Finding the nth Term of a Sequence: General Methods & Worked Examples | 寻找数列通项公式的通用方法与典型例题

📚 Finding the nth Term of a Sequence: General Methods & Worked Examples | 寻找数列通项公式的通用方法与典型例题

In the Edexcel IGCSE Mathematics syllabus, sequences appear both as pure algebra and as problem-solving tools. Being able to find the nth term of a sequence “means” you have discovered a formula that produces every term of that sequence without listing them all. This article explains the standard methods for linear, quadratic, cubic and geometric sequences, followed by typical exam-style examples.

在 Edexcel IGCSE 数学考纲中，数列既是纯代数内容，也是解决问题的重要工具。能够找到数列的通项公式（nth term）意味着你发现了一个无需逐一列举就能生成该数列每一项的公式。本文将系统讲解线性、二次、三次与等比数列的标准求法，并配以典型考试例题。

1. What Is the nth Term? | 什么是通项公式？

The nth term of a sequence is a formula expressed in terms of n (where n = 1, 2, 3, …) that gives the value of the term in position n. For example, if the nth term is 2n + 1, then the first term (n = 1) is 3, the second term is 5, the third term is 7, and so on.

数列的通项公式是一个以 n 为变量（n = 1, 2, 3, …）的表达式，它直接给出第 n 项的值。例如，若通项公式为 2n + 1，则第一项（n = 1）为 3，第二项为 5，第三项为 7，依此类推。

The key question is: given a list of terms, how do we identify the type of sequence and then construct the formula? The answer lies in the differences between consecutive terms (first differences) and, if necessary, the second differences.

关键问题是：给定一列数后，我们如何判断数列类型并构造公式？答案在于相邻两项的差（一阶差分），如有必要还要看二阶差分。

First difference: subtract each term from the one after it.
Second difference: subtract each first difference from the next first difference.

一阶差分：用后一项减去前一项。
二阶差分：用相邻两个一阶差分相减。

2. Arithmetic (Linear) Sequences | 等差数列（线性数列）

An arithmetic sequence has a constant first difference. Its general form is:

等差数列的一阶差分恒定，其一般形式为：

nth term = a + (n − 1)d

where a is the first term and d is the common difference. Equivalently, we can write it as dn + c, where c = a − d.

其中 a 为首项，d 为公差。等价地，也可写成 dn + c，其中 c = a − d。

Method: Find d by subtracting consecutive terms. Then find c by substituting n = 1 into dn + c and setting it equal to the first term.

方法：通过相邻项相减求出 d；然后将 n = 1 代入 dn + c 并令其等于首项，从而求出 c。

Example: For the sequence 5, 8, 11, 14, … the first difference is 3, so d = 3. Write nth term = 3n + c. Using n = 1: 3(1) + c = 5, so c = 2. The nth term is 3n + 2.

例如：数列 5, 8, 11, 14, … 的一阶差分为 3，故 d = 3。设通项为 3n + c。代入 n = 1：3(1) + c = 5，得 c = 2。因此通项公式为 3n + 2。

3. Geometric Sequences | 等比数列

A geometric sequence has a constant ratio between consecutive terms. Its general form is:

等比数列相邻两项的比值恒定，其一般形式为：

nth term = a × r⁽ⁿ⁻¹⁾

where a is the first term and r is the common ratio. To find r, divide any term by the previous term.

其中 a 为首项，r 为公比。求 r 的方法是用任一项除以前一项。

Example: For 2, 6, 18, 54, … we have r = 6 ÷ 2 = 3. The nth term is 2 × 3⁽ⁿ⁻¹⁾. Always check with n = 2: 2 × 3¹ = 6 ✓.

例如：对于 2, 6, 18, 54, …，r = 6 ÷ 2 = 3。通项公式为 2 × 3⁽ⁿ⁻¹⁾。务必验证 n = 2：2 × 3¹ = 6 ✓。

Note that r can be a fraction, a negative number, or even a surd (like √2). The same formula always applies.

注意 r 可以是分数、负数甚至是根式（如 √2），上述公式始终适用。

4. Quadratic Sequences | 二次数列

A quadratic sequence has a constant second difference. Its general form is:

二次数列的特征是二阶差分恒定，其一般形式为：

nth term = an² + bn + c

To find a, b and c, follow these steps:

求 a、b、c 的步骤如下：

Calculate the first differences and then the second differences.
If the second difference is constant and equals 2a, then a = (second difference) ÷ 2.
Substitute n = 1, 2, 3 into an² + bn + c to create three equations and solve for b and c.

先求一阶差分，再求二阶差分。
若二阶差分恒定且等于 2a，则 a =（二阶差分）÷ 2。
分别代入 n = 1, 2, 3，建立关于 b 和 c 的方程并求解。

Example: For 3, 7, 13, 21, …

例如：对于 3, 7, 13, 21, …

First differences: 4, 6, 8. Second difference: 2. So 2a = 2, hence a = 1. The sequence has formula n² + bn + c. Substitute n = 1: 1 + b + c = 3. Substitute n = 2: 4 + 2b + c = 7. Subtracting gives b = 1, and then c = 1. The nth term is n² + n + 1.

一阶差分为 4, 6, 8；二阶差分为 2。因此 2a = 2，得 a = 1。该数列通项形如 n² + bn + c。代入 n = 1：1 + b + c = 3；代入 n = 2：4 + 2b + c = 7。两式相减得 b = 1，进而 c = 1。故通项公式为 n² + n + 1。

5. Cubic Sequences | 三次数列（拓展）

A cubic sequence has a constant third difference. Its general form is:

三次数列的特征是三阶差分恒定，其一般形式为：

nth term = an³ + bn² + cn + d

If the third difference is constant and equals 6a, then a = (third difference) ÷ 6. After finding a, use the first four terms to set up a system of four equations and solve.

若三阶差分恒定为 6a，则 a =（三阶差分）÷ 6。求出 a 后，利用前四项建立四元一次方程组并求解。

Example: For 6, 23, 58, 117, …

例如：对于 6, 23, 58, 117, …

First differences: 17, 35, 59. Second differences: 18, 24. Third difference: 6. Hence a = 6 ÷ 6 = 1. The formula is n³ + bn² + cn + d. Substituting n = 1, 2, 3, 4 gives four equations. Solving them yields b = 0, c = −2, d = 7. Therefore the nth term is n³ − 2n + 7.

一阶差分：17, 35, 59；二阶差分：18, 24；三阶差分：6。于是 a = 6 ÷ 6 = 1。通项形如 n³ + bn² + cn + d。代入 n = 1, 2, 3, 4 得到四个方程，解得 b = 0，c = −2，d = 7。因此通项公式为 n³ − 2n + 7。

6. Recurrence Relations | 递推数列

Some sequences are defined by a recurrence relation, where each term depends on the previous term using a rule, rather than directly on n. For example: u₁ = 3 and uₙ₊₁ = 2uₙ + 1. This generates 3, 7, 15, 31, …

有些数列由递推关系定义，即每一项由前一项通过某种规则得到，而不是直接关于 n 的表达式。例如：u₁ = 3，且 uₙ₊₁ = 2uₙ + 1，可生成 3, 7, 15, 31, …

In IGCSE, you may be asked to find the first few terms from a recurrence relation, or to find a specific term. For the above example, the nth term could also be expressed in closed form as 2⁽ⁿ⁺¹⁾ − 1, but exam questions usually only require recursive calculation.

在 IGCSE 考试中，你可能会被要求根据递推关系写出数列的前几项，或求某一特定项。对于上述例子，通项公式也可写为 2⁽ⁿ⁺¹⁾ − 1，但考试通常只要求按递推式计算。

Tip: Always use the given starting value exactly; do not assume u₀ exists unless stated.

提示：务必严格按照题目给定的起始值计算；除非题目说明，否则不要假设存在 u₀。

7. Worked Example 1: Linear with Negative Difference | 例1：公差为负的等差数列

Find the nth term of the sequence: 20, 16, 12, 8, …

求数列 20, 16, 12, 8, … 的通项公式。

Solution: The first difference is 16 − 20 = −4, so d = −4. Write nth term = −4n + c. Using n = 1: −4(1) + c = 20, so c = 24. Hence nth term = −4n + 24. Check: n = 3 gives −12 + 24 = 12 ✓.

解答：一阶差分为 16 − 20 = −4，故 d = −4。设通项为 −4n + c。代入 n = 1：−4(1) + c = 20，得 c = 24。因此通项公式为 −4n + 24。验证：n = 3 时 −12 + 24 = 12 ✓。

8. Worked Example 2: Quadratic with Fractional Second Difference | 例2：二阶差分为偶数的二次数列

Find the nth term of the sequence: 2, 9, 20, 35, …

求数列 2, 9, 20, 35, … 的通项公式。

Solution: First differences: 7, 11, 15. Second difference: 4. So 2a = 4, hence a = 2. The formula is 2n² + bn + c. Substitute n = 1: 2 + b + c = 2. Substitute n = 2: 8 + 2b + c = 9. Subtract the first from the second: 6 + b = 7, so b = 1. Then c = −1. The nth term is 2n² + n − 1. Check n = 3: 18 + 3 − 1 = 20 ✓.

解答：一阶差分：7, 11, 15；二阶差分：4。因此 2a = 4，得 a = 2。通项形如 2n² + bn + c。代入 n = 1：2 + b + c = 2；代入 n = 2：8 + 2b + c = 9。两式相减：6 + b = 7，得 b = 1，进而 c = −1。通项公式为 2n² + n − 1。验证 n = 3：18 + 3 − 1 = 20 ✓。

9. Worked Example 3: Geometric with Fractions | 例3：公比为分数的等比数列

Find the nth term of the sequence: 27, 9, 3, 1, …

求数列 27, 9, 3, 1, … 的通项公式。

Solution: The common ratio is 9 ÷ 27 = ⅓. The first term is 27. Therefore the nth term is 27 × (⅓)⁽ⁿ⁻¹⁾. Check: n = 2 gives 27 × ⅓ = 9 ✓; n = 4 gives 27 × (⅓)³ = 27 ÷ 27 = 1 ✓.

解答：公比为 9 ÷ 27 = ⅓。首项为 27。因此通项公式为 27 × (⅓)⁽ⁿ⁻¹⁾。验证：n = 2 得 27 × ⅓ = 9 ✓；n = 4 得 27 × (⅓)³ = 27 ÷ 27 = 1 ✓。

10. Word Problem Application | 实际应用题

Each day, a runner increases the distance he runs by 400 metres. On day 1 he runs 2 km. What is the total distance on day 15, and on which day will he first run more than 10 km?

一名跑步者每天增加跑步距离 400 米。第 1 天他跑了 2 公里。请问第 15 天跑了多远？从第几天起他每天跑步距离首次超过 10 公里？

Solution: The daily distances form an arithmetic sequence with a = 2000 m and d = 400 m. The nth term is 2000 + (n − 1) × 400 = 400n + 1600. For day 15: 400(15) + 1600 = 7600 m = 7.6 km. To exceed 10,000 m, solve 400n + 1600 > 10000, giving 400n > 8400, so n > 21. Thus on day 22 he first runs more than 10 km.

解答：每天的跑步距离构成等差数列，a = 2000 米，d = 400 米。通项公式为 2000 + (n − 1) × 400 = 400n + 1600。第 15 天：400(15) + 1600 = 7600 米 = 7.6 公里。要超过 10000 米，解 400n + 1600 > 10000，得 400n > 8400，故 n > 21。因此从第 22 天起首次超过 10 公里。

11. Common Pitfalls and Tips | 常见错误与技巧

Using the wrong starting index: Most IGCSE questions use n = 1 as the first term. Read the question carefully.
Failing to check your answer: Always substitute n = 2 or n = 3 back into your formula to verify at least two terms.
Confusing first and second differences: A sequence with a constant second difference is quadratic, not linear.
Forgetting the formula for geometric sequences: The exponent is (n − 1), not n.

起始下标搞错：IGCSE 大多数题目以 n = 1 为首项，请仔细审题。
没有验证公式：务必代入 n = 2 或 n = 3 回去检验至少两项。
混淆一阶差分与二阶差分：二阶差分恒定说明是二次数列，而不是线性数列。
忘记等比数列的公式：指数是 (n − 1)，而不是 n。

Also remember to simplify expressions fully and to use brackets correctly when substituting negative values.

同时还要注意将表达式化到最简，并在代入负数时正确使用括号。

12. Summary of General Methods | 方法总结

Sequence type	Key feature	General form	How to find coefficients
Linear	Constant first difference	dn + c	d = first difference; c from n = 1
Quadratic	Constant second difference = 2a	an² + bn + c	Use equations from n = 1, 2, 3
Cubic	Constant third difference = 6a	an³ + bn² + cn + d	Use equations from n = 1 to 4
Geometric	Constant ratio r	a × r⁽ⁿ⁻¹⁾	r = term ÷ previous term

Once you identify the type of sequence, the rest is simple algebra. Practice by working from first principles each time, and always check your final formula with two different terms.

一旦判断出数列类型，剩下的就是简单的代数运算。请每次从基本原理出发进行练习，并始终用两个不同项检验最终公式。

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Finding the nth Term of a Sequence: General Methods & Worked Examples | 寻找数列通项公式的通用方法与典型例题