Question

Express the following surds as continued fractions, and find the sixth convergent to $3\sqrt{5}$.

Answer 1

$3\sqrt{5}=1+3\sqrt{5}-1$

$=1+\frac{44}{1+3\sqrt{5}}$

$=1+\frac{44}{1+1+\frac{44}{1+3\sqrt{5}}}$

$=1+\frac{44}{2+\frac{44}{1+3\sqrt{5}}}$

$=1+\frac{44}{2+\frac{44}{1+1+\frac{44}{1+3\sqrt{5}}}}$

$=1+\frac{44}{2+\frac{44}{2+\frac{44}{2+...}}}$

$3\sqrt{5}=[1,23,\frac{45}{23},...]$