부등식 증명 빈칸

수열 $\left\{ a _{n} \right\}$이 모든 자연수 $n$에 대하여 $1 \cdot 2 a _{1} + 3 \cdot 4 a _{2} + 5 \cdot 6 a _{3} + \cdots + ( 2 n - 1 ) \cdot 2 n a _{n} \geq n$ 을 만족시킬 때, 다음은 부등식 $a _{1} + a _{2} + a _{3} + \cdots + a _{n} \geq \square {( \text{가} )}$ 이 성립함을 증명한 것이다.‘가’형

<증명> $a _{1} + a _{2} + a _{3} + \cdots + a _{n}$ $\displaystyle = \left( 1 - \frac{1}{2} \right) ( 1 \cdot 2 a _{1} ) + \left( \frac{1}{3} - \frac{1}{4} \right) ( 3 \cdot 4 a _{2} ) + \left( \frac{1}{5} - \frac{1}{6} \right) ( 5 \cdot 6 a _{3} )$$+ \cdots$ $\displaystyle + \left( \frac{1}{2 n - 1} - \frac{1}{2 n} \right) \left\{ ( 2 n - 1 ) \cdot 2 n a _{n} \right\}$ $\displaystyle = \left( 1 - \frac{1}{2} - \frac{1}{3} + \frac{1}{4} \right) ( 1 \cdot 2 a _{1} ) + \left( \frac{1}{3} - \frac{1}{4} - \frac{1}{5} + \frac{1}{6} \right) ( 1 \cdot 2 a _{1} + 3 \cdot 4 a _{2} )$ $\displaystyle + \left( \frac{1}{5} - \frac{1}{6} - \frac{1}{7} + \frac{1}{8} \right) ( 1 \cdot 2 a _{1} + 3 \cdot 4 a _{2} + 5 \cdot 6 a _{3} )$$+ \cdots$ $\displaystyle + \left( \frac{1}{2 n - 1} - \frac{1}{2 n} \right) \left\{ 1 \cdot 2 a _{1} + 3 \cdot 4 a _{2} + \cdots + ( 2 n - 1 ) \cdot 2 n a _{n} \right\}$$\geq$ $\displaystyle \left( 1 - \frac{1}{2} - \frac{1}{3} + \frac{1}{4} \right) + 2 \left( \frac{1}{3} - \frac{1}{4} - \frac{1}{5} + \frac{1}{6} \right) + 3 \left( \frac{1}{5} - \frac{1}{6} - \frac{1}{7} + \frac{1}{8} \right)$ $+ \cdots + \square {( \text{나} )}$ $\displaystyle = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots + \frac{1}{2 n - 1} - \frac{1}{2 n}$ $\displaystyle = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{2 n - 1} + \frac{1}{2 n} - \square {( \text{다} )}$ $= \square {( \text{가} )}$

위의 과정에서 (가), (나), (다)에 알맞은 것은? [4점] (가) (나) (다) ①$\displaystyle \sum\limits _{k = 1} ^{n} \frac{1}{n + k}$ $\displaystyle \left( \frac{1}{2 n - 1} - \frac{1}{2 n} \right)$ $\displaystyle 2 \left( \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots + \frac{1}{2 n} \right)$②$\displaystyle \sum\limits _{k = 1} ^{n} \frac{1}{n + k}$ $\displaystyle n \left( \frac{1}{2 n - 1} - \frac{1}{2 n} \right)$ $\displaystyle 2 \left( \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots + \frac{1}{2 n} \right)$③$\displaystyle \sum\limits _{k = 1} ^{2 n} \frac{1}{n + k}$ $\displaystyle n \left( \frac{1}{2 n - 1} - \frac{1}{2 n} \right)$ $\displaystyle 2 \left( \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots + \frac{1}{2 n} \right)$④$\displaystyle \sum\limits _{k = 1} ^{2 n} \frac{1}{n + k}$ $\displaystyle n \left( \frac{1}{2 n - 1} - \frac{1}{2 n} \right)$ $\displaystyle \left( \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots + \frac{1}{2 n} \right)$⑤$\displaystyle \sum\limits _{k = 1} ^{2 n} \frac{1}{n + k}$ $\displaystyle \left( \frac{1}{2 n - 1} - \frac{1}{2 n} \right)$ $\displaystyle \left( \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots + \frac{1}{2 n} \right)$