Finding a Formula for a Sequence | 求数列的公式

📚 Finding a Formula for a Sequence | 求数列的公式

A sequence is simply a list of numbers that follows a particular pattern or rule. In IGCSE Mathematics, being able to find a formula for the nth term allows you to calculate any term in the sequence without listing all preceding terms. This skill is essential for solving problems involving linear and quadratic patterns, and it also lays the foundation for more advanced algebraic thinking.

数列就是按照某种特定模式或规则排列的一串数字。在IGCSE数学中，学会求第n项的公式可以让你直接计算出序列中的任意一项，而无需逐项列出。这项技能对于解决线性和二次模式的问题至关重要，同时也为更高级的代数思维打下基础。

1. Introduction to Sequences and Formulas | 数列与公式简介

its position n (n = 1, 2, 3, ...). A formula for the nth term, often written as T(n) or u_n a T(n) = 2n + 1 generates the sequence 3, 5, 7, 9, ... when n takes the values 1, 2, 3, 4.

数列是一组有序的数字，称为项。每一项通过其位置n（n = 1, 2, 3, ...）来标识。第n项的公式，通常写作T(n)或u_n，直接给出位置n处的数值。比如，公式T(n) = 2n + 1在n取值1,2,3,4时生成的数列为3,5,7,9, ...

Finding this formula from a given sequence is a core objective. The approach depends on the type of sequence, which can often be identified by examining the patterns in the terms or in the differences between consecutive terms.

从已知数列寻找该公式是核心目标。采用的方法取决于数列的类型，通常可以通过观察项本身的模式或相邻项之间的差来识别。

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2. Linear (Arithmetic) Sequences – Finding the nth Term | 线性（等差）数列 – 求第n项

A linear sequence has a constant difference between consecutive terms. This constant is called the common difference, d. The nth term formula for any linear sequence can be written as:

线性数列的相邻项之差恒定。这个常数称为公差d。任何线性数列的第n项公式都可以写成：

T(n) = a + (n – 1)d

where a is the first term and d is the common difference. To find a and d, simply look at the first term and subtract any term from the next term.

其中a是首项，d是公差。要找到a和d，只需看第一项，并用后一项减去前一项得到d。

., the first term a = 4 and the difference d = 3. Therefore, the nth term is T(n) = 4 + (n – 1) × 3, which simplifies to T(n) = 3n + 1.

例如，在数列4, 7, 10, 13, ...中，首项a = 4，公差d = 3。因此，第n项公式为T(n) = 4 + (n – 1) × 3，简化后为T(n) = 3n + 1。

= 1, 2, 3 to ensure you recover the given terms. This quick check helps avoid sign errors.

请务必将n = 1, 2, 3代入检验，以确保能还原给定的项。这一快速检查有助于避免符号错误。

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3. Worked Example: Linear Sequence | 示例：线性数列

Find the nth term of the sequence 11, 8, 5, 2, ...

求数列11, 8, 5, 2, ...的第n项公式。

Step 1: Identify the first term a = 11.

步骤1：确定首项a = 11。

Step 2: Calculate the common difference d = 8 – 11 = –3. The sequence is decreasing.

步骤2：计算公差d = 8 – 11 = –3。数列是递减的。

Step 3: Substitute into T(n) = a + (n – 1)d to get T(n) = 11 + (n – 1)(–3).

步骤3：代入T(n) = a + (n – 1)d，得到T(n) = 11 + (n – 1)(–3)。

Step 4: Simplify: T(n) = 11 – 3n + 3 = 14 – 3n.

步骤4：化简：T(n) = 11 – 3n + 3 = 14 – 3n。

Check: When n = 1, 14 – 3 = 11; n = 2 gives 8; n = 3 gives 5. The formula works.

检验：当n = 1时，14 – 3 = 11；n = 2得8；n = 3得5。公式正确。

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4. Using Differences to Confirm a Linear Pattern | 利用差值确认线性模式

If you are not told a sequence is linear, you can check by finding the first differences (the differences between consecutive terms). If these differences are all the same, the sequence is linear. This is a crucial diagnostic tool before attempting to find a formula.

如果没有明确说明数列是线性的，你可以通过计算一阶差（相邻项之间的差）来检查。如果这些差值完全相同，那么该数列就是线性的。这是在尝试寻找公式之前的重要诊断工具。

where d = 4. Then you can find c by comparing with the term when n = 1: T(1) = 4×1 + c = 5 ⇒ c = 1. So T(n) = 4n + 1.

例如，给定项5, 9, 13, 17，差值为4, 4, 4。恒定的一阶差表明第n项公式具有T(n) = dn + c的形式，其中d = 4。接着可结合n=1的项求出c：T(1) = 4×1 + c = 5 ⇒ c = 1。因此T(n) = 4n + 1。

This method is especially useful when the first term does not immediately suggest the multiplier.

当首项不能立即反映乘数时，该方法尤为有用。

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5. Quadratic Sequences – Finding the nth Term | 二次数列 – 求第n项

When the first differences are not constant, the sequence may be quadratic. In a quadratic sequence, the second differences (differences of the first differences) are constant. The nth term of a quadratic sequence has the general form:

当一阶差不恒定时，该数列可能是二次的。在二次数列中，二阶差（一阶差的差）是恒定的。二次数列的第n项通式为：

T(n) = an² + bn + c

The coefficient a is linked to the second difference: the second difference equals 2a. Once a is found, you can determine b and c using the terms themselves.

系数a与二阶差相关：二阶差等于2a。求出a后，可利用数列的项来确定b和c。

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6. Method for Quadratic Sequences: Second Difference | 二次数列的方法：二阶差分

To find the formula for a quadratic sequence, follow these systematic steps:

要找出二次数列的公式，请遵循以下系统步骤：

1. Write down the sequence and work out the first differences.

1. 写出数列并计算一阶差。

2. Work out the second differences. Confirm they are constant.

2. 计算二阶差，确认它们恒定。

3. The second difference = 2a. Divide by 2 to get a.

3. 二阶差 = 2a。除以2得到a。

substitute known terms to form equations for b and c. Often you can subtract the quadratic part an² from each term to obtain a linear remainder, then find the nth term of that linear part.

4. 使用公式 T(n) = an² + bn + c，代入已知项构造关于b和c的方程。通常可以从每一项中减去二次部分an²，得到一个线性余数，再求该线性部分的第n项。

2, 6, 12, 20, 30. First differences: 4, 6, 8, 10. Second differences: 2, 2, 2. Thus 2a = 2 ⇒ a = 1. Subtract n² from each term: 2–1=1, 6–4=2, 12–9=3, 20–16=4, 30– The remainder 1,2,3,4,5 is linear with nth term n. So the final formula is T(n) = n² + n.

例如，考虑数列2, 6, 12, 20, 30。一阶差：4, 6, 8, 10。二阶差：2, 2, 2。因此2a = 2 ⇒ a = 1。从每一项减去n²：2–1=1, 6–4=2, 12–9=3, 20–16=4, 30–25=5。余数列1,2,3,4,5是线性的，第n项为n。因此最终公式为T(n) = n² + n。

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7. Worked Example: Quadratic Sequence | 示例：二次数列

Find the nth term of the sequence 3, 9, 17, 27, 39.

求数列3, 9, 17, 27, 39的第n项公式。

Step 1: First differences: 6, 8, 10, 12.

步骤1：一阶差：6, 8, 10, 12。

Step 2: Second differences: 2, 2, 2 (constant).

步骤2：二阶差：2, 2, 2（恒定）。

Step 3: a = second difference ÷ 2 = 2 ÷ 2 = 1.

步骤3：a = 二阶差 ÷ 2 = 2 ÷ 2 = 1。

Step 4: Subtract n² from terms: 3–1=2, 9–4=5, 17–9=8, 27–16=11, 39– , 5, 8, 11, 14, which is linear with first term 2 and difference 3. Its nth term is 2 + (n–1)×3 = 3n – 1.

步骤4：从各项中减去n²：3–1=2, 9–4=5, 17–9=8, 27–16=11, 39–25=14。余数列2, 5, 8, 11, 14是线性的，首项为2，公差为3。其第n项为2 + (n–1)×3 = 3n – 1。

Step 5: Combine: T(n) = n² + 3n – 1. Check with n=1: 1+3-1=3; n=2: 4+6-1=9. Correct.

步骤5：合并：T(n) = n² + 3n – 1。检验n=1: 1+3-1=3；n=2: 4+6-1=9。正确。

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8. Sequences with a Constant Ratio: Geometric Sequences | 等比数列

Some sequences multiply by the same factor each time. These are called geometric sequences. The constant multiplier is called the common ratio, r. The nth term of a geometric sequence is:

有些数列每次乘以相同的因子。这类数列称为等比数列。这个常数乘数称为公比r。等比数列的第n项公式为：

T(n) = a × r^n–1

where a is the first term and r is the common ratio.

其中a为首项，r为公比。

For the sequence 3, 6, 12, 24, ..., a = 3 and r = 6 ÷ 3 = 2. The nth term is T(n) = 3 × 2^n–1.

对于数列3, 6, 12, 24, ...，a = 3，r = 6 ÷ 3 = 2。第n项为T(n) = 3 × 2^n–1。

Note: Geometric sequences are often included in IGCSE extension material. Even if not examined in depth, recognising the pattern can help with problem solving.

注意：等比数列通常包含在IGCSE拓展内容中。即使不深入考查，识别该模式也有助于问题解决。

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9. Finding the nth Term of a Geometric Sequence | 求等比数列的第n项

To find the formula, identify the first term and then divide any term by the previous term to obtain r. Ensure r is consistent for a few pairs to confirm the sequence is geometric. Then plug the values into the formula.

要找公式，先确定首项，然后将任意项除以前一项得到r。通过多对项验证r的一致性，以确认数列是等比数列。然后将数值代入公式。

iven the sequence 5, 15, 45, 135, ..., a = 5, r = 15 ÷ 5 = 3. The nth term is T(n) = 5 × 3^n–1. Check: n=3 gives 5 × 3² = 45. This matches.

例如，给定数列5, 15, 45, 135, ...，a=5，r=15÷5=3。第n项为T(n)=5×3^n–1。检验：n=3时5×3²=45，吻合。

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10. Recurrence Relations | 递推关系

Another way to define a sequence is by using a recurrence relation, which gives each term based on the one(s) before it. A typical first-order recurrence relation has the form:

另一种定义数列的方法是使用递推关系，它基于前一项（或前几项）给出下一项。典型的一阶递推关系具有以下形式：

u_n+1 = f(u_n) with u₁ given

For example, u_n+1 = 2u_n + 1 with u₁ = 3 generates the terms: u₂ = 7, u₃ = 15, u₄ = 31, and so on.

例如，u_n+1 = 2u_n + 1，首项u₁ = 3生成：u₂ = 7，u₃ = 15，u₄ = 31，以此类推。

Recurrence relations are highly efficient for generating terms term by term, but they do not directly give a closed-form formula. However, IGCSE questions often ask you to find the next few terms or to identify when a term exceeds a certain value.

递推关系在逐项生成时非常高效，但不能直接给出封闭形式的公式。不过，IGCSE题目常要求找出后续若干项，或者确定某项何时超过某个值。

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11. Using Recurrence Relations to Generate Terms | 使用递推关系生成项

When working with a recurrence relation, always write down the first term clearly. Then apply the relation stepwise. Keep intermediate results to avoid errors.

在使用递推关系时，务必清晰地写出首项。然后逐步应用递推关系。保留中间结果以避免错误。

Example: u_n+1 = 3u_n – 5, u₁ = 4. Find u₃.

示例：u_n+1 = 3u_n – 5，u₁ = 4。求u₃。

u₁ = 4
u₂ = 3(4) – 5 = 7
u₃ = 3(7) – 5 = 16. So u₃ = 16.

u₁ = 4
u₂ = 3(4) – 5 = 7
u₃ = 3(7) – 5 = 16。因此u₃ = 16。

Questions may also ask to find the order of a term that satisfies a condition, e.g., 'Find the smallest n such that u_n > 100'. In such cases, continue generating terms until the condition is met.

题目还可能要求找出满足某条件的项的顺序，例如“求使u_n > 100的最小n”。此时，继续生成项直至满足条件即可。

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12. Problem Solving: Mixed Sequences | 问题求解：混合数列

In IGCSE exams, problems often blend sequences with algebra or require you to decide which method to use. Always examine the sequence carefully: check the differences first. If first differences are constant, use the linear formula; if second differences are constant, use the quadratic approach; if there is a common ratio, it might be geometric.

在IGCSE考试中，问题常将数列与代数结合，或要求你决定使用哪种方法。仔细检查数列：首先看差值。如果一阶差恒定，用线性公式；如果二阶差恒定，用二次函数方法；如果存在公比，则可能是等比数列。

Sometimes a sequence is neither linear nor quadratic but follows a pattern such as alternating signs or powers. For such cases, look for a rule relating the term position to its value. Practice with a variety of sequences builds confidence.

有时，数列既非线性也非二次，而是遵循某种模式，如交替符号或幂次。对于这类情况，要寻找项位置与其值之间的规律。通过练习多种数列来建立信心。

ula by substituting n = 1, 2, 3. This simple verification catches most algebraic mistakes.

记住，始终通过代入n = 1, 2, 3来检验最终公式。这种简单的验证能发现大部分代数错误。

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