Continuing Sequences | 继续数列

📚 Continuing Sequences | 继续数列

In IGCSE Edexcel Mathematics, continuing sequences is a fundamental skill that tests your ability to recognise patterns, describe rules, and predict future terms. This article explains the key methods step by step, from simple arithmetic and geometric progressions to quadratic and more complex recursive sequences.

在 IGCSE Edexcel 数学中，继续数列是一项基本技能，考查你识别规律、描述规则并预测后续项的能力。本文将从简单的等差与等比数列，到二次数列和更复杂的递推数列，逐步讲解核心方法。

1. What Is a Sequence? | 什么是数列？

A sequence is an ordered list of numbers, called terms, usually written as \(a_1, a_2, a_3, \dots\) or simply as a list. Each term follows a rule based on its position. For example, 2, 4, 6, 8, … is a sequence where each term increases by 2.

数列是按顺序排列的一组数，称为项，通常记作 a₁, a₂, a₃, … 或直接写成一串。每一项都按照其位置遵循某个规则。例如 2, 4, 6, 8, … 就是一个每项增加 2 的数列。

2. Continuing Arithmetic Sequences | 继续等差数列

An arithmetic sequence has a constant difference \(d\) between consecutive terms. To continue such a sequence, simply add the same difference each time. For example, 5, 8, 11, 14, … has \(d = 3\), so the next term is 14 + 3 = 17.

等差数列的相邻两项之差 d 恒定。要延续这样的数列，只需每次加上同一个差值。例如 5, 8, 11, 14, … 中 d = 3，所以下一项是 14 + 3 = 17。

General form: \(a_n = a_1 + (n-1)d\)

aₙ = a₁ + (n−1)d

For the sequence 7, 12, 17, 22, … the first term \(a_1 = 7\) and \(d = 5\). The 10th term is \(7 + 9 \times 5 = 52\). To continue the sequence, keep adding 5.

对于数列 7, 12, 17, 22, …，首项 a₁ = 7，d = 5。第 10 项是 7 + 9×5 = 52。要延续该数列，只需继续加 5。

3. Continuing Geometric Sequences | 继续等比数列

A geometric sequence has a constant ratio \(r\) between consecutive terms. To continue, multiply each term by \(r\). For example, 3, 6, 12, 24, … has \(r = 2\), so the next term is 24 × 2 = 48.

等比数列的相邻两项之比 r 恒定。要延续，只需将每一项乘以 r。例如 3, 6, 12, 24, … 中 r = 2，所以下一项是 24 × 2 = 48。

General form: \(a_n = a_1 \times r^{n-1}\)

aₙ = a₁ × rⁿ⁻¹

For the sequence 100, 50, 25, 12.5, … the ratio is 0.5. The next term is 12.5 × 0.5 = 6.25. Always check whether the ratio is positive, negative, or fractional.

对于数列 100, 50, 25, 12.5, …，公比为 0.5。下一项是 12.5 × 0.5 = 6.25。始终要注意公比是正数、负数还是分数。

4. Finding the Difference or Ratio | 找出差值或公比

To continue a sequence, first identify the pattern. Subtract consecutive terms to see if the difference is constant (arithmetic) or divide consecutive terms to see if the ratio is constant (geometric).

要延续数列，首先要识别规律。用相邻两项相减看差值是否恒定（等差），或相除看比值是否恒定（等比）。

If differences are constant → arithmetic sequence
If ratios are constant → geometric sequence
If differences change in a regular way → higher-order sequences

如果差值恒定 → 等差数列
如果比值恒定 → 等比数列
如果差值的变化有规律 → 高阶数列

5. Quadratic Sequences | 二次数列

A quadratic sequence has a constant second difference. The general form is \(a_n = An^2 + Bn + C\). To continue such a sequence, use the second difference to find the next first difference, then the next term.

二次数列具有恒定的二阶差。其通项形式为 aₙ = An² + Bn + C。要延续此类数列，利用二阶差求出下一项的一阶差，再求出下一项。

Example: 2, 5, 10, 17, 26, …

示例：2, 5, 10, 17, 26, …

First differences: 3, 5, 7, 9 (increase by 2). Second difference: 2 (constant). Next first difference = 9 + 2 = 11, so next term = 26 + 11 = 37.

一阶差：3, 5, 7, 9（每次增加 2）。二阶差：2（恒定）。下一项的一阶差 = 9 + 2 = 11，所以下一项 = 26 + 11 = 37。

6. Using the nth Term Formula | 使用通项公式

Given the nth term formula, you can continue a sequence by substituting \(n = 1, 2, 3, \dots\) For example, if \(a_n = 3n + 1\), then the sequence is 4, 7, 10, 13, …

给定了通项公式，只需代入 n = 1, 2, 3, … 即可延续数列。例如若 aₙ = 3n + 1，则数列为 4, 7, 10, 13, …

Similarly, for a quadratic nth term like \(a_n = n^2 + 2n\), substitute values: \(n=1\) gives 3, \(n=2\) gives 8, \(n=3\) gives 15, and so on.

同理，对于二次通项如 aₙ = n² + 2n，代入数值：n=1 得 3，n=2 得 8，n=3 得 15，依此类推。

7. Recursive Sequences | 递推数列

Some sequences are defined by a rule that connects each term to the previous term(s). For example, \(u_{n+1} = 2u_n + 1\) with \(u_1 = 3\). Then \(u_2 = 2(3)+1 = 7\), \(u_3 = 2(7)+1 = 15\), and so on.

有些数列通过连接当前项与前一项（或前几项）的规则来定义。例如 uₙ₊₁ = 2uₙ + 1，且 u₁ = 3。则 u₂ = 2(3)+1 = 7，u₃ = 2(7)+1 = 15，依此类推。

Another famous example is the Fibonacci sequence where the next term is the sum of the two previous terms: 1, 1, 2, 3, 5, 8, 13, …

另一个著名例子是斐波那契数列，其中下一项是前两项之和：1, 1, 2, 3, 5, 8, 13, …

8. Special Patterns: Squares, Cubes, Primes | 特殊规律：平方数、立方数、质数

Recognising familiar number patterns can help you continue sequences quickly. Common patterns include perfect squares (1, 4, 9, 16, …), perfect cubes (1, 8, 27, 64, …), and prime numbers (2, 3, 5, 7, 11, …).

识别熟悉的数字规律可以帮助你快速延续数列。常见规律包括完全平方数（1, 4, 9, 16, …）、完全立方数（1, 8, 27, 64, …）和质数（2, 3, 5, 7, 11, …）。

Example: 0, 3, 8, 15, 24, … can be seen as \(n^2 – 1\) for \(n = 1, 2, 3, 4, 5\). So the next term is \(6^2 – 1 = 35\).

例如：0, 3, 8, 15, 24, … 可看作 n² − 1（n = 1, 2, 3, 4, 5）。所以下一项是 6² − 1 = 35。

9. Continuing Combined Sequences | 延续混合数列

Some sequences combine arithmetic and geometric operations. For example, 1, 3, 7, 15, 31, … follows the rule \(a_n = 2^n – 1\) or the recursive rule \(a_{n+1} = 2a_n + 1\). Always look for more than one possible rule, and choose the most natural one.

有些数列同时涉及算术和几何运算。例如 1, 3, 7, 15, 31, … 遵循规则 aₙ = 2ⁿ − 1 或递推规则 aₙ₊₁ = 2aₙ + 1。始终注意可能存在多种规则，要选择最自然的一种。

Sequence	Rule	Next term
1, 2, 4, 8, 16	× 2	32
1, 4, 9, 16, 25	n²	36
100, 93, 86, 79	− 7	72
3, 5, 9, 17, 33	× 2 − 1	65

数列	规则	下一项
1, 2, 4, 8, 16	× 2	32
1, 4, 9, 16, 25	n²	36
100, 93, 86, 79	− 7	72
3, 5, 9, 17, 33	× 2 − 1	65

10. Common Mistakes to Avoid | 常见错误与预防

One common mistake is assuming a sequence is arithmetic when it is not. Always compute both first and second differences if needed. Another mistake is misreading a decreasing geometric sequence, such as 80, 40, 20, … as arithmetic (subtracting 40, then 20). The actual pattern is dividing by 2.

一个常见错误是武断地认为数列是等差，而实际上不是。必要时既要计算一阶差也要计算二阶差。另一个错误是把递减等比数列如 80, 40, 20, … 误认为等差（因为先减 40，再减 20）。实际上规律是除以 2。

11. Practice Worked Example | 例题演练

Consider the sequence: 6, 11, 18, 27, 38, … Find the next two terms.

考虑数列：6, 11, 18, 27, 38, … 求接下来的两项。

First differences: 5, 7, 9, 11. Second difference: 2. Next first difference = 11 + 2 = 13, so term 6 = 38 + 13 = 51. Then next first difference = 13 + 2 = 15, so term 7 = 51 + 15 = 66.

一阶差：5, 7, 9, 11。二阶差：2。下一项的一阶差 = 11 + 2 = 13，所以第 6 项 = 38 + 13 = 51。再下一项的一阶差 = 13 + 2 = 15，所以第 7 项 = 51 + 15 = 66。

12. Summary | 总结

To continue a sequence successfully, always identify the type of sequence first. Check for constant differences (arithmetic), constant ratios (geometric), or constant second differences (quadratic). Use the nth term formula when available. Practise with a variety of patterns to build confidence.

要成功延续数列，首先要确定数列类型。检查是否有恒定差值（等差）、恒定公比（等比）或恒定二阶差（二次）。有通项公式时直接使用。通过练习各种规律来建立信心。

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Continuing Sequences | 继续数列