Q82. Let the first term of a series be $T _ { 1 } = 6$ and its $r ^ { \text {th } }$ term $T _ { r } = 3 T _ { r - 1 } + 6 ^ { r } , r = 2,3 , \quad n$. If the sum of the first $n$ terms of this series is $\frac { 1 } { 5 } \left( n ^ { 2 } - 12 n + 39 \right) \left( 4 \cdot 6 ^ { n } - 5 \cdot 3 ^ { n } + 1 \right)$, then $n$ is equal to $\_\_\_\_$