Mathematics - Sequence & Series Question with Solution | TestHub

$\begin{array}{rl}& {\mathrm{T}}_{\mathrm{r}}=3{\mathrm{T}}_{\mathrm{r}-1}+6^{\mathrm{r}},\mathrm{r}=2,3,4,\dots \mathrm{n}\\ & {\mathrm{T}}_2=3.{\mathrm{T}}_1+6^2\\ & {\mathrm{T}}_2=3.6+6^2...(1)\end{array}$ $\begin{array}{rl}& {\mathrm{T}}_3=3{\mathrm{T}}_2+6^3\\ & {\mathrm{T}}_3=3{\mathrm{T}}_2+6^3\\ & {\mathrm{T}}_3=3\left(3.6+6^2\right)+6^3\\ & {\mathrm{T}}_3=3^2.6+3.6^2+6^3...(2)\end{array}$ $\begin{array}{rl}& {\mathrm{T}}_{\mathrm{r}}=3^{\mathrm{r}-1}\cdot 6\left[1+\frac{6}{3}+{\left(\frac{6}{3}\right)}^2+\dots +{\left(\frac{6}{3}\right)}^{\mathrm{r}-1}\right]\\ & {\mathrm{T}}_{\mathrm{r}}=3^{\mathrm{r}-1}\cdot 6\left(1+2+2^2+\dots +2^{\mathrm{r}-1}\right)\\ & {\mathrm{T}}_{\mathrm{r}}=6\cdot 3^{\mathrm{r}-1}1\cdot \frac{\left(1-2^{\mathrm{r}}\right)}{(-1)}\\ & {\mathrm{T}}_{\mathrm{r}}=6\cdot 3^{\mathrm{r}-1}\cdot \left(2^{\mathrm{r}}-1\right)\\ & {\mathrm{T}}_{\mathrm{r}}=\frac{6\cdot 3^{\mathrm{r}}}{3}\cdot \left(2^{\mathrm{r}}-1\right)\end{array}$ $\begin{array}{rl}& {\mathrm{T}}_{\mathrm{r}}=2\cdot \left(6^{\mathrm{r}}-3^{\mathrm{r}}\right)\\ & {\mathrm{S}}_{\mathrm{n}}=2\Sigma \left(6^{\mathrm{r}}-3^{\mathrm{r}}\right)\\ & {\mathrm{S}}_{\mathrm{n}}=2\cdot \left[\frac{6.\left(6^{\mathrm{n}}-1\right)}{5}-\frac{3.\left(3^{\mathrm{n}}-1\right)}{2}\right]\\ & {\mathrm{S}}_{\mathrm{n}}=2\left[\frac{12\left(6^{\mathrm{n}}-1\right)-15\left(3^{\mathrm{n}}-1\right)}{10}\right]\\ & {\mathrm{S}}_{\mathrm{n}}=\frac{3}{5}\left[4\cdot 6^4-5\cdot 3^{\mathrm{n}}+1\right]\\ & \therefore {\mathrm{n}}^2-12\mathrm{n}+39=3\\ & {\mathrm{n}}^2-12\mathrm{n}+36=0\\ & \mathrm{n}=6\end{array}$