5점 · 2-1 2-1 [풀이] 함수 f ( x ) = log 2 x ( x ≥ 1 ) \displaystyle f ( x ) = \log_{2} x ( x \geq 1 ) f ( x ) = log 2 x ( x ≥ 1 ) g ( x ) \displaystyle g ( x ) g ( x ) g ( x ) = 2 x \displaystyle g ( x ) = 2^{x} g ( x ) = 2 x
10점 · 2-1 제시문 <가>와 <다>로부터 다음을 얻는다. b m = lim n → ∞ ∑ k = 1 n g ( n + m k − k n ) m − 1 n = lim n → ∞ ∑ k = 1 n g ( 1 + ( m − 1 ) k n ) m − 1 n = ∫ 1 m g ( x ) d x = ∫ 1 m 2 x d x = 2 x ln 2 ∣ 1 m = 2 m − 2 ln 2 \displaystyle \begin{aligned}b_{m} &= \lim_{n \to \infty} {\sum_{k = 1}^{n}} g \left( \frac{{n + m k - k}}{n} \right) \frac{{m - 1}}{n}\\&= \lim_{n \to \infty} {\sum_{k = 1}^{n}} g \left( 1 + \frac{{( m - 1 ) k}}{n} \right) \frac{{m - 1}}{n}\\&= \int_{1}^{m} {g ( x ) d x}\\&= \int_{1}^{m} {2^{x} d x} = \left. \frac{{2^{x}}}{\ln 2} \right|_{1}^{m} = \frac{{2^{m} - 2}}{\ln 2}\end{aligned} b m = n → ∞ lim k = 1 ∑ n g ( n n + mk − k ) n m − 1 = n → ∞ lim k = 1 ∑ n g ( 1 + n ( m − 1 ) k ) n m − 1 = ∫ 1 m g ( x ) d x = ∫ 1 m 2 x d x = ln 2 2 x 1 m = ln 2 2 m − 2
10점 · 2-2 2-2 [풀이] a m = c ( b m + 1 − b m ) = c ( 2 m + 1 − 2 m ) ln 2 = c 2 m ln 2 \displaystyle \begin{aligned}a_{m} &= c ( b_{m + 1} - b_{m} )\\&= \frac{{c ( 2^{m + 1} - 2^{m} )}}{\ln 2} = c \frac{{2^{m}}}{\ln 2}\end{aligned} a m = c ( b m + 1 − b m ) = ln 2 c ( 2 m + 1 − 2 m ) = c ln 2 2 m 1 a m \displaystyle \frac{{1}}{a_{m}} a m 1 ∑ 1 100 1 a m = 1 \displaystyle \sum_{1}^{100} \frac{{1}}{a_{m}} = 1 1 ∑ 100 a m 1 = 1 Correction. “1 / a m \displaystyle 1/a_m 1/ a m 1 / a m \displaystyle 1/a_m 1/ a m 1 / a m \displaystyle 1/a_m 1/ a m X \displaystyle X X m \displaystyle m m 그러나 원칙에 따라 원문 표기를 그대로 실었다. 한편, ∑ 1 100 1 a m = 1 c ∑ 1 100 ln 2 2 m = 1 c ln 2 ∑ 1 100 1 2 m = 1 c ln 2 ( 1 − 1 2 100 ) \displaystyle \begin{aligned}\sum_{1}^{100} \frac{{1}}{a_{m}} &= \frac{{1}}{c} \sum_{1}^{100} \frac{{\ln 2}}{2^{m}}\\&= \frac{{1}}{c} \ln 2 \sum_{1}^{100} \frac{{1}}{2^{m}} = \frac{{1}}{c} \ln 2 \left( 1 - \frac{{1}}{2^{100}} \right)\end{aligned} 1 ∑ 100 a m 1 = c 1 1 ∑ 100 2 m ln 2 = c 1 ln 2 1 ∑ 100 2 m 1 = c 1 ln 2 ( 1 − 2 100 1 ) c = ln 2 ( 1 − 1 2 100 ) \displaystyle c = \ln 2 \left( 1 - \frac{{1}}{2^{100}} \right) c = ln 2 ( 1 − 2 100 1 )
5점 · 2-3 2-3 [풀이] a m = ( 1 − 1 2 100 ) 2 m \displaystyle a_{m} = \left( 1 - \frac{{1}}{2^{100}} \right) 2^{m} a m = ( 1 − 2 100 1 ) 2 m 1 a m = 1 ( 1 − 1 2 100 ) 2 m \displaystyle \frac{{1}}{a_{m}} = \frac{{1}}{\left( 1 - \frac{{1}}{2^{100}} \right) 2^{m}} a m 1 = ( 1 − 2 100 1 ) 2 m 1 기댓값 E ( X ) \displaystyle E ( X ) E ( X ) E ( X ) = ∑ k = 1 100 k 1 a k \displaystyle E ( X ) = \sum_{k = 1}^{100} k \frac{{1}}{a_{k}} E ( X ) = k = 1 ∑ 100 k a k 1 ∑ k = 1 100 k 1 a k = 2 100 2 100 − 1 ∑ k = 1 100 k 1 2 k = 2 100 2 100 − 1 ∑ k = 1 100 k 2 k \displaystyle \begin{aligned}\sum_{k = 1}^{100} k \frac{{1}}{a_{k}} &= \frac{{2^{100}}}{2^{100} - 1} \sum_{k = 1}^{100} k \frac{{1}}{2^{k}}\\&= \frac{{2^{100}}}{2^{100} - 1} \sum_{k = 1}^{100} \frac{{k}}{2^{k}}\end{aligned} k = 1 ∑ 100 k a k 1 = 2 100 − 1 2 100 k = 1 ∑ 100 k 2 k 1 = 2 100 − 1 2 100 k = 1 ∑ 100 2 k k 따라서 E ( X ) = 2 100 2 100 − 1 ∑ k = 1 100 k 2 k \displaystyle E ( X ) = \frac{{2^{100}}}{2^{100} - 1} \sum_{k = 1}^{100} \frac{{k}}{2^{k}} E ( X ) = 2 100 − 1 2 100 k = 1 ∑ 100 2 k k
15점 · 2-3 한편, S = ∑ k = 1 100 k 2 k = 1 2 + 2 2 2 + 3 2 3 + 4 2 4 + ⋯ + 100 2 100 \displaystyle S = \sum_{k = 1}^{100} \frac{{k}}{2^{k}} = \frac{{1}}{2} + \frac{{2}}{2^{2}} + \frac{{3}}{2^{3}} + \frac{{4}}{2^{4}} + \cdots + \frac{{100}}{2^{100}} S = k = 1 ∑ 100 2 k k = 2 1 + 2 2 2 + 2 3 3 + 2 4 4 + ⋯ + 2 100 100 S − 1 2 S = 1 2 + 1 2 2 + 1 2 3 + 1 2 4 + ⋯ + 1 2 100 − 100 2 101 \displaystyle S - \frac{{1}}{2} S = \frac{{1}}{2} + \frac{{1}}{2^{2}} + \frac{{1}}{2^{3}} + \frac{{1}}{2^{4}} + \cdots + \frac{{1}}{2^{100}} - \frac{{100}}{2^{101}} S − 2 1 S = 2 1 + 2 2 1 + 2 3 1 + 2 4 1 + ⋯ + 2 100 1 − 2 101 100 S = 2 − 1 2 99 − 100 2 100 \displaystyle S = 2 - \frac{{1}}{2^{99}} - \frac{{100}}{2^{100}} S = 2 − 2 99 1 − 2 100 100 E ( X ) = 2 100 2 100 − 1 ( 2 − 1 2 99 − 100 2 100 ) \displaystyle E ( X ) = \frac{{2^{100}}}{2^{100} - 1} \left( 2 - \frac{{1}}{2^{99}} - \frac{{100}}{2^{100}} \right) E ( X ) = 2 100 − 1 2 100 ( 2 − 2 99 1 − 2 100 100 )
수리논술, 배움에서 논증의 완성까지.