题目

Problem

A sequence $u_1, u_2, u_3, \ldots$ is defined by

$u_1 = 3$ $u_{n+1} = 2 - \frac{4}{u_n}$

(a) Find the value of $u_2$ , the value of $u_3$ and the value of $u_4$ .

(b) Find the value of $\displaystyle\sum_{r=1}^{100} u_r$ .

(5)

题目中文翻译

数列 $u_1, u_2, u_3, \ldots$ 定义如下：

$u_1 = 3$ ， $u_{n+1} = 2 - \dfrac{4}{u_n}$

(a) 求 $u_2$ 、 $u_3$ 和 $u_4$ 的值。

(b) 求 $\displaystyle\sum_{r=1}^{100} u_r$ 的值。

（5分）

解答

递推数列问题。逐项计算几项后会发现数列以3为周期循环，利用周期性可以简化前100项的求和。

(a) Using the recurrence relation $u_{n+1} = 2 - \dfrac{4}{u_n}$ with $u_1 = 3$ :

u_2 =&\, 2 - \frac{4}{u_1} = 2 - \frac{4}{3} = \frac{2}{3} \\[2mm] u_3 =&\, 2 - \frac{4}{u_2} = 2 - \frac{4}{\frac{2}{3}} = 2 - 6 = -4 \\[2mm] u_4 =&\, 2 - \frac{4}{u_3} = 2 - \frac{4}{-4} = 2 + 1 = 3 \end{align*}$$ $$u_2 = \dfrac{2}{3}, \quad u_3 = -4, \quad u_4 = 3$$ **(b)** Since $u_4 = u_1 = 3$, the sequence is periodic with period 3. The sum of one complete period is: $$\begin{align*} u_1 + u_2 + u_3 =&\, 3 + \frac{2}{3} + (-4) = -\frac{1}{3} \end{align*}$$ 100 terms contain 33 complete periods and 1 remaining term: $$\begin{align*} \sum_{r=1}^{100} u_r =&\, 33 \times \left(-\frac{1}{3}\right) + u_{100} \\ =&\, -11 + u_1 \\ =&\, -11 + 3 \\ =&\, -8 \end{align*}$$