Question

Let $P\left(n\right)=n(n^2+20)$ is divisible by $48$, then for which of the following $P\left(n\right)$ is true

• A. $\forall n\in N$
• B. $\forall n\in$ odd natural number
• C. $\forall n\in$ even natural numbers
• D. None of these

Answer 1

The correct option is C $\forall n\in$ even natural numbers

$P\left(n\right)=n(n^2+20)$

When $n$ is even, then $n(n^2+20)=2k({2k}^2+20)=2k\times 4\times (k^2+5)$ divisible by $8$ for sure.

Let $n=8$

$P\left(8\right)=8(8^2+20)=8\left(84\right)=672$ is divisible by $48$.

Let $n=17$

$P\left(17\right)=17({17}^2+20)=17\left(309\right)=5253$ is not divisible by $48$.

Let $n=26$

$P\left(26\right)=26({26}^2+20)=26\left(696\right)=18096$ is divisible by $48$.

Let $n=9$

$P\left(9\right)=9(9^2+20)=9\left(101\right)=909$ is not divisible by $48$.

So, we can conclude that for all even $n$, it is divisible by $48$.