This is not a difficult problem, but I would like please to discuss with you how I solved it:
Solving recurrence relation $T(n) = 5T(\frac{n}{3}) + 2n$, $T(1)=2$. What is the value of $T(9)$? This can be done directly by applying $T(9)$ to get 98. However, I did it recursively as follows:
$$ \begin{align} T(n) = 5\left[5T(\frac{n-1}{3^2}) + 2(n-1)\right] + 2n \tag{1} \\ T(n) = 5\left[5\left[5T(\frac{n-2}{3^3}) + 2(n-2)\right] + 2(n-1)\right] + 2n\\ = 5^3 T(\frac{n-2}{3^3}) + 5^2\times2(n-2) + 5\times2(n-1) + 2n \tag{2} \\ \vdots\\ T(n) = 5^n \times T\left(\frac{n-(n-1)}{3^n}\right) + 5^{n-1}\times2(n-(n-1)) + \cdots 5\times2(n-1) + 2n \tag{3} \\ \end{align} $$
Question: what is the value of $T(n)$ above please as we have $n$, so we can not infinite apply geometric series I guess please?