An Odd (and Even) Conundrum

$\left( n^2 + 1 \right) \left( 4n^2 + 4n + 1 \right) = \left( n^2 + 1 \right) {\left( 2n+1 \right)}^2$

Now $2n+1$ is always odd for $n \in \mathbb{N}$ and so ${\left( 2n+1 \right)}^2$ is also odd.

And

$n^2 +1 \begin{cases} \text{odd} & \text{, if } n \text{ is even} \\ \text{even} & \text{, if } n \text{ is odd} \end{cases}$

So, we conclude that

$\left( n^2 + 1 \right) \left( 4n^2 + 4n + 1 \right) \begin{cases} \text{odd} & \text{, if } n \text{ is even} \\ \text{even} & \text{, if } n \text{ is odd} \end{cases}$

Hence, the appropriate choice is $\boxed{\text{Sometimes Odd and Sometimes Even}}$ .