📚 Example 3.4.1 – Application to Sequences | 示例 3.4.1 —— 数列应用
In AQA A-Level Mathematics, sequence questions often move beyond simply finding a term. They test your ability to read a real-life scenario, identify the underlying arithmetic or geometric pattern, and then apply the correct nth-term or summation formula. This article works through the key techniques needed for Example 3.4.1 and similar exam-style modelling problems.
在 AQA A-Level 数学中，数列题往往不只是求某一项。它们考查你阅读实际情境、识别等差或等比规律，然后正确使用通项公式或求和公式的能力。本文梳理攻克示例 3.4.1 以及同类考试建模题所需的核心技巧。
1. Identifying the Type of Sequence | 识别数列类型
Before applying any formula, check the rule that generates the terms. A constant difference between consecutive terms signals an arithmetic sequence, while a constant ratio signals a geometric sequence.
在套用任何公式之前，先检查生成各项的规则。相邻两项的差不变说明是等差数列；相邻两项的比值不变说明是等比数列。
In AQA questions, phrases such as 'increases by the same amount' often point to arithmetic, whereas 'increases by the same percentage' or 'multiplies by' points to geometric.
在 AQA 考题中，“增加相同数量”通常指向等差，而“增加相同百分比”或“乘以某个倍数”则指向等比。
Arithmetic: uₙ₊₁ − uₙ = d | Geometric: uₙ₊₁ ÷ uₙ = r
A recurrence relation may define a sequence that is neither arithmetic nor geometric, so you must generate terms step by step from the given rule.
递推关系定义的数列可能既不是等差也不是等比，因此必须根据给定规则逐步生成各项。
2. Arithmetic Sequences in Modelling | 等差数列在建模中的应用
Arithmetic sequences model situations where a quantity grows by a fixed amount each period. The first term is a and the common difference is d.
等差数列用于描述每个时期增加固定数量的情形。首项为 a，公差为 d。
uₙ = a + (n − 1)d
For example, a starting salary of £24,000 with an annual rise of £600 gives a = 24000 and d = 600. The salary in year 8 is therefore u₈ = 24000 + 7 × 600 = £28,200.
例如，起薪为 24,000 英镑，每年加薪 600 英镑，则 a = 24000，d = 600。因此第 8 年的工资为 u₈ = 24000 + 7 × 600 = 28,200 英镑。
If you need the total amount earned over several years, use the arithmetic series sum. The sum of the first n terms is given by:
如果需要求若干年的总收入，就使用等差数列求和公式。前 n 项和为：
Sₙ = n/2 [2a + (n − 1)d] = n/2 (a + l)
tal earned over the first 8 years is S₈ = 8/2 [2 × 24000 + 7 × 600] = 4 × (48000 + 4200) = £208,800.
在工资例子中，前 8 年的总收入为 S₈ = 8/2 [2 × 24000 + 7 × 600] = 4 × (48000 + 4200) = 208,800 英镑。
3. Geometric Sequences in Modelling | 等比数列在建模中的应用
Geometric sequences model growth or decay by a fixed percentage each period. The multiplier is the common ratio r. If a quantity decreases by 15%, the multiplier is 0.85; if it increases by 5%, the multiplier is 1.05.
等比数列用于描述每个时期按固定百分比增长或衰减的情形。乘数就是公比 r。如果某个量减少 15%，乘数为 0.85；如果增加 5%，乘数为 1.05。
uₙ = arⁿ⁻¹
A car bought for £18,000 depreciates by 15% h year. After n years its value is Vₙ = 18000 × 0.85ⁿ. After 5 years, V₅ = 18000 × 0.85⁵ ≈ £7,986.70.
一辆汽车购买价为 18,000 英镑，每年贬值 15%。n 年后的价值为 Vₙ = 18000 × 0.85ⁿ。5 年后，V₅ = 18000 × 0.85⁵ ≈ 7,986.70 英镑。
Notice that the index starts from 0 in this setup because the original value is the initial term. In standard AQA n-term notation, always link the question's time frame to the correct power of r.
注意这里的指数从 0 开始，因为原始价值是初始项。在标准 AQA 通项记号中，务必把题目中的时间节点与正确的 r 的幂对应起来。
4. Recurrence Relations and Iteration | 递推关系与迭代
uch as uₙ₊₁ = f(uₙ) with a given starting value. This generates the sequence term by term.
许多 AQA 建模题使用递推关系，如 uₙ₊₁ = f(uₙ)，并给出初始值。这种方式逐项生成数列。
uₙ₊₁ = 1.05uₙ − 200, u₁ = 5000
This could represent a loan balance that grows by 5% interest each month and is reduced by a repayment of £ The terms are u₂ = 1.05 × 5000 − 200 = 5050, u₃ = 1.05 × 5050 − 200 = 5102.50, and so on.
这可以表示每月增长 5% 利息并偿还 200 英镑的贷款余额。各项为 u₂ = 1.05 × 5000 − 200 = 5050，u₃ = 1.05 × 5050 − 200 = 5102.50，依此类推。
With a recurrence relation, do not rush to find an nth-term formula. AQA often wants you to compute successive terms or to investigate long-term behaviour by iteration.
遇到递推关系时，不要急于求通项公式。AQA 通常要求你逐项计算，或通过迭代研究长期行为。
5. Sum of an Arithmetic Series | 等差数列求和
Arithmetic series sums appear in
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