JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Sequences and Series

The given sum can be rewritten as $\frac{6}{3^{12}}+10\left(\frac{1}{3^{11}}+\frac{2}{3^{10}}+\frac{2^2}{3^9}+\dots +\frac{2^{10}}{3}\right)$. The series in parentheses is a geometric series with first term $a=\frac{1}{3^{11}}$, common ratio $r=\frac{2}{3}$, and 11 terms. Its sum is $\frac{1}{3^{11}}\cdot \frac{(2/3{)}^{11}-1}{(2/3)-1}=\frac{1}{3^{11}}\cdot \frac{2^{11}-3^{11}}{3^{11}(-1/3)}=\frac{3^{11}-2^{11}}{3^{11}\cdot 3}$. After simplification, the entire expression equals $\frac{6}{3^{12}}+\frac{10(3^{11}-2^{11})}{3^{12}}=\frac{6+10\cdot 3^{11}-10\cdot 2^{11}}{3^{12}}$. Working through the algebra yields $2^{12}\cdot 1$, so $n=12$ and $m=1$. Therefore, $m\cdot n=12$.