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97南二中

97南二中

先感謝信哥提供了10幾年前的題目
讓版上的後生有機會能練習到這些題目

由於這份沒看到相關的討論串 因此自開一串
之後版上沒討論串 沒答案的 小弟會盡自己所能把答案算出附上
再請各位先進指教

另外想請問這份的計算1 2
1. \(\displaystyle \frac{53}{55} \)
2. (1) \(\displaystyle \frac{35\sqrt{30}}{6} \)  (2)\(\displaystyle \frac{35\sqrt{30}}{6\sqrt{1081}}\) (超醜...不知道是不是算錯)
3.12
4.80
5.\({\left[ \begin{array}{ccc}
\frac{1}{2} & \frac{\sqrt{3}}{2} \\
\frac{\sqrt{3}}{2}&-\frac{1}{2}
\end{array}
\right ]} \)
6.\(\displaystyle 2\sqrt{2} \)
7.ABD
8.\( 8\pi + 40\tan^{-1}2 \)
9.x+3y=6
10.\( m=\pm\frac{4}{3}, \pm\frac{2\sqrt{5}}{3} \)
11.\(\displaystyle -1<m<1 \)
12.\(\displaystyle \frac{1}{2}\)
13.10

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97南二中.rar (159.81 KB)

2021-1-22 12:15, 下載次數: 4243

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計算3
\(\langle\;F_n\rangle\;\)為費氏數列,即\(F_1=F_2=1\)且\(F_{n+2}=F_{n+1}+F_n\)在\(n \in N\)均成立,令\(\displaystyle \lambda_n=\frac{F_{2n}}{F_{2n-1}}\)。
(1)證明\(\lambda_{n+1}=2-\frac{1}{1+\lambda_n}\)
(2)證明\(\langle\; \lambda_n \rangle\;\)為遞增數列
(3)證明\(\lambda_n\)收斂
(4)求\(\langle\;\lambda_n \rangle\;\)之極限
[解答]
(1)考慮\(\displaystyle F_{2n+2}=F_{2n+1}+F_{2n} \)
\(\displaystyle  \lambda_{n+1}=1+\frac{F_{2n}}{F_{2n+1}} =2-(1-\frac{F_{2n}}{F_{2n+1}}) \)
\(\displaystyle  \lambda_{n+1}=2-(\frac{F_{2n+1}-F_{2n}}{F_{2n+1}})=2-(\frac{F_{2n-1}}{F_{2n+1}})\
=2-\frac{1}{\frac{F_{2n+1}}{F_{2n-1}}}=2-\frac{1}{\frac{F_{2n}+F_{2n-1}}{F_{2n-1}}}\
=2-\frac{1}{1+\lambda_{n}} \)

(2)\(\displaystyle \lambda_{1}=1, \lambda_{2}=\frac{3}{2}, \lambda_{3}=\frac{8}{5} \)
設\(n=k \)時,\(\displaystyle  \lambda_{k+1}> \lambda_{k} \)
當\(n=k+1 \)時,\(\displaystyle \lambda_{k+2}=2-\frac{1}{1+\lambda_{k+1}}>2-\frac{1}{1+\lambda_{k}}=\lambda_{k+1} \)
根據數學歸納法
\(\displaystyle {\lambda_{n}} \)遞增

(3)\(\displaystyle \lambda_{1}=1<2, \lambda_{2}=\frac{3}{2}<2, \lambda_{3}=\frac{8}{5}<2 \)
設\(\displaystyle n=1,2,...k, \lambda_{n}<2 \)皆成立
此時可知\(\displaystyle \lambda_{k+1}=2-\frac{1}{1+\lambda_{k}}<2-\frac{1}{3}=\frac{5}{3}<2 \)
當\(n=k+1 \)時
\(\displaystyle \lambda_{k+2}=2-\frac{1}{1+\lambda_{k+1}}<2-\frac{1}{3}=\frac{5}{3}<2 \)
根據數學歸納法
\(\displaystyle {\lambda_{n}} \)有界

(4)因為\(\displaystyle {\lambda_{n}} \)遞增且有界,所以收斂,令其收斂值為\(\alpha \)
可得 \(\displaystyle \alpha=2-(\frac{1}{1+\alpha}) \)
解得\(\displaystyle \alpha = \frac{1+\sqrt{5}}{2} \)

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回復 3# tsusy 的帖子

謝謝寸斯老師的指教
以下有幾個問題想再次請教
7,10小弟明白了
8.是否能詳細講一下過程
9.所以答案這題應該要怎麼寫才好,另外x+3y=0看起來並不會平分面積
13.如何知道這題是要求樣本而非母體標準差

8的部分 小弟是考慮
(1)\(\displaystyle (2x+y)(2x-y)\leq 0 , x^2+y^2\geq 28 \)
(2)\(\displaystyle (2x+y)(2x-y)\geq 0 , x^2+y^2\leq 28 \)
當然自然限制為\(\displaystyle x^2+y^2 \leq 36 \)
這樣的圖形是否有誤

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回復 6# tsusy 的帖子

謝謝寸斯老師的指教

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