Question

If $\frac{6}{3^{12}}+\frac{10}{3^{11}}+\frac{20}{3^{10}}+\frac{40}{3^9}+\dots ..+\frac{10240}{3}=2^n\cdot m$ where $m$ is odd, then $m.n$ is equal to

Answer 1

Answer: 12

$\frac{6}{3^{12}}+10\left(\frac{1}{3^{11}}+\frac{2}{3^{10}}+\frac{2^2}{3^9}+\frac{2^3}{3^8}+\dots ..+\frac{2^{10}}{3}\right)$
$\frac{6}{3^{12}}+\frac{10}{3^{11}}\left(\frac{6^{11}-1}{6-1}\right)$
$=2^{12}\cdot 1;m\cdot n=12$