Let the first term of a series be T 1 = 6 and its r t h term T r = 3 T r - 1 + 6 r , ? r = 2 , 3 , … , n . If the sum of the first n terms of this series is 1 5 ( n 2 - 12 n + 39 ) ( 4 · 6 n - 5 · 3 n + 1 ) , then n is equal to ______ .

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Solution

Q.
Let the first term of a series be $T_{1} = 6$ and its $r^{t h}$ term $T_{r} = 3 T_{r - 1} + 6^{r}, r = 2, 3, \dots, n .$ If the sum of the first $n$ terms of this series is $\frac{1}{5} (n^{2} - 12 n + 39) (4 \cdot 6^{n} - 5 \cdot 3^{n} + 1),$ then $n$ is equal to ______ . [2024]

Ans.
(6)

We have, $T_{1} = 6$ and $T_{r} = 3 T_{r - 1} + 6^{r}$

Now, $T_{2} = 3 T_{1} + 6^{2} = 3 \times 6 + 6^{2}$

$T_{3} = 3 T_{2} + 6^{3} = 3 (3 \times 6 + 6^{2}) + 6^{3} = 3^{2} \times 6 + 3 \times 6^{2} + 6^{3}$

$T_{4} = 3 T_{3} + 6^{4} = 3 (3^{2} \times 6 + 3 \times 6^{2} + 6^{3}) + 6^{4}$

$= 3^{3} \times 6 + 3^{2} \times 6^{2} + 3 \times 6^{3} + 6^{4}$

$∴$    $T_{r} = 3^{r - 1} \times 6 + 3^{r - 2} \times 6^{2} + \dots + 6^{r}$

$= 3^{r - 1} \times 6 (1 + \frac{6}{3} + {(\frac{6}{3})}^{2} + \dots {(\frac{6}{3})}^{r - 1})$

$= 3^{r - 1} \times 6 \times (1 + 2 + 2^{2} + \dots + 2^{r - 1})$

$= 3^{r - 1} \times 6 (\frac{2^{r} - 1}{1}) = \frac{6 \times 3^{r}}{3} (2^{r} - 1)$

   $= 2 \times 3^{r} (2^{r} - 1) = 2 (6^{r} - 3^{r})$

Sum of $n$ terms $S_{n} = \sum_{n} 2 (6^{r} - 3^{r})$

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

   $= \frac{1}{5} (12 \times 6^{n} - 12 - 15 \times 3^{n} + 15)$

$= \frac{3}{5} (4 \times 6^{n} - 5 \times 3^{n} + 1) \Rightarrow n^{2} - 12 n + 39 = 3$

$\Rightarrow n^{2} - 12 n + 36 = 0 \Rightarrow n = 6$

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