Answer
The sequence is not monotonic and not bounded.
Work Step by Step
$a_{n}=(-2)^{n+1}$
{$a_{n}$} is increasing if $a_{n} \lt a_{n+1}$ for all $n \geq 1$
{$a_{n}$} is decreasing if $a_{n} \gt a_{n+1}$ for all $n \geq 1$
{$a_{n}$} is monotonic if it is either increasing or decreasing.
The odd terms of the sequence result in $4,16,64,...$ and are increasing.
The even terms of the sequence result in $-8, -32,-128,...$ and are decreasing.
Therefore {$a_{n}$} is not monotonic.
The terms of the sequence are $4,-8,16,-32,64,...$
By definition, {$a_{n}$} is bounded above if there is a number $M$ such that $a_{n} \leq M$ for all $n \geq 1$
and is bounded below if there is a number $m$ such that $m \leq a_{n}$ for all $n\geq 1$
The odd terms $4,16,64...$ are bounded below by 4 and the even terms $-8,-32,-128,...$ are bounded above by -8.
Therefore, the sequence $a_{n}=(-2)^{n+1}$ is not bounded.