Official solution
\[S = 1 + {2 \over 3} + {7 \over {{3^2}}} + {{12} \over {{3^3}}} + {{17} \over {{3^4}}} + ....\]
\[{S \over 3} = {1 \over 3} + {2 \over {{3^2}}} + {7 \over {{3^3}}} + {{12} \over {{3^4}}} + ....\]
\[2S = 1 + {1 \over 3} + {5 \over {{3^2}}} + {5 \over {{3^3}}} + {5 \over {{3^4}}} + ....\] + up to infinite terms
\[{{2S} \over 3} = {4 \over 3} + {5 \over 3}\left\{ {{{1/3} \over {1 - {1 \over 3}}}} \right\} = {5 \over 6} + {4 \over 3} = {{13} \over 6}\]
\[ \Rightarrow \] S = \[{13 \over 4}\]