4. A General Formula | 通用公式

📚 4. A General Formula | 通用公式

In A-Level Mathematics, a general formula is a powerful algebraic expression that gives every term in a sequence or every case in a pattern without listing them individually. It turns repeated effort into a single substitution.

在 A-Level 数学中，通用公式是一种强大的代数表达式，它能不逐一列举而给出数列中的每一项或模式中的每一种情况。它把重复劳动转化为一次代入。

1. What Is a General Formula? | 什么是通用公式

A general formula for the nth term, often written as uₙ or aₙ, allows you to find any term when you know its position n. For sequences, this formula depends only on n and on fixed constants determined by the starting values.

第 n 项的通用公式通常写成 uₙ 或 aₙ，使你在知道位置 n 后就能找到任意一项。对于数列，该公式只取决于 n 以及由初始值确定的常数。

General formulas also appear for sums, products, and expansions. The key is that the formula works for all positive integers n, provided the pattern continues.

通用公式也出现在求和、乘积和展开式中。关键在于，只要模式延续，该公式就对所有正整数 n 成立。

2. Arithmetic Sequences: The nth Term | 等差数列：第 n 项

In an arithmetic sequence, the difference between consecutive terms is constant. This difference is called the common difference, usually denoted by d. If the first term is a, then the nth term is:

在等差数列中，相邻两项的差是常数。这个差称为公差，通常用 d 表示。若首项为 a，则第 n 项为：

uₙ = a + (n − 1)d

For example, the sequence 5, 8, 11, 14, … has a = 5 and d = 3. To find the 20th term, substitute n = 20:

例如数列 5, 8, 11, 14, … 中 a = 5，d = 3。要求第 20 项，代入 n = 20：

u₂₀ = 5 + (20 − 1) × 3 = 62

This formula is central to many exam problems, including those involving logs when d is given as a variable.

该公式是许多考试问题的基础，包括当公差作为变量出现时需要使用对数求解的情形。

3. Geometric Sequences: The nth Term | 等比数列：第 n 项

A geometric sequence has a constant ratio between consecutive terms, called the common ratio r. If the first term is a, then:

等比数列中相邻两项之比是常数，称为公比 r。若首项为 a，则：

uₙ = arⁿ⁻¹

Notice that the exponent is n − 1, because the first term corresponds to n = 1 and r⁰ = 1. For example, the sequence 3, 6, 12, 24, … has a = 3 and r = 2.

注意指数是 n − 1，因为首项对应 n = 1，且 r⁰ = 1。例如数列 3, 6, 12, 24, … 中 a = 3，r = 2。

When the ratio is negative, the terms alternate in sign. For instance, 2, −6, 18, −54, … has r = −3. The general formula still works perfectly.

当公比为负时，各项符号交替。例如 2, −6, 18, −54, … 中 r = −3。通用公式依然完全有效。

4. Quadratic Sequences | 二次序列

Some sequences are not linear but have second differences that are constant. For a quadratic sequence, the general term has the form:

有些数列不是线性的，但其二阶差分为常数。对于二次序列，其通项形式为：

uₙ = an² + bn + c

Here a, b, and c are constants. To find them, you can use three known terms and solve simultaneous equations. For example, for the sequence 1, 4, 9, 16, …, we know uₙ = n², so a = 1, b = 0, c = 0.

这里 a、b、c 是常数。要求出它们，可以使用三个已知项并解联立方程。例如数列 1, 4, 9, 16, … 中可知 uₙ = n²，所以 a = 1，b = 0，c = 0。

Another method is to note that for any quadratic sequence, the coefficient a is half of the second difference. If the second difference is 2, then a = 1. This shortcut saves time in exams.

另一种方法是注意对于任何二次序列，系数 a 等于二阶差的一半。若二阶差为 2，则 a = 1。这种捷径在考试中节省时间。

5. Summing an Arithmetic Series | 等差数列求和

When we add the first n terms of an arithmetic sequence, we get an arithmetic series. The sum Sₙ has two equivalent formulas:

当我们将等差数列的前 n 项相加时，得到等差数列型级数。其和 Sₙ 有两个等价公式：

Sₙ = ⁿ⁄₂ [2a + (n − 1)d] = ⁿ⁄₂ (a + l)

Here l is the last term, l = a + (n − 1)d. The second version is useful when you know the first and last terms but not the difference.

这里 l 是末项，l = a + (n − 1)d。第二个版本在你只知道首项和末项而不知道公差时非常有用。

For example, the sum of the first 10 terms of 5, 8, 11, … is S₁₀ = 10⁄2 [2×5 + (10−1)×3] = 5(10 + 27) = 185.

例如，数列 5, 8, 11, … 的前 10 项和为 S₁₀ = 10⁄2 [2×5 + (10−1)×3] = 5(10 + 27) = 185。

6. Summing a Geometric Series | 等比数列求和

The sum of the first n terms of a geometric series with first term a and common ratio r is a key general formula. For r ≠ 1, we have:

对于首项为 a、公比为 r 的等比级数，其前 n 项和是一个关键的通用公式。当 r ≠ 1 时：

Sₙ = a(1 − rⁿ) ⁄ (1 − r) = a(rⁿ − 1) ⁄ (r − 1)

The two forms are identical; choose based on whether r is less than or greater than 1. If r = 1, then Sₙ = na.

两种形式完全等价；根据 r 小于 1 还是大于 1 来选择使用哪一个。若 r = 1，则 Sₙ = na。

In an exam, you should be able to derive this formula by subtracting rSₙ from Sₙ. This “trick” is often tested directly.

在考试中，你应该能够通过从 Sₙ 中减去 rSₙ 来推导该公式。这种“技巧”常常被直接考查。

7. Infinite Geometric Series | 无穷等比级数

If the common ratio r satisfies −1 < r < 1, then as n becomes very large, rⁿ approaches 0. Therefore the infinite sum converges to a simple general formula:

如果公比 r 满足 −1 < r < 1，那么当 n 变得非常大时，rⁿ 趋近于 0。因此无穷和收敛到一个简洁的通用公式：

S∞ = a ⁄ (1 − r)

This formula is only valid when |r| < 1. If |r| ≥ 1, the series either diverges or oscillates without converging.

这个公式仅在 |r| < 1 时成立。如果 |r| ≥ 1，级数要么发散，要么振荡且不收敛。

For example, the series 1 + ½ + ¼ + ⅛ + … has a = 1 and r = ½, so S∞ = 1 ⁄ (1 − ½) = 2. This result is often used in recurring decimals and geometric problems.

例如，级数 1 + ½ + ¼ + ⅛ + … 中 a = 1，r = ½，所以 S∞ = 1 ⁄ (1 − ½) = 2。这个结果常用于循环小数和几何问题。

8. Sigma Notation and the General Formula | Σ符号与通用公式

The symbol Σ is used to write a sum compactly. For example:

符号 Σ 用于简洁地表示求和。例如：

Σₖ₌₁ⁿ (2k + 1) = 3 + 5 + 7 + … + (2n + 1)

The most useful general formulas are those for simple summations:

最有用的通用公式是那些针对简单求和的公式：

Σₖ₌₁ⁿ k = n(n + 1) ⁄ 2

Σₖ₌₁ⁿ k = n(n + 1) ⁄ 2
Σₖ₌₁ⁿ k² = n(n + 1)(2n + 1) ⁄ 6

Σₖ₌₁ⁿ k² = n(n + 1)(2n + 1) ⁄ 6
Σₖ₌₁ⁿ k³ = [n(n + 1) ⁄ 2]²

Σₖ₌₁ⁿ k³ = [n(n + 1) ⁄ 2]²

These formulas allow you to evaluate sums without counting each term, and they are essential for induction proofs and series questions.

这些公式使你不必逐项相加就能算出和，并且对于归纳法证明和级数题至关重要。

9. Deriving the Formula: The Method of Differences | 推导公式：差分法

Given a sequence of terms, you can often find its general formula by looking at first, second, and higher differences. If the first difference is constant, use a linear formula. If the second difference is constant, use a quadratic formula. If the third difference is constant, use a cubic formula.

给出一个数列的若干项后，通常可以通过观察一阶、二阶和更高阶差分来找到其通项公式。如果一阶差分为常数，则使用线性公式；如果二阶差分为常数，则使用二次公式；如果三阶差分为常数，则使用三次公式。

For a quadratic sequence, once you know the second difference is 2m, the coefficient of n² is m. Then subtract mn² from each term and find the remaining linear sequence.

对于二次序列，一旦知道二阶差为 2m，则 n² 的系数就是 m。然后从每一项中减去 mn²，在剩下的数列中寻找线性规律。

This method is efficient and avoids solving three simultaneous equations. It is a favourite in AQA exam questions.

这种方法效率高，避免了求解三个联立方程。它是 AQA 考试中的常见考点。

10. Applying the General Formula to the Binomial Expansion | 二项式展开中的通用公式

The binomial theorem gives a general formula for expanding (a + b)ⁿ where n is a positive integer:

二项式定理给出了 (a + b)ⁿ（n 为正整数）展开的通用公式：

(a + b)ⁿ = Σₖ₌₀ⁿ ⁿCₖ aⁿ⁻ᵏ bᵏ

Here ⁿCₖ = n! ⁄ [k!(n − k)!] is the binomial coefficient. The general term Tₖ₊₁ is often written as ⁿCₖ aⁿ⁻ᵏ bᵏ.

这里 ⁿCₖ = n! ⁄ [k!(n − k)!] 是二项式系数。其通项 Tₖ₊₁ 常写成 ⁿCₖ aⁿ⁻ᵏ bᵏ。

This formula is indispensable when finding a particular term without writing out the whole expansion. For example, the coefficient of x³ in (1 + 2x)⁵ is ⁵C₃ × 1² × (2x)³ = 10 × 8 = 80.

这个公式在需要找到特定项而无需写出整个展开式时不可或缺。例如，(1 + 2x)⁵ 中 x³ 的系数是 ⁵C₃ × 1² × (2x)³ = 10 × 8 = 80。

11. Worked Example | 综合例题

Consider the sequence 5, 11, 21, 35, 53, … Find a general formula for uₙ and hence find u₁₀.

考虑数列 5, 11, 21, 35, 53, …。求 uₙ 的通项公式，并由此求 u₁₀。

First differences: 6, 10, 14, 18. Second differences: 4, 4, 4. Since the second difference is constant and equals 4, the sequence is quadratic with a = 4⁄2 = 2.

一阶差：6, 10, 14, 18。二阶差：4, 4, 4。由于二阶差为常数且等于 4，该数列为二次数列，且 a = 4⁄2 = 2。

So uₙ = 2n² + bn + c. Using n = 1: 2 + b + c = 5 ⇒ b + c = 3. Using n = 2: 8 + 2b + c = 11 ⇒ 2b + c = 3. Subtracting gives b = 0, c = 3.

所以 uₙ = 2n² + bn + c。利用 n = 1：2 + b + c = 5 ⇒ b + c = 3。利用 n = 2：8 + 2b + c = 11 ⇒ 2b + c = 3。相减得 b = 0，c = 3。

Therefore uₙ = 2n² + 3. Substituting n = 10 gives u₁₀ = 2(100) + 3 = 203.

因此 uₙ = 2n² + 3。代入 n = 10 得 u₁₀ = 2(100) + 3 = 203。

12. Common Errors and Exam Advice | 常见错误与考试建议

One common error is confusing the index of the first term. If the first term is n = 1, the arithmetic formula uses (n − 1)d. If the sequence starts at n = 0, the formula changes to a + nd.

一个常见错误是混淆首项的下标。如果首项对应 n = 1，等差数列公式使用 (n − 1)d。如果数列从 n = 0 开始，公式变为 a + nd。

For geometric sums, always check whether r = 1 before applying the fraction formula. Also, in an infinite geometric sum, never use the infinite formula without verifying |r| < 1.

对于等比数列求和，应用分数公式前务必检查 r = 1 的特殊情况。另外，在使用无穷等比级数公式前，必须确认 |r| < 1。

Finally, write down the general formula before substituting numbers. In AQA mark schemes, many method marks are awarded for the correct formula even if your arithmetic is wrong.

最后，先写出通用公式再代入数值。在 AQA 评分标准中，即使计算错误，只要公式正确也能获得大量方法分。

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4. A General Formula | 通用公式