행렬 곱과 수열 보기

[출제의도] 행렬의 성질을 이용하여 추론하기 ㄱ. $\displaystyle {\begin{pmatrix} a _{2} \\ b _{2} \end{pmatrix}} = {\begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}} {\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}} {\begin{pmatrix} 0 \\ 1 \end{pmatrix}} = {\begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}} {\begin{pmatrix} 1 \\ 1 \end{pmatrix}} = {\begin{pmatrix} 3 \\ 2 \end{pmatrix}}$이므로 $\displaystyle \frac{a _{2}}{b _{2}} = \frac{3}{2}$이다. (참) ㄴ. $\displaystyle {\begin{pmatrix} a _{n + 1} \\ b _{n + 1} \end{pmatrix}} = A _{n + 1} A _{n} \cdots A _{2} A _{1} {\begin{pmatrix} 0 \\ 1 \end{pmatrix}}$ $\displaystyle = A _{n + 1} {\begin{pmatrix} a _{n} \\ b _{n} \end{pmatrix}}$ (참) ㄷ. $\displaystyle \begin{aligned} {\begin{pmatrix} a _{n + 1} \\ b _{n + 1} \end{pmatrix}} & = {\begin{pmatrix} n + 1 & 1 \\ 0 & n + 1 \end{pmatrix}} {\begin{pmatrix} a _{n} \\ b _{n} \end{pmatrix}} & = {\begin{pmatrix} \left( n + 1 \right) a _{n} + b _{n} \\ \left( n + 1 \right) b _{n} \end{pmatrix}} \end{aligned}$ $\displaystyle \frac{a _{n + 1}}{b _{n + 1}} = \frac{\left( n + 1 \right) a _{n} + b _{n}}{\left( n + 1 \right) b _{n}} = \frac{a _{n}}{b _{n}} + \frac{1}{n + 1}$ $\displaystyle \frac{a _{n}}{b _{n}} = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} = \sum\limits _{k = 1} ^{n} \frac{1}{k}$ $\therefore$$\displaystyle \frac{a _{n}}{b _{n}} = \sum\limits _{k = 1} ^{n} \frac{1}{k}$ (참)