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수악중독

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)
\[\sum \limits_{k=1}^{n} \dfrac{1}{n+k}\] \[\left ( \dfrac{1}{2n-1}-\dfrac{1}{2n} \right )\] \[2 \left ( \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{6} + \cdots +\dfrac{1}{2n} \right )\]
\[\sum \limits_{k=1}^{n} \dfrac{1}{n+k}\] \[n\left ( \dfrac{1}{2n-1}-\dfrac{1}{2n} \right )\] \[2 \left ( \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{6} + \cdots +\dfrac{1}{2n} \right )\]
\[\sum \limits_{k=1}^{2n} \dfrac{1}{n+k}\] \[n\left ( \dfrac{1}{2n-1}-\dfrac{1}{2n} \right )\] \[2 \left ( \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{6} + \cdots +\dfrac{1}{2n} \right )\]
\[\sum \limits_{k=1}^{2n} \dfrac{1}{n+k}\] \[n\left ( \dfrac{1}{2n-1}-\dfrac{1}{2n} \right )\] \[\left ( \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{6} + \cdots +\dfrac{1}{2n} \right )\]
\[\sum \limits_{k=1}^{2n} \dfrac{1}{n+k}\] \[\left ( \dfrac{1}{2n-1}-\dfrac{1}{2n} \right )\] \[\left ( \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{6} + \cdots +\dfrac{1}{2n} \right )\]

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