The General Formula for Sequences | 数列的通项公式

📚 The General Formula for Sequences | 数列的通项公式

In A-Level Mathematics, one of the most powerful tools is the ability to express the terms of a sequence using a general formula. A general formula allows us to find any term of a sequence without listing all the previous terms, and it forms the foundation for solving problems involving arithmetic and geometric progressions. This article explores how to derive and apply these formulas, with a focus on the AQA specification.

在 A-Level 数学中，最强大的工具之一就是能够用通项公式来表达数列的项。通项公式使我们无需列举前面的所有项就能找到数列中的任意一项，它也是解决等差与等比数列问题的基础。本文将围绕 AQA 考纲，讲解如何推导并应用这些公式。

1. What is a Sequence? | 什么是数列？

A sequence is an ordered list of numbers, called terms. Each term of a sequence can often be written using a rule that relates the term number \(n\) to the term value \(a_n\). For example, the sequence 3, 7, 11, 15, … has each term 4 greater than the previous one, so we can write the general term as \(a_n = 4n – 1\).

数列是按顺序排列的数，这些数称为项。通常可以用一个规则将项号 \(n\) 与项值 \(a_n\) 联系起来。例如数列 3, 7, 11, 15, … 每一项都比前一项大 4，因此我们可以写出通项 \(a_n = 4n – 1\)。

Sequences are often classified by how consecutive terms are formed. The two most common types studied in A-Level Mathematics are arithmetic sequences and geometric sequences. In an arithmetic sequence, consecutive terms differ by a fixed constant. In a geometric sequence, consecutive terms are multiplied by a fixed constant.

数列通常根据相邻项的形成方式来分类。A-Level 数学中研究得最多的是等差数列和等比数列。在等差数列中，相邻两项的差是一个固定常数；在等比数列中，相邻两项的比（即公比）是一个固定常数。

Understanding the general formula of a sequence is more than memorising an equation. It involves recognising patterns, using notation correctly, and applying the formula in contexts such as financial modelling, growth and decay, and even integration approximations.

理解数列的通项公式不仅仅是背一个方程，它涉及识别规律、正确使用记号，并在金融建模、增长与衰减，甚至积分近似等情境中加以应用。

2. Notation and Terms | 记号与项

We use subscript notation to identify the position of a term in a sequence. For instance, \(a_1\) is the first term, \(a_2\) is the second term, and \(a_n\) is the \(n\)th term. The letter \(n\) is usually a positive integer, but some formulas also require \(n = 0\), depending on the convention.

我们用下标记号来标识数列中项的位置。例如，\(a_1\) 表示第一项，\(a_2\) 表示第二项，\(a_n\) 表示第 \(n\) 项。字母 \(n\) 通常为正整数，但某些公式中也可能需要取 \(n = 0\)，具体取决于约定。

In an arithmetic sequence, the first term is often denoted by \(a\) (or \(a_1\)) and the common difference by \(d\). The general term is written as

在等差数列中，首项通常用 \(a\)（或 \(a_1\)）表示，公差用 \(d\) 表示。其通项写作

aₙ = a + (n − 1)d

In a geometric sequence, the first term is still \(a\) and the common ratio is \(r\). The general term is

在等比数列中，首项仍为 \(a\)，公比为 \(r\)。其通项为

aₙ = a × rⁿ⁻¹

Notice that the index \(n\) in \(aₙ\) is a subscript, while \(r^{n-1}\) is a power. These two uses of \(n\) are different: one labels the term number, the other appears in the exponent. Keeping this distinction clear is essential for correct calculations.

注意，\(aₙ\) 中的下标 \(n\) 与 \(r^{n-1}\) 中的指数 \(n\) 意义不同：前者表示项数的位置，后者出现在指数中。明确这一区别对正确计算至关重要。

3. Arithmetic Sequence Formula | 等差数列的通项公式

An arithmetic sequence is defined by the property that the difference between consecutive terms is constant. If the first term is \(a\) and the common difference is \(d\), then the sequence is

等差数列由相邻两项之差为常数这一性质定义。若首项为 \(a\)，公差为 \(d\)，则数列为

a, a + d, a + 2d, a + 3d, …

To find the \(n\)th term, we observe that starting from \(a\), we add \(d\) exactly \((n-1)\) times. Therefore the general formula is

要求第 \(n\) 项，我们观察到从 \(a\) 开始，一共加上 \((n-1)\) 个 \(d\)。因此通项公式为

aₙ = a + (n − 1)d

For example, the arithmetic sequence with \(a = 5\) and \(d = 3\) has the general term \(aₙ = 5 + 3(n-1) = 3n + 2\). Using this formula, the 20th term is \(a₂₀ = 3(20) + 2 = 62\).

例如，首项 \(a = 5\)、公差 \(d = 3\) 的等差数列的通项为 \(aₙ = 5 + 3(n-1) = 3n + 2\)。利用这个公式，第 20 项为 \(a₂₀ = 3(20) + 2 = 62\)。

Sometimes the formula is written in an equivalent form \(aₙ = pn + q\), where \(p\) and \(q\) are constants. In fact, every arithmetic sequence has a linear general term. This is useful because we can quickly determine the common difference from the coefficient of \(n\).

有时通项公式也写成等价形式 \(aₙ = pn + q\)，其中 \(p\) 和 \(q\) 是常数。事实上，每个等差数列的通项都是 \(n\) 的一次函数。这很有用，因为我们可以直接从 \(n\) 的系数看出公差。

4. Geometric Sequence Formula | 等比数列的通项公式

A geometric sequence is defined by the property that the ratio between consecutive terms is constant. If the first term is \(a\) and the common ratio is \(r\), then the sequence is

等比数列由相邻两项之比为常数这一性质定义。若首项为 \(a\)，公比为 \(r\)，则数列为

a, ar, ar², ar³, …

To find the \(n\)th term, we multiply the first term by \(r\) exactly \((n-1)\) times. Hence the general formula is

要求第 \(n\) 项，我们将首项乘以 \(n-1\) 次 \(r\)。因此通项公式为

aₙ = a × rⁿ⁻¹

For \(a = 2\) and \(r = 3\), the sequence is 2, 6, 18, 54, … and the general term is \(2 × 3^{n-1}\). The 10th term is \(2 × 3^9 = 39366\).

当 \(a = 2\)，\(r = 3\) 时，数列为 2, 6, 18, 54, …，通项为 \(2 × 3^{n-1}\)。第 10 项为 \(2 × 3^9 = 39366\)。

Unlike arithmetic sequences, geometric sequences can increase or decrease rapidly. If \(|r| < 1\), the terms tend to zero as \(n\) becomes large; if \(|r| > 1\), the terms grow without bound (in magnitude) unless the first term is zero. If \(r\) is negative, the signs of the terms alternate.

与等差数列不同，等比数列可能迅速增大或减小。当 \(|r| < 1\) 时，随着 \(n\) 增大，项趋于 0；当 \(|r| > 1\) 时，若首项不为零，则项的绝对值无限增大。若 \(r\) 为负数，则各项正负交替。

5. Deriving the Formulas from a Recurrence | 从递推关系推导通项公式

Many sequences are defined by a recurrence relation in which each term is expressed in terms of the previous term. For an arithmetic sequence, the recurrence is

许多数列由递推关系定义，即每一项用前一项来表达。等差数列的递推关系为

aₙ₊₁ = aₙ + d, with a₁ = a

By repeatedly applying this relation, we can derive the general formula. Starting with \(a₁ = a\), we have \(a₂ = a + d\), \(a₃ = a + 2d\), and generally \(aₙ = a + (n-1)d\). This proof is often required in exam questions.

反复应用这个关系，我们可以推导出通项公式。从 \(a₁ = a\) 开始，有 \(a₂ = a + d\)，\(a₃ = a + 2d\)，一般地得到 \(aₙ = a + (n-1)d\)。这个推导过程经常在考试题中要求写出。

Similarly, a geometric sequence has the recurrence

类似地，等比数列的递推关系为

aₙ₊₁ = r × aₙ, with a₁ = a

Applying this gives \(a₂ = ar\), \(a₃ = ar²\), and in general \(aₙ = ar^{n-1}\). The key idea is that the exponent \(n-1\) comes from the number of times we multiply by the ratio after the first term.

应用此关系可得 \(a₂ = ar\)，\(a₃ = ar²\)，一般得到 \(aₙ = ar^{n-1}\)。关键在于指数 \(n-1\) 是在首项之后乘以公比的总次数。

It is important to remember that the first term \(a\) in the recurrence is the starting value, not the value of \(a₀\). If the index starts at \(n = 0\), then the formula would be \(aₙ = a × rⁿ\), because the first term is \(a₀\). Always check the convention used in the question.

需要记住，递推关系中的首项 \(a\) 是指起始值，而不是 \(a₀\)。如果下标从 \(n = 0\) 开始，公式则为 \(aₙ = a × rⁿ\)，因为首项是 \(a₀\)。一定要仔细检查题目中使用的约定。

6. Applications: Finding Terms and Number of Terms | 应用：求项数与项值

A general formula can be used in two opposite ways: we can substitute a given value of \(n\) to find the term value \(aₙ\), or we can solve an equation to find which term equals a certain value. Both skills are regularly tested in A-Level exams.

通项公式可以有两种反向用途：我们可以代入给定的 \(n\) 来求项值 \(aₙ\)，或者解方程来找出哪一项等于某个给定值。这两种技能在 A-Level 考试中经常被考查。

For example, in the arithmetic sequence \(aₙ = 15 – 2n\), we can find \(a₁₀ = 15 – 20 = -5\). To find which term is \(-7\), we set \(15 – 2n = -7\), giving \(2n = 22\), so \(n = 11\). Hence the 11th term is \(-7\).

例如，在等差数列 \(aₙ = 15 – 2n\) 中，\(a₁₀ = 15 – 20 = -5\)。要找哪一项等于 \(-7\)，我们令 \(15 – 2n = -7\)，得到 \(2n = 22\)，所以 \(n = 11\)。因此第 11 项为 \(-7\)。

Geometric sequences often lead to exponential equations. Suppose \(aₙ = 3 × 2^{n-1}\) and we want to find \(n\) such that \(aₙ = 384\). We solve \(3 × 2^{n-1} = 384\), so \(2^{n-1} = 128\). Since \(128 = 2^7\), \(n-1 = 7\), so \(n = 8\).

等比数列常常会涉及指数方程。设 \(aₙ = 3 × 2^{n-1}\)，要求 \(n\) 使 \(aₙ = 384\)。我们解 \(3 × 2^{n-1} = 384\)，得 \(2^{n-1} = 128\)。因为 \(128 = 2^7\)，所以 \(n-1 = 7\)，即 \(n = 8\)。

Sometimes two unknown parameters appear, such as \(a\) and \(d\) in an arithmetic sequence. If we are given two terms, we form a pair of simultaneous equations to solve for these unknowns. This is a common exam technique.

有时会出现两个未知参数，例如等差数列中的 \(a\) 和 \(d\)。若已知两项，我们列出关于这些未知数的方程组来求解。这是常见的考试技巧。

7. Sum of the First n Terms: Arithmetic | 等差数列的前 n 项和

In addition to the general term, a general formula exists for the sum of the first \(n\) terms of an arithmetic sequence. If the first term is \(a\), the last term is \(\ell = a + (n-1)d\), then the sum is the average of the first and last terms multiplied by the number of terms:

除了通项，等差数列前 \(n\) 项的和也有通项公式。若首项为 \(a\)，末项为 \(\ell = a + (n-1)d\)，则和为首末项的平均值乘以项数：

Sₙ = n/2 (a + ℓ)

Since \(\ell = a + (n-1)d\), we can also write

代入 \(\ell = a + (n-1)d\)，还可以写成

Sₙ = n/2 [2a + (n − 1)d]

These two forms are equivalent. The first is useful when the last term is known; the second is useful when we know the common difference. One classic proof pairs terms from the beginning and end: \(a_1 + a_n = a_2 + a_{n-1} = …\), so the sum equals \(n/2\) times that constant pair sum.

这两个形式等价。第一种在已知末项时使用；第二种在已知公差时使用。经典证法是将首尾配对：\(a_1 + a_n = a_2 + a_{n-1} = …\)，因此和等于项数 \(n\) 的一半乘以这个常数配对和。

For example, the sum of the first 20 terms of the arithmetic sequence \(aₙ = 3n + 2\) can be found using \(a = 5\), \(d = 3\): \(S₂₀ = 20/2 [2(5) + 19(3)] = 10 × (10 + 57) = 670\).

例如，等差数列 \(aₙ = 3n + 2\) 的前 20 项和可用 \(a = 5\)，\(d = 3\) 计算：\(S₂₀ = 20/2 [2(5) + 19(3)] = 10 × (10 + 57) = 670\)。

8. Sum of the First n Terms: Geometric | 等比数列的前 n 项和

The sum of the first \(n\) terms of a geometric sequence also has a closed general formula. Let \(Sₙ = a + ar + ar² + … + ar^{n-1}\). By multiplying the sum by \(r\), then subtracting \(rSₙ\) from \(Sₙ\), all intermediate terms cancel. This yields

等比数列前 \(n\) 项和同样有封闭通项。设 \(Sₙ = a + ar + ar² + … + ar^{n-1}\)。将和乘以 \(r\)，再用 \(Sₙ – rSₙ\)，所有中间项都被抵消，得到

Sₙ = a(1 − rⁿ) / (1 − r), for r ≠ 1

Equivalently, we can write \(Sₙ = a(rⁿ – 1)/(r – 1)\). The second form is convenient when \(r > 1\), to avoid a negative numerator and denominator.

等价地可写为 \(Sₙ = a(rⁿ – 1)/(r – 1)\)。当 \(r > 1\) 时用第二种形式更方便，以免分子分母都为负。

This formula is a direct generalisation of the geometric series. For example, the sum of the first 6 terms of \(2, 6, 18, …\) with \(a = 2\), \(r = 3\) is \(S₆ = 2(3^6 – 1)/(3 – 1) = 2(729 – 1)/2 = 728\).

这个公式是等比级数的直接推广。例如，数列 \(2, 6, 18, …\) 中 \(a = 2\)，\(r = 3\)，前 6 项和为 \(S₆ = 2(3^6 – 1)/(3 – 1) = 2(729 – 1)/2 = 728\)。

When \(r = 1\), the geometric sequence becomes constant with every term equal to \(a\), so \(Sₙ = an\). This special case must be handled separately because the formula above would have a zero denominator.

当 \(r = 1\) 时，等比数列变成每项都是 \(a\) 的常数数列，因此 \(Sₙ = an\)。这个特殊情形需要单独处理，因为上面的公式分母为零。

9. Infinite Geometric Series | 无穷等比级数

Another important general formula is the sum of an infinite geometric series. If \(-1 < r < 1\) and \(r \neq 0\), then as \(n\) increases, \(r^n\) tends to zero. Taking the limit of the finite sum formula gives

另一个重要的通项公式是无穷等比级数的和。当 \(-1 < r < 1\) 且 \(r \neq 0\) 时，随着 \(n\) 增大，\(r^n\) 趋向于 0。对有限和公式取极限得到

S ∞ = a / (1 − r), for |r| < 1

For example, the series \(8 + 2 + 1/2 + 1/8 + …\) has \(a = 8\) and \(r = 1/4\). Since \(|1/4| < 1\), the sum is \(8 / (1 - 1/4) = 8 / (3/4) = 32/3\).

例如，级数 \(8 + 2 + 1/2 + 1/8 + …\) 中 \(a = 8\)，公比 \(r = 1/4\)。因为 \(|1/4| < 1\)，和为 \(8 / (1 - 1/4) = 8 / (3/4) = 32/3\)。

The condition \(|r| < 1\) is essential. If \(|r| \ge 1\), the infinite sum does not have a finite limit; the series is said to diverge, and the general formula cannot be applied.

条件 \(|r| < 1\) 是必需的。若 \(|r| \ge 1\)，无穷和不存在有限极限；此时级数发散，通项公式不能使用。

This concept appears in many applications, such as converting recurring decimals to fractions. A recurring decimal like \(0.272727…\) can be written as \(27/100 + 27/10000 + …\), which is an infinite geometric series with \(a = 27/100\) and \(r = 1/100\).

这个概念在许多应用中都会出现，例如将循环小数化为分数。循环小数 \(0.272727…\) 可以写成 \(27/100 + 27/10000 + …\)，这是一个 \(a = 27/100\)、\(r = 1/100\) 的无穷等比级数。

10. Exam-Style Worked Example | 考试例题

Let us work through a typical AQA-style problem that uses several of these formulas. The fourth term of an arithmetic sequence is 20, and the tenth term is 38. Find the first term, the common difference, and the sum of the first 15 terms.

让我们解一道典型的 AQA 风格例题。一个等差数列的第 4 项为 20，第 10 项为 38。求首项、公差以及前 15 项的和。

Let \(a\) be the first term and \(d\) the common difference. From the general formula \(aₙ = a + (n-1)d\), we have

设首项为 \(a\)，公差为 \(d\)。根据通项公式 \(aₙ = a + (n-1)d\)，可得

a + 3d = 20

a + 9d = 38

Subtracting the first equation from the second gives \(6d = 18\), so \(d = 3\). Substituting back, \(a + 9 = 20\), hence \(a = 11\).

用第二个方程减去第一个方程，得 \(6d = 18\)，所以 \(d = 3\)。代回得 \(a + 9 = 20\)，因此 \(a = 11\)。

To find \(S₁₅\), we use \(Sₙ = n/2 [2a + (n-1)d]\). Thus

求 \(S₁₅\)，我们使用 \(Sₙ = n/2 [2a + (n-1)d]\)。于是

S₁₅ = 15/2 [2(11) + 14(3)] = 15/2 [22 + 42] = 15/2 × 64 = 480

This example shows how the general formula unifies the process of solving for parameters and then evaluating a sum.

这个例子展示了通项公式如何将求参数和求和的过程统一起来。

Another common question involves a geometric sequence. The second term is 12 and the fourth term is 108. Both terms are positive. Find the common ratio and the sum of the first five terms.

另一个常见问题是等比数列。第 2 项为 12，第 4 项为 108，两项均为正。求公比及前五项的和。

Using \(aₙ = a r^{n-1}\), we have \(a₂ = ar = 12\) and \(a₄ = ar³ = 108\). Dividing \(a₄\) by \(a₂\) gives \(r² = 9\), so \(r = 3\) (since terms are positive). Then \(a × 3 = 12\), so \(a = 4\).

利用 \(aₙ = a r^{n-1}\)，有 \(a₂ = ar = 12\)，\(a₄ = ar³ = 108\)。用 \(a₄\) 除以 \(a₂\) 得 \(r² = 9\)，所以 \(r = 3\)（因为项为正）。于是 \(a × 3 = 12\)，得 \(a = 4\)。

Now \(S₅ = a(r⁵ – 1)/(r – 1) = 4(3⁵ – 1)/(3 – 1) = 4(243 – 1)/2 = 484\).

因此 \(S₅ = a(r⁵ – 1)/(r – 1) = 4(3⁵ – 1)/(3 – 1) = 4(243 – 1)/2 = 484\)。

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

One of the most common mistakes is confusing the index \(n\) with the exponent. In the geometric formula \(aₙ = ar^{n-1}\), the exponent is \(n-1\), not \(n\). When the sequence is labelled starting at \(a₁\), there is always one fewer multiplication than the term number.

最常见的错误之一是混淆下标 \(n\) 和指数。在等比通项 \(aₙ = ar^{n-1}\) 中，指数应为 \(n-1\)，而不是 \(n\)。当数列从 \(a₁\) 开始编号时，乘法次数总是比项数少 1。

Another frequent error is applying the infinite geometric series formula when \(|r| \geq 1\). Always check the convergence condition before writing down the denominator \(1-r\). Similarly, the formula for \(Sₙ\) is undefined when \(r = 1\), but in that case the sequence is simply constant.

另一个常见错误是在 \(|r| \geq 1\) 时使用无穷等比级数公式。在写出分母 \(1-r\) 之前，一定要先检查收敛条件。同样地，\(Sₙ\) 的公式在 \(r = 1\) 时无定义，但此时数列为常数数列。

For arithmetic sums, students sometimes confuse the number of terms with the last term number. For example, the sum of terms from the 5th term to the 12th term has \(12 – 5 + 1 = 8\) terms, not \(12 – 5 = 7\). The \(+1\) is essential because both endpoints are included.

在等差数列求和中，学生有时会混淆项数与末项的编号。例如，从第 5 项到第 12 项的和共有 \(12 – 5 + 1 = 8\) 项，而不是 \(12 – 5 = 7\) 项。这个 \(+1\) 是必须的，因为首尾两项都要计算在内。

Finally, always make sure the units and algebraic expressions are written without LaTeX notation in your exam. Use the accepted mathematical symbols and clearly show substitutions. In this article we use Unicode symbols for clarity, but in a written exam you should follow the standard notation of your textbook.

最后，在考试中请确保不使用 LaTeX 记法书写代数和单位。应使用规范的数学符号并清楚展示代入过程。本文为了清晰而使用 Unicode 符号，但在书面考试中请遵循教材的标准记号。

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The General Formula for Sequences | 数列的通项公式