Notice
Recent Posts
«   2020/04   »
1 2 3 4
5 6 7 8 9 10 11
12 13 14 15 16 17 18
19 20 21 22 23 24 25
26 27 28 29 30
Tags more
Archives
Today
772
Total
2,130,485
관리 메뉴

## 수학1_여러 가지 수열_난이도 중 본문

(8차) 수학1 질문과 답변/수열

### 수학1_여러 가지 수열_난이도 중

수악중독 2012. 3. 13. 13:44
수열 $$\{a_n\}$$ 이 모든 자연수 $$n$$ 에 대하여 $1\cdot 2a_1 + 3 \cdot 4a_2 +5 \cdot 6 a_3 + \cdots + (2n-1)\cdot 2na_n \ge n$을 만족시킬 때, 다음은 부등식 $a_1 +a_2 +a_3 +\cdots + a_n \ge \;(가)$이 성립함을 증명한 것이다.

$$a_1 +a_2 +a_3 +\cdots +a_n$$

$$=\left (1- \dfrac{1}{2} \right ) \left (1 \cdot 2a_1 \right )+ \left (\dfrac{1}{3}- \dfrac{1}{4} \right ) \left (3 \cdot 4a_2 \right ) + \left (\dfrac{1}{5}- \dfrac{1}{6} \right ) \left (5 \cdot 6a_3 \right )$$

$$+ \cdots + \left (\dfrac{1}{2n-1}- \dfrac{1}{2n} \right ) \left \{ (2n-1) \cdot 2a_n \right \}$$

$$=\left ( 1-\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4} \right ) ( 1 \cdot 2a_1 )$$

$$+ \left ( \dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{6} \right ) \left ( 1 \cdot 2a_1 + 3 \cdot 4a_2 \right )$$

$$+\left ( \dfrac{1}{5}-\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{8} \right ) \left ( 1 \cdot 2a_1 + 3 \cdot 4a_2 +5 \cdot 6a_3 \right )$$

$$+ \cdots$$

$$+ \left ( \dfrac{1}{2n-1} - \dfrac{1}{2n} \right ) \left \{ 1 \cdot 2a_1 + 3 \cdot 4a_2 + \cdots + (2n-1) \cdot 2na_n \right \}$$

$$\ge \left (1- \dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4} \right ) + 2 \left ( \dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{6}\right ) + 3 \left ( \dfrac{1}{5} - \dfrac{1}{6} - \dfrac{1}{7} +\dfrac{1}{8} \right )$$

$$+ \cdots + \;(나)$$

$$=1 - \dfrac{1}{2} +\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}- \dfrac{1}{6}+ \cdots + \dfrac{1}{2n-1}-\dfrac{1}{2n}$$

$$=1 +\dfrac{1}{2} +\dfrac{1}{3}+\dfrac{1}{4}+ \cdots + \dfrac{1}{2n-1}+\dfrac{1}{2n}-\;(다)$$

위의 과정에서 (가), (나), (다)에 알맞은 것은?

 (가) (나) (다) ① $\sum \limits_{k=1}^{n} \dfrac{1}{n+k}$ $\left ( \dfrac{1}{2n-1}-\dfrac{1}{2n} \right )$ $2 \left ( \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{6} + \cdots +\dfrac{1}{2n} \right )$ ② $\sum \limits_{k=1}^{n} \dfrac{1}{n+k}$ $n\left ( \dfrac{1}{2n-1}-\dfrac{1}{2n} \right )$ $2 \left ( \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{6} + \cdots +\dfrac{1}{2n} \right )$ ③ $\sum \limits_{k=1}^{2n} \dfrac{1}{n+k}$ $n\left ( \dfrac{1}{2n-1}-\dfrac{1}{2n} \right )$ $2 \left ( \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{6} + \cdots +\dfrac{1}{2n} \right )$ ④ $\sum \limits_{k=1}^{2n} \dfrac{1}{n+k}$ $n\left ( \dfrac{1}{2n-1}-\dfrac{1}{2n} \right )$ $\left ( \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{6} + \cdots +\dfrac{1}{2n} \right )$ ⑤ $\sum \limits_{k=1}^{2n} \dfrac{1}{n+k}$ $\left ( \dfrac{1}{2n-1}-\dfrac{1}{2n} \right )$ $\left ( \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{6} + \cdots +\dfrac{1}{2n} \right )$