1) is true whenever P(k) is true.
Consider
(k + 1){(k + 1) + 1} {(k+1) + 5}
=(k + 1)(k + 2) {(k + 5) +1}
=(k + 1)(k + 2) (k + 5) + (k + 1) (k + 2)
={k(k+1)(k + 5) + 2(k + 1) (k + 5)} + (k + 1) (k + 2)
=3m + (k + 1){2(k + 5) + 2(k + 2)}
=3m + (k + 1){2k + 10 + k + 2}
= 3m + 3(k + 1) (3k + 12)
=3m + 3(k + 1) (k + 4)
=3{m + (k + 1) (k + 4)} = 3 ×q, where q = {m + (k + 1) (k + 4)} is some natural number
Therefore, (k + 1) {(k + 1) + 1} {(k + 1) + 5} is a multiple of 3.
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
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