Not so standard binomial

The general term of the series can be written as $\frac { \left( 5+2i \right) }{ (i+1)(i+2) } { \left( \frac { 1 }{ 3 } \right) }^{ i }$ Let us name the summation as $S$ . So, $S=\sum _{ i=1 }^{ 10 }{ \frac { \left( 5+2i \right) }{ (i+1)(i+2) } { \left( \frac { 1 }{ 3 } \right) }^{ i } }$ $S=\sum _{ i=1 }^{ 10 }{ \left( \frac { 3 }{ (i+1) } -\frac { 1 }{ (i+2) } \right) { \left( \frac { 1 }{ 3 } \right) }^{ i } }$ $=\frac { 1 }{ 2 } +\left( -\frac { 1 }{ 3(3) } +\frac { 1 }{ 3(3) } \right) +\left( -\frac { 1 }{ 4({ 3 }^{ 2 }) } +\frac { 1 }{ 4({ 3 }^{ 2 }) } \right) ...........+\left( -\frac { 1 }{ 11({ 3 }^{ 9 }) } +\frac { 1 }{ 11({ 3 }^{ 9 }) } \right) -\frac { 1 }{ 12({ 3 }^{ 10 }) } \\ S=\frac { 1 }{ 2 } -\frac { 1 }{ 12{ \times 3 }^{ 10 } }$