Ignore the integer parts of three consecutive summands with numerators $3^{3k}$, $3^{3k+1}$, $3^{3k+2}$. The sum of three such fractions is an integer; moreover it equals $3^{3k}$:
$$\frac{3^{3k}}{13} + \frac{3^{3k+1}}{13} + \frac{3^{3k+2}}{13} = \frac{3^k(1+3+3^2)}{13} = 3^{3k} \quad \text{for } 0 \le k \le 33.$$

Let $x_0, x_1, x_2$ be the fractional parts of $\frac{3^{3k}}{13}, \frac{3^{3k+1}}{13}, \frac{3^{3k+2}}{13}$. The remainders of $3^{3k}, 3^{3k+1}, 3^{3k+2}$ modulo $13$ are $1, 3, 9$ since $3^3 \equiv 1 \pmod{13}$. Hence $x_0 = \frac{1}{13}, x_1 = \frac{3}{13}, x_2 = \frac{9}{13}$ and so
$$\begin{aligned} \frac{3^{3k}}{13} + \frac{3^{3k+1}}{13} + \frac{3^{3k+2}}{13} &= \left\lfloor \frac{3^{3k}}{13} \right\rfloor + \left\lfloor \frac{3^{3k+1}}{13} \right\rfloor + \left\lfloor \frac{3^{3k+2}}{13} \right\rfloor + (x_0 + x_1 + x_2) \\ &= \left\lfloor \frac{3^{3k}}{13} \right\rfloor + \left\lfloor \frac{3^{3k+1}}{13} \right\rfloor + \left\lfloor \frac{3^{3k+2}}{13} \right\rfloor + 1. \end{aligned}$$
Therefore the given sum is equal to $\sum_{i=0}^{33} (3^{3k} - 1) = \frac{27^{34}-1}{26} - 34$.