Question

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

Answer 1

$ \frac{6}{3^{12}}+10\left(\frac{1}{3^{11}}+\frac{2}{3^{10}}+\frac{2^2}{3^9}+\frac{2^3}{3^8}+\ldots . .+\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$