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Question
Prove the statement by using the Principle of Mathematical Induction:
n3 – n is divisible by 6, for each natural number n ≥ 2.
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Solution
P(n) = n3 – n is divisible by 6.
So, substituting different values for n, we get,
P(0) = 03 – 0 = 0 Which is divisible by 6.
P(1) = 13 – 1 = 0 Which is divisible by 6.
P(2) = 23 – 2 = 6 Which is divisible by 6.
P(3) = 33 – 3 = 24 Which is divisible by 6.
Let P(k) = k3 – k be divisible by 6.
So, we get,
⇒ k3 – k = 6x
Now, we also get that,
⇒ P(k + 1) = (k + 1)3 – (k + 1)
= (k + 1)(k2+ 2k + 1 − 1)
= k3 + 3k2 + 2k
= 6x + 3k(k + 1) ......[n(n + 1) is always even and divisible by 2]
= 6x + 3 × (2y) Which is divisible by 6, where y = k(k + 1)
⇒ P(k + 1) is true when P(k) is true.
Therefore, by Mathematical Induction,
P(n) = n3 – n is divisible by 6, for each natural number n.
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