Simplify the expression: $\frac{ 3^{2n} -1} { 3^{n+1}-3} $

$$\frac{ 3^{2n} -1} { 3^{n+1}-3}=\frac { { \left( { 3 }^{ n } \right) }^{ 2 }-{ 1 }^{ 2 } }{ 3\left( { 3 }^{ n }-1 \right) } =\frac { \left( { 3 }^{ n }-1 \right) \left( { 3 }^{ n }+1 \right) }{ 3\left( { 3 }^{ n }-1 \right) } =\frac { { 3 }^{ n }+1 }{ 3 } $$

The denominator can be broken down into $3(3^n-1)$. The numerator can also be broken down into $(3^n-1)(3^n+1)$. Simplifying this gives us $(3^n+1)/3$ as the answer.