삼각형 넓이 합

[출제의도] 여러 가지 수열을 활용하여 문제해결하기 점 $\displaystyle \mathrm{P} _{\mathit{n}} \left( \mathit{\sqrt{n}} , n \right)$을 지나고 직선 $\displaystyle y = \sqrt{n} x$와 수직인 직선의 방정식은 $\displaystyle y = - \frac{1}{\sqrt{n}} \left( x - \sqrt{n} \right) + n$이므로 $\displaystyle \mathrm{Q} _{\mathit{n}} \left( \mathit{\left( n + 1 \right)} \sqrt{n} , 0 \right)$, $\mathrm{R} _{\mathit{n}} \left( 0 , \mathit{n} + 1 \right)$ $\displaystyle S _{n} = \frac{1}{2} \times \left( n + 1 \right) \sqrt{n} \times \left( n + 1 \right)$ 따라서 $\displaystyle \sum\limits _{n = 1} ^{5} \frac{2 S _{n}}{\sqrt{n}} = \sum\limits _{n = 1} ^{5} \left\{ \frac{2}{\sqrt{n}} \times \frac{\left( n + 1 \right) ^{2} \sqrt{n}}{2} \right\}$ $\displaystyle = \sum\limits _{n = 1} ^{5} \left( n + 1 \right) ^{2}$ $\displaystyle = \sum\limits _{n = 1} ^{5} \left( n ^{2} + 2 n + 1 \right)$ $\displaystyle = \frac{5 \times 6 \times 11}{6} + 2 \times \frac{5 \times 6}{2} + 5$$= 90$