241 views

1 vote

Let, $n = 2m+1; n>1;m \in \mathbb{Z},$ and $m>0$

Also let, $x = n(n^{2}-1)$

$\Rightarrow x = (2m+1)[ \,(2m+1)^{2}-1]$

$\Rightarrow x = (2m+1)(4m^{2}+1+4m-1)$

$\Rightarrow x = (2m+1)(4m^{2}+4m)$

$\Rightarrow \boxed{x = (2m+1)4m(m+1)}$

Now, we can substitute the various value of $m$

- $m = 1 \Rightarrow x = 3 \times 4 \times 2 = 24$
- $m = 2 \Rightarrow x = 5 \times 8 \times 3 = 120$
- $m = 3 \Rightarrow x = 7 \times 12 \times 4 = 336$
- $\vdots \quad \vdots \quad \vdots$

$\therefore x = n(n^{2}-1)$ is always divisible by $24.$

Correct Answer $:\text{C}$