题目

Problem

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

$u_{n+1} = b - au_n$

$u_1 = 3$

where $a$ and $b$ are constants.

(a) Find, in terms of $a$ and $b$ ,

(i) $u_2$ (ii) $u_3$

Given that $\displaystyle\sum_{n=1}^{3} u_n = 153$ and $b = a + 9$ ,

(b) show that $a^2 - 5a - 66 = 0$

(8)

题目中文翻译

数列 $u_1,u_2,u_3,\ldots$ 由 $u_{n+1}=b-au_n$ ， $u_1=3$ 定义，其中 $a,b$ 为常数。

(a) 用 $a,b$ 表示 (i) $u_2$ (ii) $u_3$

已知 $\displaystyle\sum_{n=1}^3 u_n=153$ 且 $b=a+9$ ，

(b) 证明 $a^2-5a-66=0$

（8分）

解答

逐项代入递推公式。利用 $\sum u_n = 153$ 和 $b = a + 9$ 列方程，化简得关于 $a$ 的二次方程。解方程后取使 $u_2$ 更大的那个 $a$ 值。

(a)(i)

$u_2 = b - au_1 = \boxed{b - 3a}$

(a)(ii)

u_3 =&\, b - au_2 \\[4mm] =&\, b - a(b - 3a) \\[4mm] =&\, b - ab + 3a^2 \end{align*}$$ $$\boxed{u_3 = b - ab + 3a^2}$$ **(b)** $$\begin{align*} u_1 + u_2 + u_3 =&\, 153 \\[4mm] 3 + (b - 3a) + (b - ab + 3a^2) =&\, 153 \\[4mm] 3 + 2b - 3a - ab + 3a^2 =&\, 153 \end{align*}$$ Substituting $b = a + 9$: $$\begin{align*} 3 + 2(a + 9) - 3a - a(a + 9) + 3a^2 =&\, 153 \\[4mm] 3 + 2a + 18 - 3a - a^2 - 9a + 3a^2 =&\, 153 \\[4mm] 2a^2 - 10a + 21 =&\, 153 \\[4mm] 2a^2 - 10a - 132 =&\, 0 \\[4mm] a^2 - 5a - 66 =&\, 0 \end{align*}$$ $$\boxed{a^2 - 5a - 66 = 0}$$ **(c)** Solving $(a - 11)(a + 6) = 0$: $$a = 11 \quad \text{or} \quad a = -6$$ Since $b = a + 9$: - If $a = 11$: $b = 20$, $u_2 = 20 - 3(11) = -13$ - If $a = -6$: $b = 3$, $u_2 = 3 - 3(-6) = 21$ The larger value of $u_2$ is: $$\boxed{u_2 = 21}$$