+5점 · [문제 1-1] (i) [문제 1-1]
(i)
● r = 3 : \displaystyle r=3 : r = 3 :
S 2 r = S 2 3 = S 8 = 1 + 1 2 + ( 1 3 + 1 4 ) + ( 1 5 + ⋯ + 1 8 ) \displaystyle S _{2 ^{r}} =S _{2 ^{3}} =S _{8} =1+ \frac{1}{2} + \left ( \frac{1}{3} + \frac{1}{4} \right ) + \left ( \frac{1}{5} + \cdots + \frac{1}{8} \right ) S 2 r = S 2 3 = S 8 = 1 + 2 1 + ( 3 1 + 4 1 ) + ( 5 1 + ⋯ + 8 1 )
≥ 1 + 1 2 + 2 ⋅ 1 4 + 2 2 ⋅ 1 8 = 1 + 3 ⋅ 1 2 \displaystyle \geq 1+ \frac{1}{2} +2 \cdot \frac{1}{4} +2 ^{2} \cdot \frac{1}{8} =1+3 \cdot \frac{1}{2} ≥ 1 + 2 1 + 2 ⋅ 4 1 + 2 2 ⋅ 8 1 = 1 + 3 ⋅ 2 1
● r = 7 : \displaystyle r=7 : r = 7 :
S 2 r = S 2 7 = S 128 \displaystyle S _{2 ^{r}} =S _{2 ^{7}} =S _{128} S 2 r = S 2 7 = S 128
= 1 + 1 2 + ( 1 2 1 + 1 + 1 2 2 ) + ( 1 2 2 + 1 + ⋯ + 1 2 3 ) + ⋯ + ( 1 2 6 + 1 + ⋯ + 1 2 7 ) \displaystyle =1+ \frac{1}{2} + \left ( \frac{1}{2 ^{1} +1} + \frac{1}{2 ^{2}} \right ) + \left ( \frac{1}{2 ^{2} +1} + \cdots + \frac{1}{2 ^{3}} \right ) + \cdots + \left ( \frac{1}{2 ^{6} +1} + \cdots + \frac{1}{2 ^{7}} \right ) = 1 + 2 1 + ( 2 1 + 1 1 + 2 2 1 ) + ( 2 2 + 1 1 + ⋯ + 2 3 1 ) + ⋯ + ( 2 6 + 1 1 + ⋯ + 2 7 1 )
≥ 1 + 1 2 + 2 ⋅ 1 2 2 + 2 2 ⋅ 1 2 3 + ⋯ + 2 6 ⋅ 1 2 7 = 1 + 7 ⋅ 1 2 \displaystyle \geq 1+ \frac{1}{2} +2 \cdot \frac{1}{2 ^{2}} +2 ^{2} \cdot \frac{1}{2 ^{3}} + \cdots +2 ^{6} \cdot \frac{1}{2 ^{7}} =1+7 \cdot \frac{1}{2} ≥ 1 + 2 1 + 2 ⋅ 2 2 1 + 2 2 ⋅ 2 3 1 + ⋯ + 2 6 ⋅ 2 7 1 = 1 + 7 ⋅ 2 1
+7점 · [문제 1-1] (ii) (ii)
수학적 귀납법을 이용하여 r = 0 , 1 , 2 , ⋯ \displaystyle r=0, 1, 2, \cdots r = 0 , 1 , 2 , ⋯
S 2 r ≥ 1 + r 2 ⋯ ⋯ ⋯ ( 1 ) \displaystyle S _{2 ^{r}} \geq 1+ \frac{r}{2} \cdots \cdots \cdots (1) S 2 r ≥ 1 + 2 r ⋯⋯⋯ ( 1 )
(a) r = 0 \displaystyle r=0 r = 0
S 2 r = S 2 0 = S 1 = 1 ≥ 1 = 1 + 0 2 = 1 + r 2 \displaystyle S _{2 ^{r}} =S _{2 ^{0}} =S _{1} =1 \geq 1=1+ \frac{0}{2} =1+ \frac{r}{2} S 2 r = S 2 0 = S 1 = 1 ≥ 1 = 1 + 2 0 = 1 + 2 r
(b) r = k \displaystyle r=k r = k r = k + 1 \displaystyle r=k+1 r = k + 1
S 2 k + 1 = 1 + 1 2 + ⋯ + 1 2 k + 1 2 k + 1 + ⋯ + 1 2 k + 1 \displaystyle S _{2 ^{k+1}} =1+ \frac{1}{2} + \cdots + \frac{1}{2 ^{k}} + \frac{1}{2 ^{k} +1} + \cdots + \frac{1}{2 ^{k+1}} S 2 k + 1 = 1 + 2 1 + ⋯ + 2 k 1 + 2 k + 1 1 + ⋯ + 2 k + 1 1
= S 2 k + 1 2 k + 1 + ⋯ + 1 2 k + 1 \displaystyle =S _{2 ^{k}} + \frac{1}{2 ^{k} +1} + \cdots + \frac{1}{2 ^{k+1}} = S 2 k + 2 k + 1 1 + ⋯ + 2 k + 1 1
≥ 1 + k 2 + 1 2 k + 1 + ⋯ + 1 2 k + 1 \displaystyle \geq 1+ \frac{k}{2} + \frac{1}{2 ^{k} +1} + \cdots + \frac{1}{2 ^{k+1}} ≥ 1 + 2 k + 2 k + 1 1 + ⋯ + 2 k + 1 1 r = k \displaystyle r=k r = k
≥ 1 + k 2 + 2 k ⋅ 1 2 k + 1 = 1 + k + 1 2 \displaystyle \geq 1+ \frac{k}{2} +2 ^{k} \cdot \frac{1}{2 ^{k+1}} =1+ \frac{k+1}{2} ≥ 1 + 2 k + 2 k ⋅ 2 k + 1 1 = 1 + 2 k + 1
즉, r = k + 1 \displaystyle r=k+1 r = k + 1
+3점 · [문제 1-2] [문제 1-2]
(1)식으로부터 r → ∞ \displaystyle r \to \infty r → ∞ S 2 r → ∞ \displaystyle S _{2 ^{r}} \to \infty S 2 r → ∞ ∑ k = 1 ∞ 1 k \displaystyle \sum _{k=1} ^{\infty } \frac{1}{k} k = 1 ∑ ∞ k 1
+5점 · [문제 1-3] 1단계 [문제 1-3]
<1단계> 아래 그림에서 빗금으로 표시된 직사각형 면적의 합을 L n \displaystyle L _{n} L n
L n = 1 ⋅ 1 2 + 1 ⋅ 1 3 + ⋯ + 1 ⋅ 1 n = S n − 1 \displaystyle L _{n} =1 \cdot \frac{1}{2} +1 \cdot \frac{1}{3} + \cdots +1 \cdot \frac{1}{n} =S _{n} -1 L n = 1 ⋅ 2 1 + 1 ⋅ 3 1 + ⋯ + 1 ⋅ n 1 = S n − 1 L n ≤ A n \displaystyle L _{n} \leq A _{n} L n ≤ A n
그러므로 S n ≤ A n + 1 \displaystyle S _{n} \leq A _{n} +1 S n ≤ A n + 1
+5점 · [문제 1-3] 2단계 <2단계> 아래 그림에서 빗금으로 표시된 사다리꼴 면적의 합을 U n \displaystyle U _{n} U n
U n = 1 2 ( 1 + 1 2 ) + 1 2 ( 1 2 + 1 3 ) + ⋯ + 1 2 ( 1 n − 1 + 1 n ) \displaystyle U _{n} = \frac{1}{2} \left ( 1+ \frac{1}{2} \right ) + \frac{1}{2} \left ( \frac{1}{2} + \frac{1}{3} \right ) + \cdots + \frac{1}{2} \left ( \frac{1}{n-1} + \frac{1}{n} \right ) U n = 2 1 ( 1 + 2 1 ) + 2 1 ( 2 1 + 3 1 ) + ⋯ + 2 1 ( n − 1 1 + n 1 )
= 1 2 ( 1 + 1 2 + 1 3 + ⋯ + 1 n − 1 ) + 1 2 ( 1 2 + 1 3 + 1 4 + ⋯ + 1 n ) \displaystyle = \frac{1}{2} \left ( 1+ \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n-1} \right ) + \frac{1}{2} \left ( \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{n} \right ) = 2 1 ( 1 + 2 1 + 3 1 + ⋯ + n − 1 1 ) + 2 1 ( 2 1 + 3 1 + 4 1 + ⋯ + n 1 )
= 1 2 ( S n − 1 n ) + 1 2 ( S n − 1 ) = S n − 1 2 n − 1 2 \displaystyle = \frac{1}{2} \left ( S _{n} - \frac{1}{n} \right ) + \frac{1}{2} \left ( S _{n} -1 \right ) =S _{n} - \frac{1}{2n} - \frac{1}{2} = 2 1 ( S n − n 1 ) + 2 1 ( S n − 1 ) = S n − 2 n 1 − 2 1
또한, A n ≤ U n \displaystyle A _{n} \leq U _{n} A n ≤ U n A n + 1 2 n + 1 2 ≤ S n \displaystyle A _{n} + \frac{1}{2n} + \frac{1}{2} \leq S _{n} A n + 2 n 1 + 2 1 ≤ S n
그러므로, <1단계>와 <2단계>에 의하여 A n + 1 2 + 1 2 n ≤ S n ≤ A n + 1 \displaystyle A _{n} + \frac{1}{2} + \frac{1}{2n} \leq S _{n} \leq A _{n} +1 A n + 2 1 + 2 n 1 ≤ S n ≤ A n + 1
+5점 · [문제 1-4] [문제 1-4]
[문제 1-3]의 결과로부터 S 5 − 1 ≤ A 5 ≤ S 5 − 1 10 − 1 2 \displaystyle S _{5} -1 \leq A _{5} \leq S _{5} - \frac{1}{10} - \frac{1}{2} S 5 − 1 ≤ A 5 ≤ S 5 − 10 1 − 2 1
S 5 − 1 = 1 2 + 1 3 + 1 4 + 1 5 = 1.28 3 ˙ \displaystyle S _{5} -1= \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} =1.28 \dot{3} S 5 − 1 = 2 1 + 3 1 + 4 1 + 5 1 = 1.28 3 ˙
S 5 − 1 10 − 1 2 = 1 + 1 2 + 1 3 + 1 4 + 1 5 − 0.1 − 1 2 = 1.68 3 ˙ \displaystyle S _{5} - \frac{1}{10} - \frac{1}{2} =1+ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} -0.1- \frac{1}{2} =1.68 \dot{3} S 5 − 10 1 − 2 1 = 1 + 2 1 + 3 1 + 4 1 + 5 1 − 0.1 − 2 1 = 1.68 3 ˙
위의 결과로부터 1.28 3 ˙ ≤ A 5 ≤ 1.68 3 ˙ \displaystyle 1.28 \dot{3} \leq A _{5} \leq 1.68 \dot{3} 1.28 3 ˙ ≤ A 5 ≤ 1.68 3 ˙ A 5 > 1.7 \displaystyle A _{5} >1.7 A 5 > 1.7
[문제 1-1] 수학적 귀납법을 정확히 알고 있는지 평가한다.
[문제 1-2] [문제 1-1]에서 주어진 부등식에 대한 극한 개념을 이해할 수 있는지 평가한다.
[문제 1-3] 제시문 나)에서 설명된 기호 A n \displaystyle A_n A n
[문제 1-4] [문제 1-3]에서 제공된 부등식을 활용할 수 있는지 평가한다.
수리논술, 이게 다다.
