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[ 本帖最後由 tacokao 於 2013-4-17 04:49 PM 編輯 ]

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$$a_1 = 1$$

$$a_2 = r$$

$$\displaystyle a_3 = \frac{\left(a_2\right)^2}{a_1}=r^2$$

$$a_4=2a_3-a_2 = r\left(2r-1\right)$$

$$\displaystyle a_5=\frac{\left(a_4\right)^2}{a_3}=\left(2r-1\right)^2$$

$$a_6=2a_5-a_4 = \left(2r-1\right)\left(3r-2\right)$$

$$\displaystyle a_7=\frac{\left(a_6\right)^2}{a_5}=\left(3r-2\right)^2$$

$$a_8=2a_7-a_6 = \left(3r-2\right)\left(4r-3\right)$$

$$\displaystyle a_9=\frac{\left(a_8\right)^2}{a_7}=\left(4r-3\right)^2$$

$$a_{10}=2a_9-a_8 = \left(4r-3\right)\left(5r-4\right)$$

（應該可以用數學歸納法證明這件事～：Ｐ）

$$a_{2n}=\left(\left(n-1\right)r-\left(n-2\right)\cdot\left(nr-\left(n-1\right)\right)\right)<1000\Rightarrow n\leq8$$

$$\Rightarrow k=2\times8=16$$

$$a_{2n}=\left(\left(n-1\right)r-\left(n-2\right)\cdot\left(nr-\left(n-1\right)\right)\right)<1000\Rightarrow n\leq7$$

$$\Rightarrow k=2\times7+1=15$$

...

4分而已....好少....

##### 引用:

8.有一個數列，$$a_1=1$$且$$a_9+a_{10}=646$$。此數列的第一、第二、第三項成等比數列，第二、第三、第四項成等差數列；且一般而言，對所有的$$n \ge 1$$，$$a_{2n-1},a_{2n}$$及$$a_{2n+1}$$成等比數列，$$a_{2n},a_{2n+1}$$及$$a_{2n+2}$$成等差數列。設$$a_k$$為此數列中小於1000的最大項，試求$$k=$$？

A sequence of positive integers with $$a_1=1$$ and $$a_9+a_{10}=646$$ is formed so that the first three terms are in geometric progression, the second, third, and fourth terms are in arithmetic progression, and, in general, for all $$n \ge 1$$, the terms $$a_{2n-1}$$, $$a_{2n}$$, $$a_{2n+1}$$ are in geometric progression, and the terms $$a_{2n}$$, $$a_{2n+1}$$, and $$a_{2n+2}$$ are in arithmetic progression. Let $$a_{n}$$ be the greatest term in this sequence that is less than 1000. Find $$n+a_{n}$$.
(2004AIME第九題，http://www.artofproblemsolving.c ... id=45&year=2004)

A(0,0), B(0,5), C(5,0)

a(ABC)=25/2
a(ABP)=24/2

a(ABP)是5*4/2=20/2

a(ABP)  的底 為5  高為4才對  所以面積 為10

[ 本帖最後由 王保丹 於 2013-4-20 08:51 PM 編輯 ]

$$\displaystyle \left(\left(\sqrt{a+1}\right)^2+\left(\sqrt{b+3}\right)^2\right)\left(1^2+1^2\right)\geq\left(1\cdot\sqrt{a+1}+1\cdot\sqrt{b+3}\right)^2$$

$$\displaystyle \Rightarrow \frac{\left(\sqrt{a+1}+\sqrt{b+3}\right)^2}{a+b+4}\leq 2$$

$$\displaystyle \Rightarrow \frac{\sqrt{a+1}+\sqrt{b+3}}{\sqrt{a+b+4}}\leq \sqrt{2}$$

[ 本帖最後由 shiauy 於 2013-4-25 10:09 PM 編輯 ]

https://math.pro/db/attachment.php?aid=1612&k=624fd7dbc0644d1d9732b7eef2a9874f&t=1534882839

$$\displaystyle \left(n+\frac{3}{2}\right)^2+\frac{39}{4} = 4k^2$$

$$\displaystyle \Rightarrow \left(4k\right)^2 - \left(2n+3\right)^2= 39$$

case 1: $$\displaystyle 4k-2n-3=1, 4k+2n+3=39 \Rightarrow k=5, n=8$$

case 2: $$\displaystyle 4k-2n-3=3, 4k+2n+3=13 \Rightarrow k=1, n=1$$ （不合）

99996，排列 5 種，僅 1 種為 4 的倍數
99987，排列 20 種，皆非 4 的倍數
99888，排列 10 種，有 3 種為 4 的倍數

k介於0~90度之間

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