Answer
$a_{n}= \frac{2^{n-1}}{3(n+2)},\quad n=1,2,3,...$
Work Step by Step
$n=1,\displaystyle \qquad a_{1}=\frac{2^0}{3\cdot 3}\qquad $
$n=2,:\displaystyle \qquad a_{2}=\frac{2^{1}}{3\cdot 4} $
$n=3,:\displaystyle \qquad a_{3}=\frac{2^{2}}{3\cdot 5} $
$n=4,:\displaystyle \qquad a_{4}=\frac{2^{3}}{3\cdot 6} \qquad $... a pattern emerges, $a_{n}=\displaystyle \frac{2^{n-1}}{3(n+2)}$
Testing $a_{5},$
$ a_{5}=\displaystyle \frac{2^{5-1}}{3(5+2)}=\frac{2^{4}}{21},\qquad$... yes, the pattern works.
Testing on the previous terms:
$ a_{1}=\displaystyle \frac{2^{1-1}}{3(1+2)}=\frac{1}{9},\qquad$... yes.
$ a_{2}=\displaystyle \frac{2^{2-1}}{3(2+2)}=\frac{2}{12},\qquad$... yes.
$ a_{3}=\displaystyle \frac{2^{3-1}}{3(3+2)}=\frac{2^{2}}{15},\qquad$... yes.
The proposed formula works for all the listed terms.
$a_{n}=\displaystyle \frac{2^{n-1}}{3(n+2)},\quad n=1,2,3,...$