Let $P = n(n+1) (2n+1)$
Now, we can check all of the option.
- It is even
Let’s take various values of $n$ and check.
- $n = 1: P = 1 (2)(3) = 6$
- $n = 2: P = 2 (3)(5) = 30$
- $n = 3: P = 3 (4) (7) = 84$
$\qquad \downarrow$
- $P = n[\text{Odd}] \; (n+1)[\text{Even}] \; (2n+1)[\text{always Odd}] = \text{Even} \quad [\text{Odd} \times \text{Even = Even}]$
So, option $A$ is correct.
- Divisible by $3.$
Let’s take various of $n$ and check.
- $n = 1: P = 1 (2)(3) = 6 \Rightarrow \text{Yes}$
- $n = 2: P = 2 (3)(5) = 30 \Rightarrow \text{Yes}$
- $n = 3: P = 3 (4) (7) = 84 \Rightarrow \text{Yes}$
$\qquad \downarrow$
- $P = n(n+1) (2n+1) \Rightarrow \text{Yes}$
So, option $B$ is correct.
Divisible by the sum of the square of square of first $n$ natural numbers.
The sum of the square of first $n$ natural numbers $ = 1^{2} + 2^{2} + 3^{2} + n^{2} = \frac{n(n+1)(2n+1)}{6}$
$ = \frac{P}{6}$
So, $\frac{P}{\frac{P}{6}} = 6.$
So, option $C$ is correct.
- Never divisible by $237.$
Let take $n = 236$
$P = n(n+1) (2n+1)$
$P = 236 (237) [2(236)+1]$
It is divisible by $237$
We can also take $n = 118$
$P = 118 (119)(237)$
So, option $D$ is not correct.
Correct Answer$: \text{D}$