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

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:
$n = 6.$

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:
$n = 6.$