+30점 · 답 a 2 = 3 \displaystyle a_2=3 a 2 = 3 a 3 = 15 \displaystyle a_3=15 a 3 = 15 a 4 = 105 \displaystyle a_4=105 a 4 = 105 【1-1】 집합 X = { 1 , 2 , ⋯ , 2 n } \displaystyle X=\{1,2,\cdots,2n\} X = { 1 , 2 , ⋯ , 2 n } k \displaystyle k k f ( k ) = l \displaystyle f(k)=l f ( k ) = l f ( x ) \displaystyle f(x) f ( x ) X \displaystyle X X x \displaystyle x x f ( x ) ≠ x \displaystyle f(x)\ne x f ( x ) = x k ≠ l \displaystyle k\ne l k = l f ( x ) \displaystyle f(x) f ( x ) X \displaystyle X X x \displaystyle x x f ∘ f ( x ) = x \displaystyle f\circ f(x)=x f ∘ f ( x ) = x f ( l ) = k \displaystyle f(l)=k f ( l ) = k a n \displaystyle a_n a n 2 n \displaystyle 2n 2 n 2 \displaystyle 2 2 a 2 = 4 C 2 × 2 C 2 2 ! = 3 , a 3 = 6 C 2 × 4 C 2 × 2 C 2 3 ! = 15 , a 4 = 8 C 2 × 6 C 2 × 4 C 2 × 2 C 2 4 ! = 105 \displaystyle a_2=\dfrac{{}_{4}\mathrm C_2\times{}_{2}\mathrm C_2}{2!}=3,\quad a_3=\dfrac{{}_{6}\mathrm C_2\times{}_{4}\mathrm C_2\times{}_{2}\mathrm C_2}{3!}=15,\quad a_4=\dfrac{{}_{8}\mathrm C_2\times{}_{6}\mathrm C_2\times{}_{4}\mathrm C_2\times{}_{2}\mathrm C_2}{4!}=105 a 2 = 2 ! 4 C 2 × 2 C 2 = 3 , a 3 = 3 ! 6 C 2 × 4 C 2 × 2 C 2 = 15 , a 4 = 4 ! 8 C 2 × 6 C 2 × 4 C 2 × 2 C 2 = 105 이다.
+15점 · ln ( n ! ) = ∑ k = 1 n ln k \displaystyle \ln(n!)=\sum_{k=1}^n\ln k ln ( n !) = k = 1 ∑ n ln k +15점 · ∫ 1 n ln x d x ≤ ∑ k = 1 n ln k \displaystyle \int_1^n\ln x\,dx\le\sum_{k=1}^n\ln k ∫ 1 n ln x d x ≤ k = 1 ∑ n ln k 【1-2】 ln ( n ! ) = ∑ k = 1 n ln k \displaystyle \ln(n!)=\sum_{k=1}^n\ln k ln ( n !) = k = 1 ∑ n ln k ∫ 1 n ln x d x ≤ ln ( n ! ) \displaystyle \int_1^n\ln x\,dx\le\ln(n!) ∫ 1 n ln x d x ≤ ln ( n !)
+20점 · 수열 { a n } \displaystyle \{a_n\} { a n } a n = ( 2 n ) ! 2 n n ! \displaystyle a_n=\dfrac{(2n)!}{2^nn!} a n = 2 n n ! ( 2 n )! +10점 · 부등식 ln ( n ! 3 n ) ln ( n ! ) ≤ ln c n ln b n ≤ ln ( n ! 3 n ) ln ( n ! ) \displaystyle \dfrac{\ln\left(\sqrt{\dfrac{n!}{3^n}}\right)}{\ln(n!)}\le\dfrac{\ln c_n}{\ln b_n}\le\dfrac{\ln\left(\sqrt{n!3^n}\right)}{\ln(n!)} ln ( n !) ln ( 3 n n ! ) ≤ ln b n ln c n ≤ ln ( n !) ln ( n ! 3 n ) 【1-3】 수열 { a n } \displaystyle \{a_n\} { a n } a n = 2 n C 2 × ⋯ × 2 C 2 n ! = ( 2 n ) ! 2 n n ! \displaystyle a_n=\dfrac{{}_{2n}\mathrm C_2\times\cdots\times{}_{2}\mathrm C_2}{n!}=\dfrac{(2n)!}{2^nn!} a n = n ! 2 n C 2 × ⋯ × 2 C 2 = 2 n n ! ( 2 n )! b n = ( 2 n ) ! 2 n a n \displaystyle b_n=\dfrac{(2n)!}{2^na_n} b n = 2 n a n ( 2 n )! b n = n ! \displaystyle b_n=n! b n = n ! 모든 자연수 n \displaystyle n n n ! 3 n ≤ c n ≤ n ! 3 n \displaystyle \sqrt{\dfrac{n!}{3^n}}\le c_n\le\sqrt{n!3^n} 3 n n ! ≤ c n ≤ n ! 3 n ln ( n ! 3 n ) ln ( n ! ) ≤ ln c n ln b n ≤ ln ( n ! 3 n ) ln ( n ! ) \displaystyle \dfrac{\ln\left(\sqrt{\dfrac{n!}{3^n}}\right)}{\ln(n!)}\le\dfrac{\ln c_n}{\ln b_n}\le\dfrac{\ln\left(\sqrt{n!3^n}\right)}{\ln(n!)} ln ( n !) ln ( 3 n n ! ) ≤ ln b n ln c n ≤ ln ( n !) ln ( n ! 3 n ) 이다. 한편, ln ( n ! 3 n ) ln ( n ! ) = ln ( n ! ) − n ln 3 2 ln ( n ! ) = 1 2 − n ln 3 2 ln ( n ! ) \displaystyle \dfrac{\ln\left(\sqrt{\dfrac{n!}{3^n}}\right)}{\ln(n!)}=\dfrac{\ln(n!)-n\ln3}{2\ln(n!)}=\dfrac12-\dfrac{n\ln3}{2\ln(n!)} ln ( n !) ln ( 3 n n ! ) = 2 ln ( n !) ln ( n !) − n ln 3 = 2 1 − 2 ln ( n !) n ln 3 ln ( n ! 3 n ) ln ( n ! ) = ln ( n ! ) + n ln 3 2 ln ( n ! ) = 1 2 + n ln 3 2 ln ( n ! ) \displaystyle \dfrac{\ln\left(\sqrt{n!3^n}\right)}{\ln(n!)}=\dfrac{\ln(n!)+n\ln3}{2\ln(n!)}=\dfrac12+\dfrac{n\ln3}{2\ln(n!)} ln ( n !) ln ( n ! 3 n ) = 2 ln ( n !) ln ( n !) + n ln 3 = 2 1 + 2 ln ( n !) n ln 3 이다.
+10점 · 부등식 0 ≤ n ln 3 2 ln ( n ! ) ≤ ln 3 2 ( ln n − 1 + 1 n ) \displaystyle 0\le\dfrac{n\ln3}{2\ln(n!)}\le\dfrac{\ln3}{2\left(\ln n-1+\dfrac1n\right)} 0 ≤ 2 ln ( n !) n ln 3 ≤ 2 ( ln n − 1 + n 1 ) ln 3 +5점 · 극한값 lim n → ∞ n ln 3 2 ln ( n ! ) = 0 \displaystyle \lim_{n\to\infty}\dfrac{n\ln3}{2\ln(n!)}=0 n → ∞ lim 2 ln ( n !) n ln 3 = 0 +10점 · lim n → ∞ ln ( n ! 3 n ) ln ( n ! ) = 1 2 \displaystyle \lim_{n\to\infty}\dfrac{\ln\left(\sqrt{\dfrac{n!}{3^n}}\right)}{\ln(n!)}=\dfrac12 n → ∞ lim ln ( n !) ln ( 3 n n ! ) = 2 1 lim n → ∞ ln ( n ! 3 n ) ln ( n ! ) = 1 2 \displaystyle \lim_{n\to\infty}\dfrac{\ln\left(\sqrt{n!3^n}\right)}{\ln(n!)}=\dfrac12 n → ∞ lim ln ( n !) ln ( n ! 3 n ) = 2 1 +5점 · 답 lim n → ∞ ln c n ln b n = 1 2 \displaystyle \lim_{n\to\infty}\dfrac{\ln c_n}{\ln b_n}=\dfrac12 n → ∞ lim ln b n ln c n = 2 1 이때, 【1-2】와 제시문 (나)에 의해 0 ≤ n ln 3 2 ln ( n ! ) ≤ n ln 3 2 ∫ 1 n ln x d x = n ln 3 2 ( n ln n − n + 1 ) = ln 3 2 ( ln n − 1 + 1 n ) \displaystyle 0\le\dfrac{n\ln3}{2\ln(n!)}\le\dfrac{n\ln3}{2\int_1^n\ln x\,dx}=\dfrac{n\ln3}{2(n\ln n-n+1)}=\dfrac{\ln3}{2\left(\ln n-1+\dfrac1n\right)} 0 ≤ 2 ln ( n !) n ln 3 ≤ 2 ∫ 1 n ln x d x n ln 3 = 2 ( n ln n − n + 1 ) n ln 3 = 2 ( ln n − 1 + n 1 ) ln 3 이고 제시문 (다)에 의해 lim n → ∞ n ln 3 2 ln ( n ! ) = 0 \displaystyle \lim_{n\to\infty}\dfrac{n\ln3}{2\ln(n!)}=0 n → ∞ lim 2 ln ( n !) n ln 3 = 0 lim n → ∞ ln ( n ! 3 n ) ln ( n ! ) = lim n → ∞ ( 1 2 − n ln 3 2 ln ( n ! ) ) = 1 2 \displaystyle \lim_{n\to\infty}\dfrac{\ln\left(\sqrt{\dfrac{n!}{3^n}}\right)}{\ln(n!)}=\lim_{n\to\infty}\left(\dfrac12-\dfrac{n\ln3}{2\ln(n!)}\right)=\dfrac12 n → ∞ lim ln ( n !) ln ( 3 n n ! ) = n → ∞ lim ( 2 1 − 2 ln ( n !) n ln 3 ) = 2 1 lim n → ∞ ln ( n ! 3 n ) ln ( n ! ) = lim n → ∞ ( 1 2 + n ln 3 2 ln ( n ! ) ) = 1 2 \displaystyle \lim_{n\to\infty}\dfrac{\ln\left(\sqrt{n!3^n}\right)}{\ln(n!)}=\lim_{n\to\infty}\left(\dfrac12+\dfrac{n\ln3}{2\ln(n!)}\right)=\dfrac12 n → ∞ lim ln ( n !) ln ( n ! 3 n ) = n → ∞ lim ( 2 1 + 2 ln ( n !) n ln 3 ) = 2 1 이다. 따라서, 제시문 (다)에 의해 lim n → ∞ ln c n ln b n = 1 2 \displaystyle \lim_{n\to\infty}\dfrac{\ln c_n}{\ln b_n}=\dfrac12 n → ∞ lim ln b n ln c n = 2 1
수리논술, 배움에서 논증의 완성까지.