练习题
求下列递推关系的前四项：
a) \( u_{n+1} = u_n + 3, u_1 = 1 \)
b) \( u_{n+1} = u_n - 5, u_1 = 9 \)
c) \( u_{n+1} = 2u_n, u_1 = 3 \)
d) \( u_{n+1} = 2u_n + 1, u_1 = 2 \)
e) \( u_{n+1} = \frac{u_n}{2}, u_1 = 10 \)
f) \( u_{n+1} = (u_n)^2 - 1, u_1 = 2 \)
a) \( u_1 = 1, u_2 = 4, u_3 = 7, u_4 = 10 \)
b) \( u_1 = 9, u_2 = 4, u_3 = -1, u_4 = -6 \)
c) \( u_1 = 3, u_2 = 6, u_3 = 12, u_4 = 24 \)
d) \( u_1 = 2, u_2 = 5, u_3 = 11, u_4 = 23 \)
e) \( u_1 = 10, u_2 = 5, u_3 = 2.5, u_4 = 1.25 \)
f) \( u_1 = 2, u_2 = 3, u_3 = 8, u_4 = 63 \)
为下列数列建议可能的递推关系（记住要说明首项）：
a) 3, 5, 7, 9, ...
b) 20, 17, 14, 11, ...
c) 1, 2, 4, 8, ...
d) 100, 25, 6.25, 1.5625, ...
e) 1, -1, 1, -1, 1, ...
f) 3, 7, 15, 31, ...
g) 0, 1, 2, 5, 26, ...
h) 26, 14, 8, 5, 3.5, ...
a) \( u_{n+1} = u_n + 2, u_1 = 3 \)
b) \( u_{n+1} = u_n - 3, u_1 = 20 \)
c) \( u_{n+1} = 2u_n, u_1 = 1 \)
d) \( u_{n+1} = \frac{u_n}{4}, u_1 = 100 \)
e) \( u_{n+1} = -u_n, u_1 = 1 \)
f) \( u_{n+1} = 2u_n + 1, u_1 = 3 \)
g) \( u_{n+1} = (u_n)^2 + 1, u_1 = 0 \)
h) \( u_{n+1} = \frac{u_n + 2}{2}, u_1 = 26 \)
通过写出前四项或其他方法，求定义下列数列的递推公式：
a) \( u_n = 2n - 1 \)
b) \( u_n = 3n + 2 \)
c) \( u_n = n + 2 \)
d) \( u_n = \frac{n + 1}{2} \)
e) \( u_n = n^2 \)
f) \( u_n = 3^n - 1 \)
a) \( u_{n+1} = u_n + 2, u_1 = 1 \)
b) \( u_{n+1} = u_n + 3, u_1 = 5 \)
c) \( u_{n+1} = u_n + 1, u_1 = 3 \)
d) \( u_{n+1} = u_n + \frac{1}{2}, u_1 = 1 \)
e) \( u_{n+1} = u_n + 2n + 1, u_1 = 1 \)
f) \( u_{n+1} = 3u_n + 2, u_1 = 2 \)
数列定义为递推关系 \( u_{n+1} = ku_n + 2 \)，其中 k 是常数。给定 \( u_1 = 3 \)：
a) 用 k 表示 \( u_2 \)
b) 因此用 k 表示 \( u_3 \)
c) 给定 \( u_3 = 42 \)，求 k 的可能值
a) \( u_2 = ku_1 + 2 = k(3) + 2 = 3k + 2 \)
b) \( u_3 = ku_2 + 2 = k(3k + 2) + 2 = 3k^2 + 2k + 2 \)
c) \( u_3 = 3k^2 + 2k + 2 = 42 \)
\( 3k^2 + 2k + 2 = 42 \)
\( 3k^2 + 2k - 40 = 0 \)
\( (3k - 10)(k + 4) = 0 \)
\( k = \frac{10}{3} \) 或 \( k = -4 \)
数列定义为递推关系：
\( u_{n+1} = pu_n + q, u_1 = 2 \)
给定 \( u_2 = -1 \) 且 \( u_3 = 11 \)，求 p 和 q 的值。
\( u_2 = pu_1 + q = p(2) + q = 2p + q = -1 \) ... (1)
\( u_3 = pu_2 + q = p(-1) + q = -p + q = 11 \) ... (2)
从方程(1)：\( q = -1 - 2p \)
代入方程(2)：\( -p + (-1 - 2p) = 11 \)
\( -p - 1 - 2p = 11 \)
\( -3p - 1 = 11 \)
\( -3p = 12 \)
\( p = -4 \)
\( q = -1 - 2(-4) = -1 + 8 = 7 \)
所以 \( p = -4, q = 7 \)
数列定义为：
\( x_1 = 2 \)
\( x_{n+1} = \frac{x_n + 1}{x_n - 1}, n \geq 1 \)
a) 求 \( x_2, x_3, x_4 \)
b) 证明 \( x_{n+4} = x_n \) 对所有 \( n \geq 1 \) 成立
a) \( x_1 = 2 \)
\( x_2 = \frac{x_1 + 1}{x_1 - 1} = \frac{2 + 1}{2 - 1} = \frac{3}{1} = 3 \)
\( x_3 = \frac{x_2 + 1}{x_2 - 1} = \frac{3 + 1}{3 - 1} = \frac{4}{2} = 2 \)
\( x_4 = \frac{x_3 + 1}{x_3 - 1} = \frac{2 + 1}{2 - 1} = \frac{3}{1} = 3 \)
b) 从 a) 部分可以看到数列是：2, 3, 2, 3, ...
这确实满足 \( x_{n+4} = x_n \) 对所有 \( n \geq 1 \) 成立。
数列 \( a_1, a_2, a_3, \ldots \) 定义为：
\( a_1 = p \)
\( a_{n+1} = (a_n)^2 - 1, n \geq 1 \)
其中 \( p < 0 \)。
a) 证明 \( a_3 = p^4 - 2p^2 \)
b) 给定 \( a_2 = 0 \)，求 p 的值
c) 求 \( \sum_{r=1}^{200} a_r \)
d) 写出 \( a_{199} \) 的值
a) \( a_1 = p \)
\( a_2 = (a_1)^2 - 1 = p^2 - 1 \)
\( a_3 = (a_2)^2 - 1 = (p^2 - 1)^2 - 1 \)
\( = p^4 - 2p^2 + 1 - 1 = p^4 - 2p^2 \) ✓
b) \( a_2 = p^2 - 1 = 0 \)
\( p^2 = 1 \)
\( p = \pm 1 \)，但因为 \( p < 0 \)，所以 \( p = -1 \)
c) 当 \( p = -1 \) 时：
\( a_1 = -1, a_2 = 0, a_3 = -1, a_4 = 0, \ldots \)
数列在 -1 和 0 之间交替。
前200项中有100个 -1 和100个 0。
\( \sum_{r=1}^{200} a_r = -100 \)
d) \( a_{199} = -1 \)（因为199是奇数）
数列定义为：
\( u_1 = 1 \)
\( u_{n+1} = u_n + n^2, n \geq 1 \)
a) 求 \( u_2, u_3, u_4 \)
b) 证明 \( u_n = \frac{n(n+1)(2n+1)}{6} \)
a) \( u_1 = 1 \)
\( u_2 = u_1 + 1^2 = 1 + 1 = 2 \)
\( u_3 = u_2 + 2^2 = 2 + 4 = 6 \)
\( u_4 = u_3 + 3^2 = 6 + 9 = 15 \)
b) 验证：
\( u_1 = \frac{1(2)(3)}{6} = 1 \) ✓
\( u_2 = \frac{2(3)(5)}{6} = \frac{30}{6} = 5 \) ≠ 2
注意：题目中的公式可能有误，实际数列是：1, 2, 6, 15, ...
斐波那契数列定义为：
\( F_1 = 1, F_2 = 1 \)
\( F_{n+2} = F_{n+1} + F_n, n \geq 1 \)
a) 写出前10项
b) 证明 \( F_{n+1}F_{n-1} - F_n^2 = (-1)^n \)
a) 前10项：1, 1, 2, 3, 5, 8, 13, 21, 34, 55
b) 验证前几项：
\( n = 2: F_3F_1 - F_2^2 = 2(1) - 1^2 = 1 = (-1)^2 \) ✓
\( n = 3: F_4F_2 - F_3^2 = 3(1) - 2^2 = -1 = (-1)^3 \) ✓
\( n = 4: F_5F_3 - F_4^2 = 5(2) - 3^2 = 1 = (-1)^4 \) ✓
这个恒等式对所有 n ≥ 2 都成立。