다음은 A = ( 1 1 0 2 ) \displaystyle A = \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} A = ( 1 0 1 2 ) A n A ^{n} A n
모든 자연수 n n n A n = ( a n b n c n d n ) \displaystyle A ^{n} = \begin{pmatrix} a _{n} & b _{n} \\ c _{n} & d _{n} \end{pmatrix} A n = ( a n c n b n d n ) A n + 1 = A ⋅ A n = A n ⋅ A A ^{n + 1} = A \cdot A ^{n} = A ^{n} \cdot A A n + 1 = A ⋅ A n = A n ⋅ A ( a n + 1 b n + 1 c n + 1 d n + 1 ) = ( 1 1 0 2 ) ( a n b n c n d n ) = ( a n + c n b n + d n 2 c n 2 d n ) \displaystyle \begin{aligned} \begin{pmatrix} a _{n + 1} & b _{n + 1} \\ c _{n + 1} & d _{n + 1} \end{pmatrix} & = \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} a _{n} & b _{n} \\ c _{n} & d _{n} \end{pmatrix} = \begin{pmatrix} a _{n} + c _{n} & b _{n} + d _{n} \\ 2 c _{n} & 2 d _{n} \end{pmatrix} \end{aligned} ( a n + 1 c n + 1 b n + 1 d n + 1 ) = ( 1 0 1 2 ) ( a n c n b n d n ) = ( a n + c n 2 c n b n + d n 2 d n ) = ( a n b n c n d n ) ( 1 1 0 2 ) = ( a n a n + 2 b n c n c n + 2 d n ) \displaystyle = \begin{pmatrix} a _{n} & b _{n} \\ c _{n} & d _{n} \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} a _{n} & a _{n} + 2 b _{n} \\ c _{n} & c _{n} + 2 d _{n} \end{pmatrix} = ( a n c n b n d n ) ( 1 0 1 2 ) = ( a n c n a n + 2 b n c n + 2 d n ) { a n + 1 = a n + c n = a n b n + 1 = b n + d n = a n + 2 b n c n + 1 = 2 c n = c n d n + 1 = 2 d n = c n + 2 d n \displaystyle \begin{cases} a _{n + 1} = a _{n} + c _{n} = a _{n} {}_{} {}_{} \\ b _{n + 1} = b _{n} + d _{n} = a _{n} + 2 b _{n} {}_{} {}_{} \\ c _{n + 1} = 2 c _{n} = c _{n} {}_{} {}_{} \\ d _{n + 1} = 2 d _{n} = c _{n} + 2 d _{n} \end{cases} ⎩ ⎨ ⎧ a n + 1 = a n + c n = a n b n + 1 = b n + d n = a n + 2 b n c n + 1 = 2 c n = c n d n + 1 = 2 d n = c n + 2 d n b n + 1 = 2 b n + b _{n + 1} = 2 b _{n} + b n + 1 = 2 b n + d n + 1 = d _{n + 1} = d n + 1 = ∴ \therefore ∴ A n = A ^{n} = A n =
이 과정에서 (가), (나), (다)에 알맞은 것은? [3점]
(가) (나) (다) ① 1 2 \displaystyle \frac{1}{2} 2 1 2 n 2 ^{n} 2 n ( 1 2 n − 1 0 2 n + 1 ) \displaystyle \begin{pmatrix} 1 & 2 ^{n} - 1 \\ 0 & 2 ^{n + 1} \end{pmatrix} ( 1 0 2 n − 1 2 n + 1 ) ② 1 2 \displaystyle \frac{1}{2} 2 1 2 n + 1 2 ^{n + 1} 2 n + 1 ( 1 2 n + 1 0 2 n − 1 ) \displaystyle \begin{pmatrix} 1 & 2 ^{n + 1} \\ 0 & 2 ^{n} - 1 \end{pmatrix} ( 1 0 2 n + 1 2 n − 1 ) ③ 1 1 1 2 n 2 ^{n} 2 n ( 1 2 n − 1 0 2 n + 1 ) \displaystyle \begin{pmatrix} 1 & 2 ^{n} - 1 \\ 0 & 2 ^{n + 1} \end{pmatrix} ( 1 0 2 n − 1 2 n + 1 ) ④ 1 1 1 2 n + 1 2 ^{n + 1} 2 n + 1 ( 1 2 n − 1 0 2 n ) \displaystyle \begin{pmatrix} 1 & 2 ^{n} - 1 \\ 0 & 2 ^{n} \end{pmatrix} ( 1 0 2 n − 1 2 n ) ⑤ 1 1 1 2 n + 1 2 ^{n + 1} 2 n + 1 ( 1 2 n 0 2 n − 1 ) \displaystyle \begin{pmatrix} 1 & 2 ^{n} \\ 0 & 2 ^{n} - 1 \end{pmatrix} ( 1 0 2 n 2 n − 1 )