If the $n^{th}$ partial sum of ${\displaystyle \underset{n=1}{\overset{}{}}}{a}_n$ is given by

$a_n=\frac{8}{(n+3)(n+2)}$

$S_n=\frac{3n+1}{n+3}$

$a_n=\frac{8}{n(n+3)}$

what is $a_n$ when $n2?$

$a_n=\frac{10}{(n+3)(n+4)}$

$a_n=\frac{10}{n(n+3)}$

$a_n=\frac{10}{(n+3)(n+2)}$

$a_n=\frac{8}{(n+3)(n+4)}$