[tex]\frac { 1 }{ 3 } \left( n+1 \right) +2=\frac { 1 }{ 6 } \left( 3n-5 \right)[/tex]
[tex]\\ \\ 6\left\{ \frac { 1 }{ 3 } \left( n+1 \right) +2 \right\} =6\cdot \frac { 1 }{ 6 } \left( 3n-5 \right)[/tex]
[tex]\\ \\ 2\left( n+1 \right) +12=3n-5\\ \\ 2n+2+12=3n-5\\ \\ 2+12+5=3n-2n\\ \\ 19=n[/tex]
