In honor of the number $3$, Albert decides to evaluate $\displaystyle\sum_{k=0}^{n}\frac{1}{3^k}(F_{3k})^3$, where $F_j$ is the $j^{\text{th}}$ Fibonacci number. He finds that it can be expressed in the form $\frac{1}{100\cdot 3^n}(a\cdot F_{9n+9}+b\cdot F_{9n}+c\cdot F_{3n+3}+d\cdot F_{3n}+e)$, where $a$, $b$, $c$, $d$, and $e$ are real numbers that may or may not be in terms of $n$. Find the ordered quintuple $(a,b,c,d,e)$.