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Question
Prove by direct method that for any integer ‘n’, n3 – n is always even.
Solution
We have two cases:
Case I: If n is even
Let n = 2k where k ∈ N
∴ n3 – n = (2k)3 – (2k)
= 2k(4k2 – 1)
= 2m
Where m = k(4k2 – 1)
Therefore (n3 – n) is even.
Case II: If 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 + 4k + 1 – 1]
= (2k + 1) (4k2 + 4k)
= 4k(2k + 1)(k + 1)
= 2.2k(2k + 1)(k + 1)
= 2λ
Where λ = 2k(2k + 1)(k + 1)
Therefore n3 – n is even.
Hence, n3 – n is always even.
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