연분수 간단히

[출제의도] 규칙성을 찾아 분모의 유리화를 한다. $\displaystyle \frac{1}{2 + \left( \sqrt{2} - 1 \right)}$의 분모를 유리화하면 $\displaystyle \frac{1}{2 + \left( \sqrt{2} - 1 \right)}$$\displaystyle = \frac{1}{\sqrt{2} + 1}$$\displaystyle = \frac{\sqrt{2} - 1}{\left( {\sqrt{2} + 1} \right) \left( {\sqrt{2} - 1} \right)}$$\displaystyle = \sqrt{2} - 1$ 이와 같은 방법으로 주어진 식을 차례대로 유리화하면 $\displaystyle 2 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2 + \left( \sqrt{2} - 1 \right)}}}$$\displaystyle = 2 + \frac{1}{2 + \frac{1}{2 + \left( \sqrt{2} - 1 \right)}}$ $\displaystyle = 2 + \frac{1}{2 + \left( \sqrt{2} - 1 \right)}$ $\displaystyle = 2 + \left( \sqrt{2} - 1 \right)$ $\displaystyle = \sqrt{2} + 1$ [다른 풀이] $\displaystyle 2 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2 + \left( \sqrt{2} - 1 \right)}}}$$\displaystyle = 2 + \frac{1}{2 + \frac{1}{2 + \frac{1}{\sqrt{2} + 1}}}$ $\displaystyle = 2 + \frac{1}{2 + \frac{\sqrt{2} + 1}{2 \left( \sqrt{2} + 1 \right) + \frac{1}{\sqrt{2} + 1} \left( \sqrt{2} + 1 \right)}}$ $\displaystyle = 2 + \frac{1}{2 + \frac{\sqrt{2} + 1}{2 \sqrt{2} + 2 + 1}}$ $\displaystyle = 2 + \frac{1}{2 + \frac{\sqrt{2} + 1}{3 + 2 \sqrt{2}}}$ $\displaystyle = 2 + \frac{3 + 2 \sqrt{2}}{2 \left( 3 + 2 \sqrt{2} \right) + \frac{\sqrt{2} + 1}{3 + 2 \sqrt{2}} \left( 3 + 2 \sqrt{2} \right)}$ $\displaystyle = 2 + \frac{3 + 2 \sqrt{2}}{6 + 4 \sqrt{2} + \sqrt{2} + 1}$ $\displaystyle = 2 + \frac{3 + 2 \sqrt{2}}{7 + 5 \sqrt{2}}$ $\displaystyle = 2 + \frac{\left( 3 + 2 \sqrt{2} \right) \left( 7 - 5 \sqrt{2} \right)}{\left( 7 + 5 \sqrt{2} \right) \left( 7 - 5 \sqrt{2} \right)}$ $\displaystyle = 2 + \frac{21 + 14 \sqrt{2} - 15 \sqrt{2} - 20}{49 - 50}$ $\displaystyle = 2 + \frac{1 - \sqrt{2}}{- 1}$ $\displaystyle = 2 - \left( 1 - \sqrt{2} \right)$ $\displaystyle = 2 - 1 + \sqrt{2}$ $\displaystyle = \sqrt{2} + 1$