Let the first term of a series be $T_1=6$ and its $r^{ ext {th }}$ term $T_r=3 T_{r-1}+6^r, r=2,3$, $\qquad$ $n$. If the sum of the first $n$ terms of this series is $\frac{1}{5}\left(n^2-12 n+39 ight)\left(4 \cdot 6^n-5 \cdot 3^n+1 ight)$, then $n$ is equal to______

Question

Let the first term of a series be $T_1=6$ and its $r^{	ext {th }}$ term $T_r=3 T_{r-1}+6^r, r=2,3$, $\qquad$ $n$. If the sum of the first $n$ terms of this series is $\frac{1}{5}\left(n^2-12 n+39ight)\left(4 \cdot 6^n-5 \cdot 3^n+1ight)$, then $n$ is equal to______

Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{T}_{\mathrm{r}}=3 \mathrm{~T}_{\mathrm{r}-1}+6^{\mathrm{r}}, \mathrm{r}=2,3,4, \ldots \mathrm{n} \ & \mathrm{T}_2=3 . \mathrm{T}_1+6^2 \ & \mathrm{~T}_2=3.6+6^2...(1)\end{aligned}$ $\begin{aligned} & \mathrm{T}_3=3 \mathrm{~T}_2+6^3 \ & \mathrm{~T}_3=3 \mathrm{~T}_2+6^3 \ & \mathrm{~T}_3=3\left(3.6+6^2ight)+6^3 \ & \mathrm{~T}_3=3^2 .6+3.6^2+6^3...(2)\end{aligned}$ $\begin{aligned} & \mathrm{T}_{\mathrm{r}}=3^{\mathrm{r}-1} \cdot 6\left[1+\frac{6}{3}+\left(\frac{6}{3}ight)^2+\ldots+\left(\frac{6}{3}ight)^{\mathrm{r}-1}ight] \ & \mathrm{T}_{\mathrm{r}}=3^{\mathrm{r}-1} \cdot 6\left(1+2+2^2+\ldots+2^{\mathrm{r}-1}ight) \ & \mathrm{T}_{\mathrm{r}}=6 \cdot 3^{\mathrm{r}-1} 1 \cdot \frac{\left(1-2^{\mathrm{r}}ight)}{(-1)} \ & \mathrm{T}_{\mathrm{r}}=6 \cdot 3^{\mathrm{r}-1} \cdot\left(2^{\mathrm{r}}-1ight) \ & \mathrm{T}_{\mathrm{r}}=\frac{6 \cdot 3^{\mathrm{r}}}{3} \cdot\left(2^{\mathrm{r}}-1ight)\end{aligned}$ $\begin{aligned} & \mathrm{T}_{\mathrm{r}}=2 \cdot\left(6^{\mathrm{r}}-3^{\mathrm{r}}ight) \ & \mathrm{S}_{\mathrm{n}}=2 \Sigma\left(6^{\mathrm{r}}-3^{\mathrm{r}}ight) \ & \mathrm{S}_{\mathrm{n}}=2 \cdot\left[\frac{6 .\left(6^{\mathrm{n}}-1ight)}{5}-\frac{3 .\left(3^{\mathrm{n}}-1ight)}{2}ight] \ & \mathrm{S}_{\mathrm{n}}=2\left[\frac{12\left(6^{\mathrm{n}}-1ight)-15\left(3^{\mathrm{n}}-1ight)}{10}ight] \ & \mathrm{S}_{\mathrm{n}}=\frac{3}{5}\left[4 \cdot 6^4-5 \cdot 3^{\mathrm{n}}+1ight] \ & 	herefore \mathrm{n}^2-12 \mathrm{n}+39=3 \ & \mathrm{n}^2-12 \mathrm{n}+36=0 \ & \mathrm{n}=6\end{aligned}$

Let the first term of a series be $T_1=6$ and its $r^{\text {th }}$ term $T_r=3 T_{r-1}+6^r, r=2,3$,…

Solution