To find the sum of the rational terms in the expansion of \( (2 + \sqrt{3})^8 \), we use the binomial theorem:
\[
(2 + \sqrt{3})^8 = \sum_{k=0}^{8} \binom{8}{k} 2^{8-k} (\sqrt{3})^k
\]
For terms with rational numbers, \( k \) must be even, as only even powers of \( \sqrt{3} \) will yield rational terms. The even values of \( k \) are \( k = 0, 2, 4, 6, 8 \). The rational terms will be:
\[
\binom{8}{0} 2^8 (\sqrt{3})^0, \quad \binom{8}{2} 2^6 (\sqrt{3})^2, \quad \binom{8}{4} 2^4 (\sqrt{3})^4, \quad \binom{8}{6} 2^2 (\sqrt{3})^6, \quad \binom{8}{8} 2^0 (\sqrt{3})^8
\]
Simplifying and adding the rational terms, we get a total sum of 18280.
