Let the first term of a series be $T_1=6$ and its $r^{\text {th }}$ term $T_r=3 T_{r-1}+6^r, r=2,3$, ............ $n$. 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 ___________.

<p>$$\begin{aligned} & T_r=3 T_{r-1}+6^r \\ & \Rightarrow \text { solving homogenous part } \\ & T_r=3 T_{r-1} \\ & \Rightarrow x=3 \text { is the root } \end{aligned}$$</p> <p>$\therefore T_r=a .3^r$</p> <p>Solving for particular part</p> <p>$$\begin{aligned} & T_r=b .6^r \\ & b .6^r=3 b 6^{r-1}+6^r \\ & \Rightarrow 6 b=3 b+6 \\ & \Rightarrow 3 b=6 \\ & \Rightarrow b=2 \\ & T_r=a^n+a^p \\ & T_r=a 3^{b r}+2.6^r \quad \text{.... (i)} \\ & T_r=3 T_{r-1}+6^r \end{aligned}$$</p> <p>Putting $r=2$</p> <p>$T_2=18+36=54 \quad \text{.... (ii)}$</p> <p>Using equation (i) and (ii)</p> <p>$$\begin{aligned} & 54=9 a+72 \Rightarrow-18=9 a \Rightarrow a=-2 \\ & \therefore T_r=2 \cdot 6^r-2 \cdot 3^r=2\left(6^r-3^r\right) \\ & \sum_{r=1}^n T_r=2 \sum 6^r-2 \sum 3^r \\ & =2 \cdot 6 \frac{\left(6^n-1\right)}{5}-2 \cdot 3 \frac{\left(3^n-1\right)}{2} \\ & =\frac{3}{5}\left(4 \cdot 6^n-5 \cdot 3^n+1\right) \\ & \therefore n^2-12 n+39=3 \\ & n^2-12 n+36=0 \\ & (n-6)^2=0 \\ & \therefore n=6 \end{aligned}$$</p>