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

We have the series $$\dfrac{6}{3^{12}} + \dfrac{10}{3^{11}} + \dfrac{20}{3^{10}} + \dfrac{40}{3^{9}} + \ldots + \dfrac{10240}{3} = 2^n \cdot m$$, where $$m$$ is odd.

The numerators are $$6, 10, 20, 40, 80, \ldots, 10240$$. We can write these as: $$a_0 = 6$$ and $$a_k = 5 \cdot 2^k$$ for $$k = 1, 2, \ldots, 11$$. The denominators are $$3^{12}, 3^{11}, 3^{10}, \ldots, 3^1$$.

Multiplying the entire sum by $$3^{12}$$, we get:

$$3^{12} \cdot S = 6 + 10 \cdot 3 + 20 \cdot 3^2 + 40 \cdot 3^3 + \ldots + 10240 \cdot 3^{11}$$

$$= 6 + 5 \cdot 2 \cdot 3 + 5 \cdot 2^2 \cdot 3^2 + 5 \cdot 2^3 \cdot 3^3 + \ldots + 5 \cdot 2^{11} \cdot 3^{11}$$

$$= 6 + 5\sum_{k=1}^{11} (2 \cdot 3)^k = 6 + 5\sum_{k=1}^{11} 6^k$$

Now, $$\sum_{k=1}^{11} 6^k = \dfrac{6(6^{11} - 1)}{6 - 1} = \dfrac{6^{12} - 6}{5}$$

So, $$3^{12} \cdot S = 6 + 5 \cdot \dfrac{6^{12} - 6}{5} = 6 + 6^{12} - 6 = 6^{12}$$

Therefore, $$S = \dfrac{6^{12}}{3^{12}} = \left(\dfrac{6}{3}\right)^{12} = 2^{12}$$

Comparing with $$2^n \cdot m$$, we get $$n = 12$$ and $$m = 1$$ (which is odd).

Hence, $$m \cdot n = 1 \times 12 = 12$$.