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

We are given a series with the first term \( T_1 = 6 \) and a recurrence relation \( T_r = 3T_{r-1} + 6^r \) for \( r \geq 2 \). We are also given the formula for the sum of the first \( n \) terms, \( S_n \), and we need to find the value of \( n \).

Concept Used:

1. Solving Linear Recurrence Relations: The given relation is a linear first-order non-homogeneous recurrence relation. A common method to solve relations of the form \( T_r = aT_{r-1} + f(r) \) is to divide the entire equation by an appropriate factor (here, \( a^r \)) to transform it into 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 given by:

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

Step-by-Step Solution:

Step 1: Manipulate the given recurrence relation to find a general form for the term \( T_r \).

The relation is \( T_r = 3T_{r-1} + 6^r \). To simplify this, we divide both sides by \( 3^r \):

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

Step 2: Let's define a new sequence \( a_r = \frac{T_r}{3^r} \). The relation becomes \( a_r = a_{r-1} + 2^r \), which can be written as \( a_r - a_{r-1} = 2^r \).

This is a telescoping sum. We can write \( a_r \) in terms of \( a_1 \):

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

Step 3: Calculate \( a_1 \) and find the explicit formula for \( a_r \).

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

The summation is a geometric series \( \sum_{k=2}^{r} 2^k = 2^2 + 2^3 + \dots + 2^r \). The first term is \( 4 \), the common ratio is \( 2 \), and there are \( r-1 \) 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 back into the expression for \( a_r \):

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

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

Since \( a_r = \frac{T_r}{3^r} \), we have \( T_r = 3^r \cdot a_r \).

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

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 back into the expression for \( S_n \).

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

Bringing to 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 our derived \( S_n \) with the given formula for \( S_n \).

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

We can cancel \( \frac{1}{5} \) from both sides. Also, notice that the term on the left can be factored:

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

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 the term \( (4 \cdot 6^n - 5 \cdot 3^n + 1) \) is non-zero for \( n \ge 1 \), we can cancel it from both sides.

Final Computation & Result:

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

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

Rearranging the terms gives a quadratic equation:

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

This is a perfect square trinomial:

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

Solving for \( n \), we get:

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

Therefore, the value of \( n \) is 6.

Answer 2

For the given series:

\( T_r = 3T_{r-1} + 6^r, \quad \text{for } r \geq 2. \)

For \( r = 2 \):
\( T_2 = 3T_1 + 6^2 = 3 \times 6 + 6^2 = 18 + 36 = 54. \) $\hspace{1cm}$ \((1)\)

For \( r = 3 \):
\( T_3 = 3T_2 + 6^3 = 3 \times 54 + 216 = 162 + 216 = 378. \) $\hspace{1cm}$ \((2)\)

General term:
\( T_r = 3^{r-1} \times 6 + 3^{r-2} \times 6^2 + \cdots + 6^r. \)

Taking \( 6 \) common:
\( T_r = 6 \times 3^{r-1} \left( 1 + \frac{6}{3} + \frac{6^2}{3^{r-1}} \right). \)

Simplify:
\( T_r = 6^{r-1} \times \left( 1 + 2 + 2^2 + \cdots + 2^{r-1} \right), \) which forms a geometric progression.

Sum of GP:
\( T_r = \frac{6^r - 3^r}{3}. \)

Sum of the series \( S_n \):
\( S_n = \sum_{r=1}^n T_r = \sum_{r=1}^n \frac{6^r - 3^r}{3}. \)

Separate the terms:
\( S_n = \frac{1}{3} \left( \sum_{r=1}^n 6^r - \sum_{r=1}^n 3^r \right). \)

Sum of \( 6^r \):
\( \sum_{r=1}^n 6^r = 6 \frac{6^n - 1}{5}. \)

Sum of \( 3^r \):
\( \sum_{r=1}^n 3^r = 3 \frac{3^n - 1}{2}. \)

Substitute:
\( S_n = \frac{1}{3} \left( \frac{6(6^n - 1)}{5} - \frac{3(3^n - 1)}{2} \right). \)

Simplify:
\( S_n = \frac{1}{5} \left( 4.6^n - 5.3^n + 1 \right). \)

Equating with the given sum:
\( \frac{1}{5} \left( n^2 - 12n + 39 \right) \left( 4.6^n - 5.3^n + 1 \right) = S_n. \)

Equating coefficients:
\( n^2 - 12n + 36 = 0. \)

Solve:
\( n = 6. \)