Let \( S = (2 + \sqrt{3})^8 \). The sum of the rational terms in the binomial expansion is obtained by calculating \( (2 + \sqrt{3})^8 + (2 - \sqrt{3})^8 \). This operation eliminates all irrational terms due to cancellation in the symmetric expansion. Thus, the sum of the rational terms is \( \frac{(2 + \sqrt{3})^8 + (2 - \sqrt{3})^8}{2} \). Alternatively, we can select terms from the binomial expansion where the exponent of \( \sqrt{3} \) is even, ensuring rationality. These terms are: \[ \binom{8}{0}(2)^8 + \binom{8}{2}(2)^6(\sqrt{3})^2 + \binom{8}{4}(2)^4(\sqrt{3})^4 + \binom{8}{6}(2)^2(\sqrt{3})^6 + \binom{8}{8}(\sqrt{3})^8 \] Which simplifies to: \[ 2^8 + 28 \cdot 2^6 \cdot 3 + 70 \cdot 2^4 \cdot 9 + 28 \cdot 2^2 \cdot 27 + 1 \cdot 81 \] And the final sum is: \[ 256 + 5376 + 10080 + 3024 + 81 = 18817 \]
If for \( 3 \leq r \leq 30 \), \( ^{30}C_{30-r} + 3 \left( ^{30}C_{31-r} \right) + 3 \left( ^{30}C_{32-r} \right) + ^{30}C_{33-r} = ^m C_r \), then \( m \) equals to_________