두 직선 사이 사각형 넓이

[출제의도] 직선의 방정식을 이용하여 수열의 합을 구하는 문제를 해결한다. 두 직선 $l$, $m$의 방정식을 각각 $y = a \left( x + 1 \right) + 1 = ax + a + 1$ ($a$는 상수) $y = b \left( x + 1 \right) + 1 = bx + b + 1$ ($b$는 상수) 로 놓으면 ${\mathrm{A}} _{n} \left( n , an + a + 1 \right)$, ${\mathrm{B}} _{n} \left( n , bn + b + 1 \right)$이다. $\displaystyle \overline{\mathrm{A} _{\mathit{n}} \mathrm{B} _{\mathit{n}}}$$= \left| \left( an + a + 1 \right) - \left( bn + b + 1 \right) \right|$$= \left| a - b \right| \left( n + 1 \right)$ 이므로 $S _{n}$$\displaystyle \left. = \frac{1}{2} \left( {\overline{\mathrm{A} _{\mathit{n}} \mathrm{B} _{\mathit{n}}}} + {\overline{\mathrm{A} _{\mathit{n} \mathrm{+} 1} \mathrm{B} _{\mathit{n} + 1}}} \right) \times \left\{ \left( n + 1 \right) - n \right\} \right.$ $\displaystyle = \frac{1}{2} \left\{ \left| a - b \right| \left( n + 1 \right) + \left| a - b \right| \left( n + 2 \right) \right\}$ $\displaystyle = \frac{1}{2} \left| a - b \right| \left( 2 n + 3 \right)$ $\displaystyle \sum\limits _{k = 1} ^{10} S _{2 k - 1}$$\displaystyle = \sum\limits _{k = 1} ^{10} \frac{1}{2} \left| a - b \right| \left\{ 2 \left( 2 k - 1 \right) + 3 \right\}$ $\displaystyle = \frac{1}{2} \left| a - b \right| \sum\limits _{k = 1} ^{10} \left( 4 k + 1 \right)$ $\displaystyle = \frac{1}{2} \left| a - b \right| \left( 4 \times \frac{10 \times 11}{2} + 10 \right)$ $= 115 \left| a - b \right|$ $\displaystyle \sum\limits _{k = 1} ^{10} S _{2 k - 1} = 115$에서 $\left| a - b \right| = 1$이므로 $\displaystyle S _{n} = \frac{1}{2} \left( 2 n + 3 \right)$ 이다. $S _{2k}$$\displaystyle = \frac{1}{2} \left( 4 k + 3 \right)$이므로 $\displaystyle \sum\limits _{k = 1} ^{10} S _{2 k}$$\displaystyle = \frac{1}{2} \sum\limits _{k = 1} ^{10} \left( 4 k + 3 \right)$ $\displaystyle = \frac{1}{2} \left( 4 \times \frac{10 \times 11}{2} + 30 \right)$ $= 125$