25.
坐标平面中的点列 ( a 1 , b 1 ) , ( a 2 , b 2 ) , ( a 3 , b 3 ) , … (a_1, b_1), (a_2, b_2), (a_3, b_3), \ldots ( a 1 , b 1 ) , ( a 2 , b 2 ) , ( a 3 , b 3 ) , … ( a 100 , b 100 ) = ( 2 , 4 ) (a_{100}, b_{100}) = (2, 4) ( a 100 , b 100 ) = ( 2 , 4 ) a 1 + b 1 a_1 + b_1 a 1 + b 1 ( a n + 1 , b n + 1 ) = ( 3 a n − b n , 3 b n + a n ) ( n = 1 , 2 , 3 , … ) \begin{aligned} &(a_{n+1}, b_{n+1}) \\ &= \left(\sqrt{3}\,a_n - b_n,\; \sqrt{3}\,b_n + a_n\right) \\ &\quad (n = 1, 2, 3, \ldots) \end{aligned} ( a n + 1 , b n + 1 ) = ( 3 a n − b n , 3 b n + a n ) ( n = 1 , 2 , 3 , … )
A sequence ( a 1 , b 1 ) , ( a 2 , b 2 ) , ( a 3 , b 3 ) , … (a_1, b_1), (a_2, b_2), (a_3, b_3), \ldots ( a 1 , b 1 ) , ( a 2 , b 2 ) , ( a 3 , b 3 ) , … ( a n + 1 , b n + 1 ) = ( 3 a n − b n , 3 b n + a n ) ( n = 1 , 2 , 3 , … ) \begin{aligned} &(a_{n+1}, b_{n+1}) \\ &= \left(\sqrt{3}\,a_n - b_n,\; \sqrt{3}\,b_n + a_n\right) \\ &\quad (n = 1, 2, 3, \ldots) \end{aligned} ( a n + 1 , b n + 1 ) = ( 3 a n − b n , 3 b n + a n ) ( n = 1 , 2 , 3 , … ) ( a 100 , b 100 ) = ( 2 , 4 ) . (a_{100}, b_{100}) = (2, 4). ( a 100 , b 100 ) = ( 2 , 4 ) . a 1 + b 1 ? a_1 + b_1? a 1 + b 1 ?
a − 1 2 97 -\dfrac{1}{2^{97}} − 2 97 1
b − 1 2 99 -\dfrac{1}{2^{99}} − 2 99 1
d 1 2 98 \dfrac{1}{2^{98}} 2 98 1
e 1 2 96 \dfrac{1}{2^{96}} 2 96 1
答案：D 解答： 令 z n = a n + b n i z_n = a_n + b_n i z n = a n + b n i z n + 1 = z n ( 3 + i ) z_{n+1} = z_n(\sqrt{3} + i) z n + 1 = z n ( 3 + i ) z 100 = z 1 ( 3 + i ) 99 z_{100} = z_1(\sqrt{3} + i)^{99} z 100 = z 1 ( 3 + i ) 99 z n + 1 = ( 3 a n − b n ) + ( 3 b n + a n ) i = ( a n + b n i ) ( 3 + i ) , \begin{aligned} z_{n+1} &= (\sqrt{3}\,a_n - b_n) \\ &\quad {}+ (\sqrt{3}\,b_n + a_n)i \\ &= (a_n + b_n i)(\sqrt{3} + i), \end{aligned} z n + 1 = ( 3 a n − b n ) + ( 3 b n + a n ) i = ( a n + b n i ) ( 3 + i ) ,
因为 3 + i = 2 ( cos 3 0 ∘ + i sin 3 0 ∘ ) \sqrt{3} + i = 2(\cos 30^\circ + i\sin 30^\circ) 3 + i = 2 ( cos 3 0 ∘ + i sin 3 0 ∘ ) ( 3 + i ) 99 (\sqrt{3} + i)^{99} ( 3 + i ) 99 = 2 99 ( cos 297 0 ∘ + i sin 297 0 ∘ ) = 2^{99}(\cos 2970^\circ + i\sin 2970^\circ) = 2 99 ( cos 297 0 ∘ + i sin 297 0 ∘ ) 297 0 ∘ 2970^\circ 297 0 ∘ 9 0 ∘ 90^\circ 9 0 ∘ 2 99 i 2^{99} i 2 99 i
于是 2 + 4 i = z 1 ⋅ 2 99 i 2 + 4i = z_1 \cdot 2^{99} i 2 + 4 i = z 1 ⋅ 2 99 i z 1 = 2 + 4 i 2 99 i = 4 − 2 i 2 99 . z_1 = \dfrac{2 + 4i}{2^{99} i} = \dfrac{4 - 2i}{2^{99}}. z 1 = 2 99 i 2 + 4 i = 2 99 4 − 2 i .
因此 a 1 = 4 2 99 a_1 = \tfrac{4}{2^{99}} a 1 = 2 99 4 b 1 = − 2 2 99 b_1 = -\tfrac{2}{2^{99}} b 1 = − 2 99 2 a 1 + b 1 = 2 2 99 = 1 2 98 . a_1 + b_1 = \dfrac{2}{2^{99}} = \dfrac{1}{2^{98}}. a 1 + b 1 = 2 99 2 = 2 98 1 .
所以正确答案是 D 。
Let z n = a n + b n i . z_n = a_n + b_n i. z n = a n + b n i . z n + 1 = ( 3 a n − b n ) + ( 3 b n + a n ) i = ( a n + b n i ) ( 3 + i ) , \begin{aligned} z_{n+1} &= (\sqrt{3}\,a_n - b_n) \\ &\quad {}+ (\sqrt{3}\,b_n + a_n)i \\ &= (a_n + b_n i)(\sqrt{3} + i), \end{aligned} z n + 1 = ( 3 a n − b n ) + ( 3 b n + a n ) i = ( a n + b n i ) ( 3 + i ) , z n + 1 = z n ( 3 + i ) z_{n+1} = z_n(\sqrt{3} + i) z n + 1 = z n ( 3 + i ) z 100 = z 1 ( 3 + i ) 99 . z_{100} = z_1(\sqrt{3} + i)^{99}. z 100 = z 1 ( 3 + i ) 99 .
Since 3 + i = 2 ( cos 3 0 ∘ + i sin 3 0 ∘ ) , \sqrt{3} + i = 2(\cos 30^\circ + i\sin 30^\circ), 3 + i = 2 ( cos 3 0 ∘ + i sin 3 0 ∘ ) , ( 3 + i ) 99 (\sqrt{3} + i)^{99} ( 3 + i ) 99 = 2 99 ( cos 297 0 ∘ + i sin 297 0 ∘ ) . = 2^{99}(\cos 2970^\circ + i\sin 2970^\circ). = 2 99 ( cos 297 0 ∘ + i sin 297 0 ∘ ) . 297 0 ∘ 2970^\circ 297 0 ∘ 9 0 ∘ , 90^\circ, 9 0 ∘ , 2 99 i . 2^{99} i. 2 99 i .
Thus 2 + 4 i = z 1 ⋅ 2 99 i , 2 + 4i = z_1 \cdot 2^{99} i, 2 + 4 i = z 1 ⋅ 2 99 i , z 1 = 2 + 4 i 2 99 i = 4 − 2 i 2 99 . z_1 = \dfrac{2 + 4i}{2^{99} i} = \dfrac{4 - 2i}{2^{99}}. z 1 = 2 99 i 2 + 4 i = 2 99 4 − 2 i .
Then a 1 = 4 2 99 a_1 = \tfrac{4}{2^{99}} a 1 = 2 99 4 b 1 = − 2 2 99 , b_1 = -\tfrac{2}{2^{99}}, b 1 = − 2 99 2 , a 1 + b 1 = 2 2 99 = 1 2 98 . a_1 + b_1 = \dfrac{2}{2^{99}} = \dfrac{1}{2^{98}}. a 1 + b 1 = 2 99 2 = 2 98 1 .
Thus, D is the correct answer.
