[출제의도] 등비수열의 극한을 이용하여 수열의 합 문제 해결하기
a k = lim n → ∞ 2 × ( k 10 ) 2 n + 1 + ( k 10 ) n ( k 10 ) 2 n + ( k 10 ) n + 1 \displaystyle a _{k} = \lim\limits _{n \rightarrow \infty} {\frac{2 \times \left( \frac{k}{10} \right) ^{2 n + 1} + \left( \frac{k}{10} \right) ^{n}}{\left( \frac{k}{10} \right) ^{2 n} + \left( \frac{k}{10} \right) ^{n} + 1}} a k = n → ∞ lim ( 10 k ) 2 n + ( 10 k ) n + 1 2 × ( 10 k ) 2 n + 1 + ( 10 k ) n 0 < k 10 < 1 \displaystyle 0 < \frac{k}{10} < 1 0 < 10 k < 1 0 < k < 10 0 < k < 10 0 < k < 10 a k = lim n → ∞ 2 × ( k 10 ) 2 n + 1 + ( k 10 ) n ( k 10 ) 2 n + ( k 10 ) n + 1 \displaystyle a _{k} = \lim\limits _{n \rightarrow \infty} {\frac{2 \times \left( \frac{k}{10} \right) ^{2 n + 1} + \left( \frac{k}{10} \right) ^{n}}{\left( \frac{k}{10} \right) ^{2 n} + \left( \frac{k}{10} \right) ^{n} + 1}} a k = n → ∞ lim ( 10 k ) 2 n + ( 10 k ) n + 1 2 × ( 10 k ) 2 n + 1 + ( 10 k ) n = 2 × 0 + 0 0 + 0 + 1 = 0 \displaystyle = \frac{2 \times 0 + 0}{0 + 0 + 1} = 0 = 0 + 0 + 1 2 × 0 + 0 = 0 k 10 = 1 \displaystyle \frac{k}{10} = 1 10 k = 1 k = 10 k = 10 k = 10 a k = lim n → ∞ 2 × 1 2 n + 1 + 1 n 1 2 n + 1 n + 1 = 3 3 = 1 \displaystyle a _{k} = \lim\limits _{n \rightarrow \infty} {\frac{2 \times 1 ^{2 n + 1} + 1 ^{n}}{1 ^{2 n} + 1 ^{n} + 1}} = \frac{3}{3} = 1 a k = n → ∞ lim 1 2 n + 1 n + 1 2 × 1 2 n + 1 + 1 n = 3 3 = 1 k 10 > 1 \displaystyle \frac{k}{10} > 1 10 k > 1 k > 10 k > 10 k > 10 a k = lim n → ∞ 2 × ( k 10 ) 2 n + 1 + ( k 10 ) n ( k 10 ) 2 n + ( k 10 ) n + 1 \displaystyle a _{k} = \lim\limits _{n \rightarrow \infty} {\frac{2 \times \left( \frac{k}{10} \right) ^{2 n + 1} + \left( \frac{k}{10} \right) ^{n}}{\left( \frac{k}{10} \right) ^{2 n} + \left( \frac{k}{10} \right) ^{n} + 1}} a k = n → ∞ lim ( 10 k ) 2 n + ( 10 k ) n + 1 2 × ( 10 k ) 2 n + 1 + ( 10 k ) n = lim n → ∞ 2 × ( k 10 ) + 1 ( k 10 ) n 1 + 1 ( k 10 ) n + 1 ( k 10 ) 2 n \displaystyle = \lim\limits _{n \rightarrow \infty} {\frac{2 \times \left( \frac{k}{10} \right) + \frac{1}{\left( \frac{k}{10} \right) ^{n}}}{1 + \frac{1}{\left( \frac{k}{10} \right) ^{n}} + \frac{1}{\left( \frac{k}{10} \right) ^{2 n}}}} = n → ∞ lim 1 + ( 10 k ) n 1 + ( 10 k ) 2 n 1 2 × ( 10 k ) + ( 10 k ) n 1 = k 5 + 0 1 + 0 + 0 \displaystyle = \frac{\frac{k}{5} + 0}{1 + 0 + 0} = 1 + 0 + 0 5 k + 0 = k 5 \displaystyle = \frac{k}{5} = 5 k a k = { 0 1 ( k < 10 ) ( k = 10 ) k 5 ( k > 10 ) \displaystyle a _{k} = {\begin{cases} \begin{matrix} 0 \\ 1 \end{matrix} & \begin{matrix} ( k < 10 ) \\ ( k = 10 ) \end{matrix} \\ \frac{k}{5} & ( k > 10 ) \end{cases}} a k = ⎩ ⎨ ⎧ 0 1 5 k ( k < 10 ) ( k = 10 ) ( k > 10 ) ∑ k = 1 20 a k = ∑ k = 1 9 a k + a 10 + ∑ k = 11 20 a k \displaystyle \sum\limits _{k = 1} ^{20} a _{k} = \sum\limits _{k = 1} ^{9} a _{k} + a _{10} + \sum\limits _{k = 11} ^{20} a _{k} k = 1 ∑ 20 a k = k = 1 ∑ 9 a k + a 10 + k = 11 ∑ 20 a k = ∑ k = 1 9 0 + 1 + ∑ k = 11 20 k 5 \displaystyle = \sum\limits _{k = 1} ^{9} 0 + 1 + \sum\limits _{k = 11} ^{20} \frac{k}{5} = k = 1 ∑ 9 0 + 1 + k = 11 ∑ 20 5 k = 1 + ∑ k = 1 10 ( 2 + k 5 ) \displaystyle = 1 + \sum\limits _{k = 1} ^{10} \left( 2 + \frac{k}{5} \right) = 1 + k = 1 ∑ 10 ( 2 + 5 k ) = 1 + 20 + 1 5 × 10 × 11 2 \displaystyle = 1 + 20 + \frac{1}{5} \times \frac{10 \times 11}{2} = 1 + 20 + 5 1 × 2 10 × 11 = 32 = 32 = 32