$$\dfrac{6}{3^{26}}+\dfrac{10.1}{3^{25}}+\dfrac{10.2}{3^{24}}+\dfrac{10.2^{2}}{3^{23}}+...+\dfrac{10.2^{24}}{3}$$ is equal to :

The given series is:

$$\dfrac{6}{3^{26}} + \dfrac{10 \cdot 1}{3^{25}} + \dfrac{10 \cdot 2}{3^{24}} + \dfrac{10 \cdot 2^{2}}{3^{23}} + \ldots + \dfrac{10 \cdot 2^{24}}{3}$$

There are 26 terms in total, as the denominator exponents decrease from 26 to 1. Let the term index be $$k$$, starting from $$k=1$$. The general term $$a_k$$ is:

For $$k=1$$: $$a_1 = \dfrac{6}{3^{26}}$$

For $$k \geq 2$$: $$a_k = \dfrac{10 \cdot 2^{k-2}}{3^{27-k}}$$

First, simplify $$a_1$$:

$$a_1 = \dfrac{6}{3^{26}} = \dfrac{2 \cdot 3}{3^{26}} = \dfrac{2}{3^{25}}$$

For $$k \geq 2$$, simplify $$a_k$$:

$$a_k = \dfrac{10 \cdot 2^{k-2}}{3^{27-k}} = 10 \cdot 2^{k-2} \cdot 3^{k-27}$$

Rewrite as:

$$a_k = 10 \cdot \dfrac{2^{k-2}}{3^{27-k}} = 10 \cdot \dfrac{2^{k-2} \cdot 3^{k}}{3^{27}} = \dfrac{10}{3^{27}} \cdot 2^{k-2} \cdot 3^{k}$$

Since $$2^{k-2} \cdot 3^{k} = 2^{k-2} \cdot 3^{k-2} \cdot 3^{2} = 3^{2} \cdot (2 \cdot 3)^{k-2} = 9 \cdot 6^{k-2}$$, so:

$$a_k = \dfrac{10}{3^{27}} \cdot 9 \cdot 6^{k-2} = \dfrac{90}{3^{27}} \cdot 6^{k-2}$$

Simplify $$\dfrac{90}{3^{27}} = 90 \cdot 3^{-27}$$, and $$90 = 10 \cdot 9 = 10 \cdot 3^2$$, so:

$$a_k = 10 \cdot 3^2 \cdot 3^{-27} \cdot 6^{k-2} = 10 \cdot 3^{-25} \cdot 6^{k-2}$$

Alternatively, using $$6^{k-2} = \dfrac{6^k}{6^2} = \dfrac{6^k}{36}$$:

$$a_k = \dfrac{10}{36} \cdot 3^{-25} \cdot 6^k = \dfrac{5}{18} \cdot 3^{-25} \cdot 6^k$$

But earlier expression is sufficient.

Now, sum the terms from $$k=2$$ to $$k=26$$:

$$\sum_{k=2}^{26} a_k = \sum_{k=2}^{26} \dfrac{10}{3^{27}} \cdot 2^{k-2} \cdot 3^{k} = \dfrac{10}{3^{27}} \sum_{k=2}^{26} 2^{k-2} \cdot 3^{k}$$

Since $$2^{k-2} \cdot 3^{k} = 2^{k-2} \cdot 3^{k-2} \cdot 3^{2} = 9 \cdot (6)^{k-2}$$, so:

$$\sum_{k=2}^{26} a_k = \dfrac{10}{3^{27}} \sum_{k=2}^{26} 9 \cdot 6^{k-2} = \dfrac{90}{3^{27}} \sum_{k=2}^{26} 6^{k-2}$$

Let $$m = k - 2$$, so when $$k=2$$, $$m=0$$; when $$k=26$$, $$m=24$$. Thus:

$$\sum_{k=2}^{26} 6^{k-2} = \sum_{m=0}^{24} 6^m$$

This is a geometric series with first term $$a = 6^0 = 1$$, common ratio $$r = 6$$, and number of terms $$n = 25$$. The sum is:

$$\sum_{m=0}^{24} 6^m = \dfrac{6^{25} - 1}{6 - 1} = \dfrac{6^{25} - 1}{5}$$

Therefore:

$$\sum_{k=2}^{26} a_k = \dfrac{90}{3^{27}} \cdot \dfrac{6^{25} - 1}{5} = \dfrac{90}{5} \cdot \dfrac{6^{25} - 1}{3^{27}} = 18 \cdot \dfrac{6^{25} - 1}{3^{27}}$$

Simplify $$6^{25} = (2 \cdot 3)^{25} = 2^{25} \cdot 3^{25}$$, so:

$$\sum_{k=2}^{26} a_k = 18 \cdot \dfrac{2^{25} \cdot 3^{25} - 1}{3^{27}} = 18 \left( \dfrac{2^{25} \cdot 3^{25}}{3^{27}} - \dfrac{1}{3^{27}} \right) = 18 \left( 2^{25} \cdot 3^{-2} - 3^{-27} \right) = 18 \left( \dfrac{2^{25}}{9} - \dfrac{1}{3^{27}} \right)$$

Distribute:

$$= 18 \cdot \dfrac{2^{25}}{9} - 18 \cdot \dfrac{1}{3^{27}} = 2 \cdot 2^{25} - \dfrac{18}{3^{27}} = 2^{26} - \dfrac{18}{3^{27}}$$

Simplify $$\dfrac{18}{3^{27}} = 18 \cdot 3^{-27}$$. Since $$18 = 2 \cdot 3^2$$,

$$\dfrac{18}{3^{27}} = 2 \cdot 3^2 \cdot 3^{-27} = 2 \cdot 3^{-25}$$

Thus:

$$\sum_{k=2}^{26} a_k = 2^{26} - 2 \cdot 3^{-25} = 2^{26} - \dfrac{2}{3^{25}}$$

Now, add the first term $$a_1 = \dfrac{2}{3^{25}}$$ to the sum:

$$S = a_1 + \sum_{k=2}^{26} a_k = \dfrac{2}{3^{25}} + \left( 2^{26} - \dfrac{2}{3^{25}} \right) = 2^{26}$$

The terms $$\dfrac{2}{3^{25}}$$ and $$-\dfrac{2}{3^{25}}$$ cancel, leaving $$2^{26}$$.

Therefore, the sum of the series is $$2^{26}$$.

The correct option is D. $$2^{26}$$.