Let the first term of a series be $$T_1 = 6$$ and its $$r^{th}$$ term $$T_r = 3T_{r-1} + 6^r$$, $$r = 2, 3, \ldots, n$$. If the sum of the first $$n$$ terms of this series is $$\frac{1}{5}(n^2 - 12n + 39)(4 \cdot 6^n - 5 \cdot 3^n + 1)$$, then $$n$$ is equal to ______

We have the series with $$T_1 = 6$$ and $$T_r = 3T_{r-1} + 6^r$$ for $$r = 2, 3, \ldots, n$$. The sum is given as $$S_n = \frac{1}{5}(n^2 - 12n + 39)(4 \cdot 6^n - 5 \cdot 3^n + 1)$$ and we need to determine the value of $$n$$.

Since the recurrence defines each term based on its predecessor, we compute successive values: $$T_1 = 6$$, $$T_2 = 3(6) + 6^2 = 18 + 36 = 54$$, $$T_3 = 3(54) + 6^3 = 162 + 216 = 378$$, $$T_4 = 3(378) + 6^4 = 1134 + 1296 = 2430$$, $$T_5 = 3(2430) + 6^5 = 7290 + 7776 = 15066$$, and $$T_6 = 3(15066) + 6^6 = 45198 + 46656 = 91854$$.

Adding these terms yields the partial sums: $$S_1 = 6$$, $$S_2 = 6 + 54 = 60$$, $$S_3 = 60 + 378 = 438$$, $$S_4 = 438 + 2430 = 2868$$, $$S_5 = 2868 + 15066 = 17934$$, and $$S_6 = 17934 + 91854 = 109788$$.

To verify the given formula, we substitute specific values of $$n$$. For $$n = 6$$:

$$S_6 = \frac{1}{5}(36 - 72 + 39)(4 \times 46656 - 5 \times 729 + 1)$$

$$= \frac{1}{5}(3)(186624 - 3645 + 1) = \frac{3 \times 182980}{5} = \frac{548940}{5} = 109788 \quad \checkmark$$

Next, for $$n = 5$$:

$$S_5^{\text{formula}} = \frac{1}{5}(25 - 60 + 39)(4 \times 7776 - 5 \times 243 + 1) = \frac{4 \times 29890}{5} = 23912 \neq 17934 \quad \times$$

Finally, for $$n = 7$$:

$$S_7^{\text{formula}} = \frac{1}{5}(49 - 84 + 39)(4 \times 279936 - 5 \times 2187 + 1) = \frac{4 \times 1108810}{5} = 887048$$

$$S_7^{\text{actual}} = 109788 + 3(91854) + 6^7 = 109788 + 275562 + 279936 = 665286 \neq 887048 \quad \times$$

Since the formula agrees with the actual sum only when $$n = 6$$, The answer is $$n = \mathbf{6}$$.