Divisibility time

We note that:

$\begin{aligned} 9 & \equiv 3 \pmod{6} \\ 9^2 & \equiv 3 \times 3 \equiv 9 \equiv 3 \pmod{6} \\ 9^3 & \equiv 3 \times 3 \equiv 9 \equiv 3 \pmod{6} \\ ... & \quad ... \\ \implies 9^k & \equiv 3 \pmod{6} \quad \text{where } k=1,2,3... \end{aligned}$

$\begin{aligned} \implies \sum_{k=1}^n 9^k & \equiv \begin{cases} 3 \pmod{6} & \text{if } n \text{ is odd.} \\ 0 \pmod{6} & \text{if } n \text{ is even.} \end{cases} \end{aligned}$

Since $n$ is odd, the remainder is $\boxed{3}$ .

The sum is divisible by 3 but not divisible by 2. Hence, it will leave a remainder of 3 when divided by 6.