Prove by direct method that for any integer ‘n’, n3 - n is always even. [Hint: Two cases (i) n is even, (ii) n is odd.]

Case I: When n s even. Let n = 2k, k ^3-n=(2k)^3-(2k)=2k(4k^2-1)=2 , Where = k(4k^2-1) Thus, (n3 - n) is even when n is even. Case II: When n is odd. Let n = 2k+1, k ^3-n=(2k+1)^3 -(2k+1) =(2k+1)[(2k+1)^2-1] =(2k+1)[4k^2+1+4k-1] =(2k+1)[4k^2+4k] =4k(2k+1)(k+1)=2 , Where =2k(k+1)(2k+1) Thus, n3 - n is even when n is odd. So, n3 - n is always even.

Prove by direct method that for any integer ‘n’, n3 - n is always…

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Prove by direct method that for any integer ‘n’, n³ - n is always even.
[Hint: Two cases (i) n is even, (ii) n is odd.]

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Answer

Case I: When n s even.
Let $\text{n} = 2\text{k}, \text{k}\in\text{N}$
$\Rightarrow\text{n}^3-\text{n}=(2\text{k})^3-(2\text{k})=2\text{k}(4\text{k}^2-1)=2\lambda,$ Where $\lambda = \text{k}(4\text{k}^2-1)$
Thus, (n³ - n) is even when n is even.
Case II: When n is odd.
Let $\text{n} = 2\text{k}+1, \text{k}\in\text{N}$
$\Rightarrow\text{n}^3-\text{n}=(2\text{k}+1)^3 -(2\text{k}+1)$
$=(2\text{k}+1)[(2\text{k}+1)^2-1]$
$=(2\text{k}+1)[4\text{k}^2+1+4\text{k}-1]$
$=(2\text{k}+1)[4\text{k}^2+4\text{k}]$
$=4\text{k}(2\text{k}+1)(\text{k}+1)=2\mu,$ Where $\mu=2\text{k}(\text{k}+1)(2\text{k}+1)$
Thus, n³ - n is even when n is odd.
So, n³ - n is always even.

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