Question 12 Marks
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.$]$
$[$Hint: Two cases $(i)$ n is even, $(ii)$ n is odd.$]$
Answer
View full question & 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.
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.