Mathematical Olympiad Rioplatense

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 $3^{3k}$: $$\frac{3^{3k}}{13}+\frac{3^{3k+1}}{13}+\frac{3^{3k+2}}{13}=\frac{3^k(1+3+3^2)}{13}=3^{3k}\text{for }0\le k\le 33.$$

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 $3^3\equiv 1(\mathrm{mod}13)$. Hence $x_0=\frac{1}{13},x_1=\frac{3}{13},x_2=\frac{9}{13}$ and so $$\begin{array}{rl}\frac{3^{3k}}{13}+\frac{3^{3k+1}}{13}+\frac{3^{3k+2}}{13}& =\lfloor \frac{3^{3k}}{13}\rfloor +\lfloor \frac{3^{3k+1}}{13}\rfloor +\lfloor \frac{3^{3k+2}}{13}\rfloor +(x_0+x_1+x_2)\\ & =\lfloor \frac{3^{3k}}{13}\rfloor +\lfloor \frac{3^{3k+1}}{13}\rfloor +\lfloor \frac{3^{3k+2}}{13}\rfloor +1.\end{array}$$ Therefore the given sum is equal to ${\sum }_{i=0}^{33}(3^{3k}-1)=\frac{27^{34}-1}{26}-34$.