Question

Show that $n(n^2-1)(3n+2)$ is divisible by $24$.

Answer 1

$n(n^2-1)(3n+2)$

$n$ should be greater than equal to $2$

$ifn=2$

$p\left(n\right)=2\cdot (2^2-1)\cdot (3\cdot 2+2)$

$2\times 3\times 8$

$=48$ which is divisible by $24$

So, $p\left(m\right)=m\cdot (m^2-1)\cdot (3m+2)isalsodivisibleby24$

Now, $p\left(m\right)=(m+1)\cdot [{(m+1)}^2-1]\cdot [3(m+1)+2]$

$=(m+1)\cdot \left[(m^2+2m)(3m+5)\right]$

$=m(m+1)(m+2)(3m+5)$

$=m(m+1)(m+2)(3m+2+3)$

Since $p\left(m\right)$ is divisible by $24,$ then $p(m+1)$ is also true