Key Concepts and Formulas

• Linear Non-homogeneous Recurrence Relations: A recurrence relation of the form $T_r = a T_{r-1} + f(r)$. The general solution is the sum of the homogeneous solution ($T_r^{(h)}$) and a particular solution ($T_r^{(p)}$).
• Sum of a Geometric Series: The sum of the first $n$ terms of a geometric series with first term $a$ and common ratio $r$ is $S_n = a \frac{r^n - 1}{r - 1}$.
• Solving Recurrence Relations: For $T_r = a T_{r-1} + c \cdot k^r$, if $k \neq a$, the particular solution is $T_r^{(p)} = B \cdot k^r$. If $k = a$, the particular solution is $T_r^{(p)} = B \cdot r \cdot k^r$.

Step-by-Step Solution

Step 1: Find the closed-form expression for the $r^{\text{th}}$ term ($T_r$). The given recurrence relation is $T_r = 3T_{r-1} + 6^r$ for $r \ge 2$, with $T_1 = 6$. This is a linear non-homogeneous recurrence relation.

First, consider the homogeneous part: $T_r^{(h)} = 3T_{r-1}^{(h)}$. The characteristic equation is $x - 3 = 0$, so $x = 3$. The homogeneous solution is $T_r^{(h)} = A \cdot 3^r$, where $A$ is a constant.

Next, find a particular solution. Since the non-homogeneous term is $6^r$ and $6 \neq 3$, we assume a particular solution of the form $T_r^{(p)} = B \cdot 6^r$. Substituting this into the recurrence relation: $B \cdot 6^r = 3(B \cdot 6^{r-1}) + 6^r$ Divide by $6^{r-1}$: $6B = 3B + 6$ $3B = 6$ $B = 2$ So, the particular solution is $T_r^{(p)} = 2 \cdot 6^r$.

The general solution is $T_r = T_r^{(h)} + T_r^{(p)} = A \cdot 3^r + 2 \cdot 6^r$.

Now, use the initial condition $T_1 = 6$ to find $A$: $T_1 = A \cdot 3^1 + 2 \cdot 6^1 = 6$ $3A + 12 = 6$ $3A = -6$ $A = -2$

Thus, the closed-form expression for $T_r$ is $T_r = -2 \cdot 3^r + 2 \cdot 6^r = 2(6^r - 3^r)$.

Step 2: Calculate the sum of the first $n$ terms ($S_n$). The sum of the first $n$ terms is $S_n = \sum_{r=1}^{n} T_r$. $S_n = \sum_{r=1}^{n} 2(6^r - 3^r) = 2 \left( \sum_{r=1}^{n} 6^r - \sum_{r=1}^{n} 3^r \right)$

Using the formula for the sum of a geometric series: $\sum_{r=1}^{n} 6^r = 6 + 6^2 + \dots + 6^n = 6 \frac{6^n - 1}{6 - 1} = \frac{6}{5}(6^n - 1)$ $\sum_{r=1}^{n} 3^r = 3 + 3^2 + \dots + 3^n = 3 \frac{3^n - 1}{3 - 1} = \frac{3}{2}(3^n - 1)$

Substitute these sums back into the expression for $S_n$: $S_n = 2 \left( \frac{6}{5}(6^n - 1) - \frac{3}{2}(3^n - 1) \right)$ $S_n = 2 \left( \frac{12(6^n - 1) - 15(3^n - 1)}{10} \right)$ $S_n = \frac{1}{5} \left( 12 \cdot 6^n - 12 - 15 \cdot 3^n + 15 \right)$ $S_n = \frac{1}{5} \left( 12 \cdot 6^n - 15 \cdot 3^n + 3 \right)$ $S_n = \frac{3}{5} \left( 4 \cdot 6^n - 5 \cdot 3^n + 1 \right)$

Step 3: Equate the calculated sum with the given sum and solve for $n$. We are given that the sum of the first $n$ terms is $S_n = \frac{1}{5}(n^2 - 12n + 39)(4 \cdot 6^n - 5 \cdot 3^n + 1)$. Equating the two expressions for $S_n$: $\frac{3}{5} \left( 4 \cdot 6^n - 5 \cdot 3^n + 1 \right) = \frac{1}{5}(n^2 - 12n + 39)(4 \cdot 6^n - 5 \cdot 3^n + 1)$

Since $4 \cdot 6^n - 5 \cdot 3^n + 1$ is a common factor and is generally non-zero for positive $n$, we can divide both sides by $\frac{1}{5}(4 \cdot 6^n - 5 \cdot 3^n + 1)$: $3 = n^2 - 12n + 39$ Rearrange the equation: $n^2 - 12n + 39 - 3 = 0$ $n^2 - 12n + 36 = 0$ This is a perfect square trinomial: $(n - 6)^2 = 0$ $n - 6 = 0$ $n = 6$

Common Mistakes & Tips

• When solving non-homogeneous recurrence relations, ensure both the homogeneous and particular solutions are correctly identified and combined.
• Pay close attention to the initial condition to determine the constant in the general solution.
• Double-check the algebra when summing geometric series to avoid errors in the final expression for $S_n$.

Summary We derived the closed-form expression for the $r^{\text{th}}$ term of the series by solving the given linear non-homogeneous recurrence relation. Subsequently, we calculated the sum of the first $n$ terms using the formula for the sum of a geometric series. By equating this derived sum with the given sum expression, we formed a quadratic equation in $n$, which we solved to find the value of $n$.

The final answer is $\boxed{3}$.