If 6 3 ^ 12 + 10 3 ^ 11 + 20 3 ^ 10 + 40 3 ^ 9 + . . + (10240)/(3)...

Given,

$\dfrac{6}{{3}^{12}} + \dfrac{10}{{3}^{11}} + \dfrac{20}{{3}^{10}} + \dfrac{40}{{3}^{9}} + \ldots . . + \dfrac{10240}{3} = {2}^{n} \cdot m$

On rearranging we get,

$⇒ \dfrac{1}{{3}^{12}} + 5\left(\dfrac{{2}^{0}}{{3}^{12}} + \dfrac{{2}^{1}}{{3}^{11}} + \dfrac{{2}^{2}}{{3}^{10}} + \ldots \ldots + \dfrac{{2}^{11}}{3}\right) = {2}^{n} \cdot m$

$⇒ \dfrac{1}{{3}^{12}} + 5\left(\dfrac{1}{{3}^{12}}\dfrac{\left({\left(6\right)}^{2} - 1\right)}{\left(6 - 1\right)}\right) = {2}^{n} \cdot m$

$⇒ \dfrac{1}{{3}^{12}} + \dfrac{5}{5}\left(\dfrac{1}{{3}^{12}} \cdot {2}^{12} \cdot {3}^{12} - \dfrac{1}{{3}^{12}}\right) = {2}^{n} \cdot m$

$⇒ \dfrac{1}{{3}^{12}} + {2}^{12} - \dfrac{1}{{3}^{12}} = {2}^{n} \cdot m$

$⇒ {2}^{n} \cdot m = {2}^{12}$

On comparing both side we get,

$⇒ m = 1$ and $n = 12$

So, $m \cdot n = 12$