How to find and write the first four terms of the sequences below with the...

${\displaystyle a_n={(-1)}^{n+1}(n+4)}$
5,6,7, and 8 are the first four terms in the sequence with the general term (n+4).
${\displaystyle {(-1)}^{n+1}}$ simply means that terms will be positive when they are odd and negative when they are even.
Therefore, ${\displaystyle a_{n=1\to 4}={(-1)}^{n+1}(n+4)=+5,-6,+7,-8}$
${\displaystyle a_n={(-1)}^{n-1}{n}^2}$
The first 4 terms of the sequence with general term ${\displaystyle n^2}$ are: 1,4,9,16
${\displaystyle {(-1)}^{n-1}}$ also means that odd terms will be positive and even terms will be negative. [Since, ${\displaystyle {(-1)}^0=1}$]
Thus, ${\displaystyle a_{n=1\to 4}={(-1)}^{n-1}{n}^2=+1,-4,+9,-16}$