Answer
$a_{n}$ = $\frac{2^{n-1}}{3^{n-2}}$ * $-1^n$
Work Step by Step
We isolate the problem by looking at the numerator and denominator separately. While this is challenging to see in the first few terms, the latter terms make it easier to see how the numerator and denominator change.
We can see the pattern occur in the numerator as $a_{3}$ $a_{4}$ and $a_{5}$ have 4, 8 and 16 in the numerator, hinting an exponential change by 2.
The denominator shows a similar trend as $a_{3}$ $a_{4}$ and $a_{5}$ have 3, 9 and 27 in the numerator, hinting an exponential change by 3. However, the numerator and denominator change at different exponential rates, as the fifth term has $2^4$ in the numerator, but $3^3$ in the denominator. Thus, we know the exponential term of $2$ will be greater by $1$.
Tracing the problem back to the first term, $-3$, allows us to see that the first term should be $2^0$ or $1$ in the numerator and $3^-1$ or $1/3$ in the denominator. So, we get the fraction $\frac{2^{n-1}}{3^{n-2}}$.
Lastly, we have to factor in a $-1$ term, as the terms of the sequence alternate between positive and negative. Since the first term is negative, we can make the exponent that $-1$ is raised to $n$.
