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Question
Prove the statement by using the Principle of Mathematical Induction:
n(n2 + 5) is divisible by 6, for each natural number n.
Solution
P(n) = n(n2 + 5) is divisible by 6.
So, substituting different values for n, we get,
P(0) = 0(02 + 5) = 0 Which is divisible by 6.
P(1) = 1(12 + 5) = 6 Which is divisible by 6.
P(2) = 2(22 + 5) = 18 Which is divisible by 6.
P(3) = 3(32 + 5) = 42 Which is divisible by 6.
Let P(k) = k(k2 + 5) be divisible by 6.
So, we get,
⇒ k(k2 + 5) = 6x
Now, we also get that,
⇒ P(k + 1) = (k + 1)((k + 1)2 + 5) = (k + 1)(k2 + 2k + 6)
= k3 + 3k2 + 8k + 6
= 6x + 3k2 +3k + 6
= 6x + 3k(k + 1) + 6[n(n + 1) is always even and divisible by 2]
= 6x + 3 × 2y + 6 Which is divisible by 6.
⇒ P(k + 1) is true when P(k) is true.
Therefore, by Mathematical Induction,
P(n) = n(n2 + 5) is divisible by 6, for each natural number n.
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