Prove that by direct method for any integer “n”, n3 – n is always even.
Case 1: When n is even.
Let n = 2 k, k ∈ N
⇒ n3 – n = (2k)3 – (2k)
= 2k(4k2 – 1)
= 2α
Where α = k(4k2 – 1)
Thus n3 – n is even when n is even.
Case 2: When n is odd.
Let n = 2k + 1, k ∈ N
⇒ n3 – n = (2k + 1)3 – (2k + 1)
= (2k + 1)[ (2k + 1)2 – 1]
= (2k + 1)[ (4k2 + 1 + 4k – 1]
= (2k + 1)[ (4k2 + 4k ]
= 4k(2k + 1)(k + 1)
= 2 β
Where β = 2k(2k + 1)(k + 1)
Thus n3 – n is even when n is odd.
Hence n3 – n is always even.
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