Question

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

Answer 1

Correct Answer: 6

Given a series with first term \( T_1 = 6 \) and recurrence relation \( T_r = 3T_{r-1} + 6^r \) for \( r \geq 2 \), and the formula for the sum of the first \( n \) terms \( S_n \), determine the value of \( n \).

Concepts Employed:

1. Solving Linear Recurrence Relations: The provided relation is a first-order, non-homogeneous linear recurrence relation. A standard technique for relations of the form \( T_r = aT_{r-1} + f(r) \) involves dividing by an appropriate factor (in this case, \( a^r \)) to facilitate a telescoping sum.

2. Sum of a Geometric Progression (GP): The sum of the first \( n \) terms of a GP with first term \( a \) and common ratio \( R \) is calculated as:

\[ S_n = \frac{a(R^n - 1)}{R - 1} \]

Detailed Solution:

Step 1: Transform the given recurrence relation to derive a general expression for \( T_r \).

The relation is \( T_r = 3T_{r-1} + 6^r \). Dividing both sides by \( 3^r \) yields:

\[ \frac{T_r}{3^r} = \frac{3T_{r-1}}{3^r} + \frac{6^r}{3^r} \] \[ \frac{T_r}{3^r} = \frac{T_{r-1}}{3^{r-1}} + \left(\frac{6}{3}\right)^r = \frac{T_{r-1}}{3^{r-1}} + 2^r \]

Step 2: Define a new sequence \( a_r = \frac{T_r}{3^r} \). The relation simplifies to \( a_r = a_{r-1} + 2^r \), or \( a_r - a_{r-1} = 2^r \).

This forms a telescoping sum. The expression for \( a_r \) in terms of \( a_1 \) is:

\[ a_r = a_1 + \sum_{k=2}^{r} (a_k - a_{k-1}) = a_1 + \sum_{k=2}^{r} 2^k \]

Step 3: Compute \( a_1 \) and establish the explicit formula for \( a_r \).

Given \( T_1 = 6 \), \( a_1 = \frac{T_1}{3^1} = \frac{6}{3} = 2 \).

The summation \( \sum_{k=2}^{r} 2^k = 2^2 + 2^3 + \dots + 2^r \) is a geometric series with a first term of \( 4 \), a common ratio of \( 2 \), and \( r-1 \) terms:

\[ \sum_{k=2}^{r} 2^k = \frac{4(2^{r-1} - 1)}{2 - 1} = 4(2^{r-1} - 1) = 2^2 \cdot 2^{r-1} - 4 = 2^{r+1} - 4 \]

Substituting this into the expression for \( a_r \):

\[ a_r = 2 + (2^{r+1} - 4) = 2^{r+1} - 2 \]

Step 4: Derive the general formula for \( T_r \).

Since \( a_r = \frac{T_r}{3^r} \), it follows that \( T_r = 3^r \cdot a_r \).

\[ T_r = 3^r (2^{r+1} - 2) = 3^r \cdot (2 \cdot 2^r - 2) = 2 \cdot (3^r \cdot 2^r) - 2 \cdot 3^r \] \[ T_r = 2 \cdot 6^r - 2 \cdot 3^r \]

Step 5: Calculate the sum of the first \( n \) terms, \( S_n = \sum_{r=1}^{n} T_r \).

\[ S_n = \sum_{r=1}^{n} (2 \cdot 6^r - 2 \cdot 3^r) = 2 \sum_{r=1}^{n} 6^r - 2 \sum_{r=1}^{n} 3^r \]

Both sums are geometric progressions. For the first sum, \( a=6, R=6 \). For the second, \( a=3, R=3 \).

\[ \sum_{r=1}^{n} 6^r = \frac{6(6^n - 1)}{6 - 1} = \frac{6}{5}(6^n - 1) \] \[ \sum_{r=1}^{n} 3^r = \frac{3(3^n - 1)}{3 - 1} = \frac{3}{2}(3^n - 1) \]

Step 6: Substitute these sums into the expression for \( S_n \).

\[ S_n = 2 \left( \frac{6}{5}(6^n - 1) \right) - 2 \left( \frac{3}{2}(3^n - 1) \right) \] \[ S_n = \frac{12}{5}(6^n - 1) - 3(3^n - 1) \]

Combining terms over a common denominator:

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

Step 7: Equate the derived \( S_n \) with the given formula for \( S_n \).

\[ \frac{1}{5} (12 \cdot 6^n - 15 \cdot 3^n + 3) = \frac{1}{5} \left(n^2 - 12n + 39\right) \left(4 \cdot 6^n - 5 \cdot 3^n + 1\right) \]

Canceling \( \frac{1}{5} \) from both sides and factoring the left side:

\[ 12 \cdot 6^n - 15 \cdot 3^n + 3 = 3(4 \cdot 6^n - 5 \cdot 3^n + 1) \]

Substituting this back into the equation:

\[ 3(4 \cdot 6^n - 5 \cdot 3^n + 1) = \left(n^2 - 12n + 39\right) \left(4 \cdot 6^n - 5 \cdot 3^n + 1\right) \]

Since \( (4 \cdot 6^n - 5 \cdot 3^n + 1) eq 0 \) for \( n \geq 1 \), it can be canceled from both sides.

Final Calculation and Outcome:

Step 8: Solve the simplified equation for \( n \).

\[ 3 = n^2 - 12n + 39 \]

Rearranging into a quadratic equation:

\[ n^2 - 12n + 39 - 3 = 0 \] \[ n^2 - 12n + 36 = 0 \]

This quadratic is a perfect square trinomial:

\[ (n - 6)^2 = 0 \]

Solving for \( n \):

\[ n - 6 = 0 \implies n = 6 \]

The value of \( n \) is thus determined to be 6.