1. Given recurrence

The terms satisfy $T_1=6, \qquad T_r=3T_{r-1}+6^r \quad (r\ge 2).$

We first find a general expression for $T_r$.

2. Solve the recurrence for $T_r$

We have $T_r-3T_{r-1}=6^r.$

Let us compute a few terms to identify the pattern:

• $T_1=6$
• $T_2=3\cdot 6+6^2=18+36=54$
• $T_3=3\cdot 54+6^3=162+216=378$

Now try the form $T_r=A\cdot 3^r+B\cdot 6^r.$

Substitute into $T_r-3T_{r-1}=6^r.$

Then $A3^r+B6^r-3\left(A3^{r-1}+B6^{r-1}\right)=6^r.$ This gives $A3^r+B6^r-A3^r-\frac{B}{2}6^r=6^r,$ so $\frac{B}{2}6^r=6^r \implies B=2.$

Thus $T_r=A3^r+2\cdot 6^r.$

Using $T_1=6$: $3A+12=6 \implies 3A=-6 \implies A=-2.$

Hence $\boxed{T_r=2\left(6^r-3^r\right)}.$

3. Find the sum of first $n$ terms

$S_n=\sum_{r=1}^n T_r=2\sum_{r=1}^n 6^r-2\sum_{r=1}^n 3^r.$

Using geometric series sums, $\sum_{r=1}^n 6^r=\frac{6(6^n-1)}{5}, \qquad \sum_{r=1}^n 3^r=\frac{3(3^n-1)}{2}.$

So $S_n=2\cdot \frac{6(6^n-1)}{5}-2\cdot \frac{3(3^n-1)}{2}.$

Simplify: $S_n=\frac{12(6^n-1)}{5}-3(3^n-1)$ $=\frac{12\cdot 6^n-12-15\cdot 3^n+15}{5}$ $=\frac{12\cdot 6^n-15\cdot 3^n+3}{5}$ $=\frac{3}{5}\left(4\cdot 6^n-5\cdot 3^n+1\right).$

Thus, $\boxed{S_n=\frac{3}{5}\left(4\cdot 6^n-5\cdot 3^n+1\right)}.$

4. Compare with the given sum

Given in the question: $S_n=\frac{1}{5}(n^2-12n+39)\left(4\cdot 6^n-5\cdot 3^n+1\right).$

Since both expressions represent the same sum, $\frac{3}{5}\left(4\cdot 6^n-5\cdot 3^n+1\right)=\frac{1}{5}(n^2-12n+39)\left(4\cdot 6^n-5\cdot 3^n+1\right).$

Now, $4\cdot 6^n-5\cdot 3^n+1 \neq 0$ for positive integer $n$ (for example at $n=1$, it is $24-15+1=10>0$, and it grows thereafter), so we cancel it: $3=n^2-12n+39.$

Therefore, $n^2-12n+36=0$ $(n-6)^2=0.$

Hence, $\boxed{n=6}.$

5. Verification with stored answer

Stored correct answer: $6$

Our derived answer: $6$

They match.