The Difference Method | 差分法

📚 The Difference Method | 差分法

The Difference Method, also known as the method of finite differences, is a powerful tool for finding the nth term of a sequence. By examining how the terms change from one position to the next, we can determine whether a sequence is linear, quadratic, or even cubic, and then construct an explicit formula for the nth term.

差分法（又称有限差分法）是寻找数列第 n 项公式的有力工具。通过观察相邻项之间的变化规律，我们可以判断一个数列是线性、二次还是三次数列，并据此构造出第 n 项的显式公式。

1. What Are First Differences? | 什么是一阶差分？

Consider a sequence of terms u₁, u₂, u₃, u₄, … The first differences are obtained by subtracting each term from the next one: Δuₙ = uₙ₊₁ − uₙ. For example, in the sequence 3, 7, 11, 15, the first differences are 7−3 = 4, 11−7 = 4, 15−11 = 4. When the first differences are constant, the sequence is linear and the nth term takes the form an + b.

考虑数列 u₁, u₂, u₃, u₄, …。一阶差分是指相邻两项相减：Δuₙ = uₙ₊₁ − uₙ。例如数列 3, 7, 11, 15 的一阶差分为 7−3 = 4, 11−7 = 4, 15−11 = 4。当一阶差分恒为常数时，数列是线性的，第 n 项可写成 an + b 的形式。

To find the formula, note that the common difference d equals a. Then use the first term u₁ = a + b to solve for b. For 3, 7, 11, 15, we have d = 4, so a = 4, and 3 = 4×1 + b giving b = −1. Thus uₙ = 4n − 1.

要找到公式，注意公差 d 等于 a。然后用首项 u₁ = a + b 解出 b。对于 3, 7, 11, 15，d = 4，所以 a = 4，且 3 = 4×1 + b，得 b = −1。因此 uₙ = 4n − 1。

2. When First Differences Are Not Constant | 当一阶差分不恒定时

If the first differences increase or decrease by a constant amount, then the sequence is quadratic. The second differences – the differences between consecutive first differences – are constant. For a quadratic sequence uₙ = an² + bn + c, the second difference is always 2a. This gives us a direct way to find a.

如果一阶差分按固定量增大或减小，则该数列是二次的。二阶差分（相邻一阶差分之间的差）恒为常数。对于二次数列 uₙ = an² + bn + c，二阶差分恒为 2a。这为我们提供了求 a 的直接方法。

Example: 2, 6, 12, 20. First differences: 4, 6, 8 (not constant). Second differences: 2, 2 (constant). Since the second difference is 2, we know 2a = 2, so a = 1.

例如：2, 6, 12, 20。一阶差分：4, 6, 8（不恒定）；二阶差分：2, 2（恒定）。因为二阶差分为 2，所以 2a = 2，得 a = 1。

3. Building a Difference Table | 构造差分表

It is helpful to organise terms and their differences in a table. For a sequence labelled by n = 1, 2, 3, 4, …, write the terms uₙ in one row, then the first differences below, then the second differences below that. A clear table prevents careless mistakes.

将数列项和各项差分整理成表格会很有帮助。对于标记为 n = 1, 2, 3, 4, … 的数列，把各项 uₙ 写在一行，下面写一阶差分，再下面写二阶差分。清晰的表格能避免粗心错误。

n	1	2	3	4	5
uₙ	2	6	12	20	30
First diff		4	6	8	10
Second diff			2	2	2

Notice that the second differences are placed starting from the third column. They represent differences between the first differences 4→6 and 6→8, etc.

注意二阶差分从第三列开始，它们表示一阶差分之间的差：4→6 和 6→8 等。

4. Finding a, b and c for a Quadratic | 求二次公式中的 a、b、c

Once we know a, we need two more equations to find b and c. Substitute n = 1 and n = 2 into uₙ = an² + bn + c. This gives:

求出 a 之后，我们需要两个方程来求 b 和 c。将 n = 1 和 n = 2 代入 uₙ = an² + bn + c，可得：

u₁ = a + b + c

u₂ = 4a + 2b + c

Subtract the first equation from the second to eliminate c:

用第二个方程减去第一个方程以消去 c：

u₂ − u₁ = 3a + b

Thus b = (u₂ − u₁) − 3a. Finally, c = u₁ − a − b. This systematic method works for any quadratic sequence.

因此 b = (u₂ − u₁) − 3a。最后 c = u₁ − a − b。这个系统方法对任何二次数列都适用。

5. Worked Example 1: 2, 6, 12, 20 | 例1：2, 6, 12, 20

We already found a = 1 from the second difference. Now compute b and c using the equations above.

我们已从二阶差分得到 a = 1。现在用上述方程计算 b 和 c。

u₁ = 2, u₂ = 6. So b = (6 − 2) − 3×1 = 4 − 3 = 1. Then c = 2 − 1 − 1 = 0. Therefore the nth term is:

u₁ = 2, u₂ = 6。所以 b = (6 − 2) − 3×1 = 4 − 3 = 1。然后 c = 2 − 1 − 1 = 0。因此第 n 项为：

uₙ = n² + n

Check: for n = 3, 3² + 3 = 12; for n = 4, 16 + 4 = 20. The formula matches the sequence.

检验：n = 3 时，3² + 3 = 12；n = 4 时，16 + 4 = 20。公式与数列完全一致。

6. Worked Example 2: 7, 12, 19, 28 | 例2：7, 12, 19, 28

Let us apply the difference method to the sequence 7, 12, 19, 28. First differences: 5, 7, 9. Second differences: 2, 2. Hence a = 2 ÷ 2 = 1.

我们对数列 7, 12, 19, 28 应用差分法。一阶差分：5, 7, 9；二阶差分：2, 2。因此 a = 2 ÷ 2 = 1。

Using u₁ = 7 and u₂ = 12: b = (12 − 7) − 3×1 = 5 − 3 = 2. Then c = 7 − 1 − 2 = 4. The nth term is:

使用 u₁ = 7 和 u₂ = 12：b = (12 − 7) − 3×1 = 5 − 3 = 2。然后 c = 7 − 1 − 2 = 4。第 n 项为：

uₙ = n² + 2n + 4

Verify: n = 3 gives 9 + 6 + 4 = 19; n = 4 gives 16 + 8 + 4 = 28. Correct.

验证：n = 3 得 9 + 6 + 4 = 19；n = 4 得 16 + 8 + 4 = 28。正确。

7. Extension to Cubic Sequences | 扩展到三次数列

If the second differences are also not constant, look at the third differences. For a cubic sequence uₙ = an³ + bn² + cn + d, the third difference is constant and equals 6a. Thus a = (third difference) ÷ 6.

如果二阶差分也不是常数，则观察三阶差分。对于三次数列 uₙ = an³ + bn² + cn + d，三阶差分恒为常数且等于 6a。因此 a = 三阶差分 ÷ 6。

Example: 1, 8, 27, 64, 125 (cubes). First differences: 7, 19, 37, 61. Second differences: 12, 18, 24. Third differences: 6, 6. So third difference = 6, giving a = 1. The sequence is indeed n³.

例如：1, 8, 27, 64, 125（立方数）。一阶差分：7, 19, 37, 61；二阶差分：12, 18, 24；三阶差分：6, 6。所以三阶差分 = 6，得到 a = 1。该数列确实为 n³。

After finding a, substitute n = 1, 2, 3 to obtain three equations and solve for b, c and d. This is a standard substitution-elimination process.

求出 a 后，代入 n = 1, 2, 3 得到三个方程，解出 b、c 和 d。这是标准的代入消元过程。

8. The Power of the Difference Table | 差分表的威力

A well-constructed difference table reveals the degree of the polynomial sequence at a glance. If the kth differences are the first to become constant, the sequence has a polynomial formula of degree k. This is a key insight for IGCSE examinations.

一张构造良好的差分表能一眼看出多项式数列的次数。如果第 k 阶差分首次变为常数，则该数列具有 k 次多项式公式。这是 IGCSE 考试的关键洞察。

Sequence type	First duplicate diff.	Formula leading coefficient
Linear	First difference	a = first difference
Quadratic	Second difference	a = second difference ÷ 2
Cubic	Third difference	a = third difference ÷ 6

Memorise these three rules; they are the backbone of the difference method in IGCSE Mathematics.

请记住这三条规则；它们是 IGCSE 数学中差分法的核心。

9. Common Mistakes and How to Avoid Them | 常见错误及避免方法

Students often make three mistakes. First, they forget to divide by 2 or 6 when finding a. Second, they use the wrong term for u₁ or u₂ when substituting. Third, they misalign differences in the table, leading to incorrect constants. Always write the table neatly and check your final formula with a known term.

学生常犯三个错误。第一，求 a 时忘记除以 2 或 6。第二，代入 u₁ 或 u₂ 时用错项。第三，表格中的差分错位，导致常数错误。务必整齐地写表格，并用已知项检验最终公式。

Example of an error: For the sequence 5, 8, 13, 20, first differences are 3, 5, 7; second differences are 2, 2. A careless student might take a = 2 instead of a = 1. Correctly, a = 2 ÷ 2 = 1, giving uₙ = n² + 2n + 2.

错误示例：对于数列 5, 8, 13, 20，一阶差分为 3, 5, 7；二阶差分为 2, 2。粗心的学生可能取 a = 2 而不是 a = 1。正确做法是 a = 2 ÷ 2 = 1，得到 uₙ = n² + 2n + 2。

10. Practice Question | 练习题

Find the nth term of the sequence 4, 9, 16, 25. Use the difference method. You should recognise these as squares, but the method will confirm it.

用差分法求数列 4, 9, 16, 25 的第 n 项。你应该能认出这些是平方数，但差分法会验证它。

Solution: First differences: 5, 7, 9. Second differences: 2, 2. So a = 1. With u₁ = 4 and u₂ = 9, b = (9 − 4) − 3 = 2, c = 4 − 1 − 2 = 1. Hence uₙ = n² + 2n + 1 = (n + 1)². For n = 3, 4² = 16; for n = 4, 5² = 25. Perfect.

解答：一阶差分：5, 7, 9；二阶差分：2, 2。所以 a = 1。由 u₁ = 4, u₂ = 9，得 b = (9 − 4) − 3 = 2，c = 4 − 1 − 2 = 1。因此 uₙ = n² + 2n + 1 = (n + 1)²。当 n = 3 时，4² = 16；当 n = 4 时，5² = 25。完美匹配。

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The Difference Method | 差分法