[출제의도] 등비수열의 극한을 이용하여 수열의 합 문제 해결하기 함수 f ( x ) = x − 1 2 x − 6 \displaystyle f \left( x \right) = \frac{x - 1}{2 x - 6} f ( x ) = 2 x − 6 x − 1 ∣ f ( 3 − a ) ∣ = ∣ ( 3 − a ) − 1 2 ( 3 − a ) − 6 ∣ = ∣ 2 − a − 2 a ∣ = ∣ a − 2 2 a ∣ \displaystyle \left| f \left( 3 - a \right) \right| = \left| \frac{\left( 3 - a \right) - 1}{2 \left( 3 - a \right) - 6} \right| = \left| \frac{2 - a}{- 2 a} \right| = \left| \frac{a - 2}{2 a} \right| ∣ f ( 3 − a ) ∣ = 2 ( 3 − a ) − 6 ( 3 − a ) − 1 = − 2 a 2 − a = 2 a a − 2 ∣ 1 − f ( 3 + a ) ∣ = ∣ 1 − ( 3 + a ) − 1 2 ( 3 + a ) − 6 ∣ = ∣ a − 2 2 a ∣ \displaystyle \left| 1 - f \left( 3 + a \right) \right| = \left| 1 - \frac{\left( 3 + a \right) - 1}{2 \left( 3 + a \right) - 6} \right| = \left| \frac{a - 2}{2 a} \right| ∣ 1 − f ( 3 + a ) ∣ = 1 − 2 ( 3 + a ) − 6 ( 3 + a ) − 1 = 2 a a − 2 h ( a ) = a − 2 2 a \displaystyle h \left( a \right) = \frac{a - 2}{2 a} h ( a ) = 2 a a − 2 ( a ≠ 0 ) \left( a \ne 0 \right) ( a = 0 ) lim n → ∞ ∣ f ( 3 − a ) ∣ n + 1 2 n + ∣ 1 − f ( 3 + a ) ∣ n = lim n → ∞ ∣ h ( a ) ∣ n + 1 2 n + ∣ h ( a ) ∣ n \displaystyle \lim\limits _{n \rightarrow \infty} {\frac{\left| f \left( 3 - a \right) \right| ^{n + 1}}{2 ^{n} + \left| 1 - f \left( 3 + a \right) \right| ^{n}}} = \lim\limits _{n \rightarrow \infty} {\frac{\left| h \left( a \right) \right| ^{n + 1}}{2 ^{n} + \left| h \left( a \right) \right| ^{n}}} n → ∞ lim 2 n + ∣ 1 − f ( 3 + a ) ∣ n ∣ f ( 3 − a ) ∣ n + 1 = n → ∞ lim 2 n + ∣ h ( a ) ∣ n ∣ h ( a ) ∣ n + 1 ∣ h ( a ) ∣ < 2 \left| h \left( a \right) \right| < 2 ∣ h ( a ) ∣ < 2 ∣ h ( a ) 2 ∣ < 1 \displaystyle \left| \frac{h \left( a \right)}{2} \right| < 1 2 h ( a ) < 1 lim n → ∞ ∣ h ( a ) 2 ∣ n = 0 \displaystyle \lim\limits _{n \rightarrow \infty} {\left| \frac{h \left( a \right)}{2} \right|} ^{n} = 0 n → ∞ lim 2 h ( a ) n = 0 lim n → ∞ ∣ h ( a ) ∣ n + 1 2 n + ∣ h ( a ) ∣ n = lim n → ∞ 2 ∣ h ( a ) 2 ∣ n + 1 1 + ∣ h ( a ) 2 ∣ n = 0 \displaystyle \lim\limits _{n \rightarrow \infty} {\frac{\left| h \left( a \right) \right| ^{n + 1}}{2 ^{n} + \left| h \left( a \right) \right| ^{n}}} = \lim\limits _{n \rightarrow \infty} {\frac{2 \left| \frac{h \left( a \right)}{2} \right| ^{n + 1}}{1 + \left| \frac{h \left( a \right)}{2} \right| ^{n}} = 0} n → ∞ lim 2 n + ∣ h ( a ) ∣ n ∣ h ( a ) ∣ n + 1 = n → ∞ lim 1 + 2 h ( a ) n 2 2 h ( a ) n + 1 = 0 k = 0 k = 0 k = 0 ∣ h ( a ) ∣ = 2 \left| h \left( a \right) \right| = 2 ∣ h ( a ) ∣ = 2 lim n → ∞ ∣ h ( a ) ∣ n + 1 2 n + ∣ h ( a ) ∣ n = lim n → ∞ 2 n + 1 2 n + 2 n = 1 \displaystyle \lim\limits _{n \rightarrow \infty} {\frac{\left| h \left( a \right) \right| ^{n + 1}}{2 ^{n} + \left| h \left( a \right) \right| ^{n}}} = \lim\limits _{n \rightarrow \infty} {\frac{2 ^{n + 1}}{2 ^{n} + 2 ^{n}} = 1} n → ∞ lim 2 n + ∣ h ( a ) ∣ n ∣ h ( a ) ∣ n + 1 = n → ∞ lim 2 n + 2 n 2 n + 1 = 1 k = 1 k = 1 k = 1 ∣ h ( a ) ∣ > 2 \left| h \left( a \right) \right| > 2 ∣ h ( a ) ∣ > 2 ∣ 2 h ( a ) ∣ < 1 \displaystyle \left| \frac{2}{h \left( a \right)} \right| < 1 h ( a ) 2 < 1 lim n → ∞ ∣ 2 h ( a ) ∣ n = 0 \displaystyle \lim\limits _{n \rightarrow \infty} {\left| \frac{2}{h \left( a \right)} \right|} ^{n} = 0 n → ∞ lim h ( a ) 2 n = 0 lim n → ∞ ∣ h ( a ) ∣ n + 1 2 n + ∣ h ( a ) ∣ n = lim n → ∞ ∣ h ( a ) ∣ ∣ 2 h ( a ) ∣ n + 1 = ∣ h ( a ) ∣ \displaystyle \lim\limits _{n \rightarrow \infty} {\frac{\left| h \left( a \right) \right| ^{n + 1}}{2 ^{n} + \left| h \left( a \right) \right| ^{n}}} = \lim\limits _{n \rightarrow \infty} {\frac{\left| h \left( a \right) \right|}{\left| \frac{2}{h \left( a \right)} \right| ^{n} + 1} = \left| h \left( a \right) \right|} n → ∞ lim 2 n + ∣ h ( a ) ∣ n ∣ h ( a ) ∣ n + 1 = n → ∞ lim h ( a ) 2 n + 1 ∣ h ( a ) ∣ = ∣ h ( a ) ∣ ∣ h ( a ) ∣ = ∣ a − 2 2 a ∣ = ∣ 1 a − 1 2 ∣ = k \displaystyle \left| h \left( a \right) \right| = \left| \frac{a - 2}{2 a} \right| = \left| \frac{1}{a} - \frac{1}{2} \right| = k ∣ h ( a ) ∣ = 2 a a − 2 = a 1 − 2 1 = k k ≥ 3 k {\geq} 3 k ≥ 3 a a a 1 a − 1 2 = k \displaystyle \frac{1}{a} - \frac{1}{2} = k a 1 − 2 1 = k a = 2 2 k + 1 \displaystyle a = \frac{2}{2 k + 1} a = 2 k + 1 2 1 a − 1 2 = − k \displaystyle \frac{1}{a} - \frac{1}{2} = - k a 1 − 2 1 = − k a = − 2 2 k − 1 \displaystyle a = - \frac{2}{2 k - 1} a = − 2 k − 1 2 g ( k ) = − 2 ( 1 2 k − 1 − 1 2 k + 1 ) \displaystyle g \left( k \right) = - 2 \left( \frac{1}{2 k - 1} - \frac{1}{2 k + 1} \right) g ( k ) = − 2 ( 2 k − 1 1 − 2 k + 1 1 ) ∑ k = 3 17 g ( k ) = − 2 ∑ k = 3 17 ( 1 2 k − 1 − 1 2 k + 1 ) \displaystyle \sum\limits _{k = 3} ^{17} g \left( k \right) = - 2 \sum\limits _{k = 3} ^{17} \left( \frac{1}{2 k - 1} - \frac{1}{2 k + 1} \right) k = 3 ∑ 17 g ( k ) = − 2 k = 3 ∑ 17 ( 2 k − 1 1 − 2 k + 1 1 ) = − 2 ( 1 5 − 1 7 + 1 7 − 1 9 + ⋯ + 1 33 − 1 35 ) \displaystyle = - 2 \left( \frac{1}{5} - \frac{1}{7} + \frac{1}{7} - \frac{1}{9} + \cdots + \frac{1}{33} - \frac{1}{35} \right) = − 2 ( 5 1 − 7 1 + 7 1 − 9 1 + ⋯ + 33 1 − 35 1 ) = − 2 ( 1 5 − 1 35 ) = − 12 35 \displaystyle = - 2 \left( \frac{1}{5} - \frac{1}{35} \right) = - \frac{12}{35} = − 2 ( 5 1 − 35 1 ) = − 35 12