The sum of all rational terms in the expansion of $$(2 + \sqrt{3})^8$$ is

Write the binomial expansion of $$(2+\sqrt{3})^{8}$$:

$$ (2+\sqrt{3})^{8} = \sum_{k=0}^{8} \binom{8}{k}\,2^{\,8-k}\,(\sqrt{3})^{\,k} $$

The factor $$(\sqrt{3})^{\,k}$$ is rational only when $$k$$ is even, because

$$ (\sqrt{3})^{\,k} = \begin{cases} 3^{\,k/2}, & k \text{ even}\\[4pt] 3^{\,(k-1)/2}\sqrt{3}, & k \text{ odd} \end{cases} $$

Hence we keep only even values of $$k$$ (0, 2, 4, 6, 8) and evaluate each term.

Case 1: $$k = 0$$
$$ \binom{8}{0}\,2^{\,8}\,3^{\,0} = 1 \times 256 \times 1 = 256 $$

Case 2: $$k = 2$$
$$ \binom{8}{2}\,2^{\,6}\,3^{\,1} = 28 \times 64 \times 3 = 5376 $$

Case 3: $$k = 4$$
$$ \binom{8}{4}\,2^{\,4}\,3^{\,2} = 70 \times 16 \times 9 = 10080 $$

Case 4: $$k = 6$$
$$ \binom{8}{6}\,2^{\,2}\,3^{\,3} = 28 \times 4 \times 27 = 3024 $$

Case 5: $$k = 8$$
$$ \binom{8}{8}\,2^{\,0}\,3^{\,4} = 1 \times 1 \times 81 = 81 $$

Add the five rational contributions:

$$ 256 + 5376 + 10080 + 3024 + 81 = 18817 $$

The sum of all rational terms in the expansion of $$(2+\sqrt{3})^{8}$$ is $$18817$$.

Therefore, the correct option is Option D.