$S = \frac{3t^4+3t^3}{2} \cdot \frac{t^{3n}-(t+2)^{n}}{t^3-t-2}$
$ t>1, n>1$
Intuitively, it feels like this expression behaves "like" $t^{3n+1}$, but it seems that some constant is lacking, but I don't see how to formally come up with some asymptotics.
Your intuition is good $$S = \frac32(t^4+t^3) \times \frac{t^{3n}-(t+2)^{n}}{t^3-t-2}$$ $$S= \frac32 t^4 \left(1+\frac{1}{t}\right)\times t^{3n-3}\times\frac {t^3-\frac{(t+2)^n}{ t^{3n-3} } }{t^3-t-2}$$ So, when $t$ is large and $n>1$ the last term tends to $1$ and we are left with $$S =\frac32 t^{3n+1} \left(1+\frac{1}{t}+O\left(\frac{1}{t^2} \right)\right)$$