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Mathematical Reasoning — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsMathematical Reasoning2 Marks
Question
Prove by direct method that for any integer $‘n’, n^3 - n$ 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, $(n^3 - n)$ 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, $n^3 - n$ is even when n is odd. So, $n^3 - n$ is always even.

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