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

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:
\( n = 6. \)