Solution
3+$$\frac{1}{4+\frac{1}{5+\frac{1}{6}}}$$
3+$$\frac{1}{4+\frac{1}{\frac{31}{6}}}$$
3+$$\frac{1}{4+\frac{6}{31}}$$
3+$$\frac{1}{\frac{124+6}{31}}$$
3+$$\frac{31}{130}$$
$$\frac{421}{130}$$
Hence, option A is correct.
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