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 are given the recurrence relation:

$$T_r = 3T_{r-1} + 6^r$$

To solve this, divide the entire equation by $$3^r$$ to simplify and group the terms:

$$\frac{T_r}{3^r} = \frac{3T_{r-1}}{3^r} + \frac{6^r}{3^r}$$

$$\frac{T_r}{3^r} - \frac{T_{r-1}}{3^{r-1}} = 2^r$$

Now, substitute values from $$r = 2$$ to $$n$$ and add them up to create a telescoping series:

For $$r=2$$: $$\frac{T_2}{3^2} - \frac{T_1}{3^1} = 2^2$$

For $$r=3$$: $$\frac{T_3}{3^3} - \frac{T_2}{3^2} = 2^3$$

$$\dots$$

For $$r=n$$: $$\frac{T_n}{3^n} - \frac{T_{n-1}}{3^{n-1}} = 2^n$$

Summing all these equations vertically, the intermediate terms cancel out perfectly:

$$\frac{T_n}{3^n} - \frac{T_1}{3} = 2^2 + 2^3 + \dots + 2^n$$

We know $$T_1 = 6$$, so $$\frac{T_1}{3} = 2$$. The right side is a Geometric Progression with $$(n-1)$$ terms, a first term $$a=4$$, and a common ratio $$r=2$$. Apply the GP sum formula $$\frac{a(r^k - 1)}{r - 1}$$:

$$\frac{T_n}{3^n} - 2 = \frac{4(2^{n-1} - 1)}{2 - 1}$$

$$\frac{T_n}{3^n} = 2^{n+1} - 4 + 2$$

$$\frac{T_n}{3^n} = 2(2^n) - 2$$

Multiply by $$3^n$$ to get the general term $$T_n$$:

$$T_n = 2(6^n) - 2(3^n)$$

Now, find the sum of the first $$n$$ terms, $$S_n = \sum T_r$$:

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

Apply the standard sum formula for a GP to both parts:

$$S_n = 2 \left[ \frac{6(6^n - 1)}{6 - 1} \right] - 2 \left[ \frac{3(3^n - 1)}{3 - 1} \right]$$

$$S_n = \frac{12}{5}(6^n - 1) - 3(3^n - 1)$$

$$S_n = \frac{12}{5} \cdot 6^n - \frac{12}{5} - 3 \cdot 3^n + 3$$

Take $$\frac{1}{5}$$ common to match the format of the given expression:

$$S_n = \frac{1}{5} [12 \cdot 6^n - 12 - 15 \cdot 3^n + 15]$$

$$S_n = \frac{1}{5} [12 \cdot 6^n - 15 \cdot 3^n + 3]$$

Pull out a common factor of 3 from the bracket:

$$S_n = \frac{3}{5} (4 \cdot 6^n - 5 \cdot 3^n + 1)$$

We are given that $$S_n = \frac{1}{5}(n^2 - 12n + 39)(4 \cdot 6^n - 5 \cdot 3^n + 1)$$.

Equating our derived sum to the given sum:

$$\frac{3}{5} = \frac{1}{5}(n^2 - 12n + 39)$$

$$3 = n^2 - 12n + 39$$

$$n^2 - 12n + 36 = 0$$

$$(n - 6)^2 = 0$$

$$n = 6$$