QJEEM24Maad75: JEE Main 2024 Mathematics - Class 11 JEEM24Maad75 Solution: Let the first term of a series be and its term , ............ . …

$\begin{aligned}& T_r=3 T_{r-1}+6^r \\& \Rightarrow \text { solving homogenous part } \\& T_r=3 T_{r-1} \\& \Rightarrow x=3 \text { is the root }\end{aligned}$

$\therefore T_r=a .3^r$

Solving for particular part

$\begin{aligned}& T_r=b .6^r \\& b .6^r=3 b 6^{r-1}+6^r \\& \Rightarrow 6 b=3 b+6 \\& \Rightarrow 3 b=6 \\& \Rightarrow b=2 \\& T_r=a^n+a^p \\& T_r=a 3^{b r}+2.6^r \quad \text{.... (i)} \\& T_r=3 T_{r-1}+6^r\end{aligned}$

Putting $r=2$

$T_2=18+36=54 \quad \text{.... (ii)}$

Using equation (i) and (ii)

$\begin{aligned}& 54=9 a+72 \Rightarrow-18=9 a \Rightarrow a=-2 \\& \therefore T_r=2 \cdot 6^r-2 \cdot 3^r=2\left(6^r-3^r\right) \\& \sum_{r=1}^n T_r=2 \sum 6^r-2 \sum 3^r \\& =2 \cdot 6 \frac{\left(6^n-1\right)}{5}-2 \cdot 3 \frac{\left(3^n-1\right)}{2} \\& =\frac{3}{5}\left(4 \cdot 6^n-5 \cdot 3^n+1\right) \\& \therefore n^2-12 n+39=3 \\& n^2-12 n+36=0 \\& (n-6)^2=0 \\& \therefore n=6\end{aligned}$