Show that $n(n^2-1)$ is divisible by 24, if $n$ is an odd integer greater than $2$. How can I show that $n(n^2-1)$ is divisible by 24, if $n$ is an odd integer greater than $2$?
I am thinking that since odd numbers have the form of $2n-1$ in which if it is to be more than $2$, it will be $2n-1+1 = 2n+1$. So would it be correct to use this and try solving through induction?

Induction on odd numbers is easier when expressed as $P(n_0)$ is true, $P(n) 
\implies P(n+2)$.
Let $f(n)=n(n^2-1)$. Then $f(1)=24$ and $f(n+2)-f(n)=6 (n + 1)^2$.
If $n$ is odd, then $n+1$ is even and so $(n+1)^2$ is a multiple of $4$. 
Therefore, $f(n+2)-f(n)$ is a multiple of $24$ and so is $f(n+2)$ by induction.
For fun, here is a different approach using finite differences.
Write $n=2t+1$. Then
$$
n(n^2-1) = 8 t^3 + 12 t^2 + 4 t
= 24 \binom{t}{1} + 72 \binom{t}{2} + 48 \binom{t}{3} 
$$
clearly a multiple of $24$.
