Modelling with Series | 用级数建模

📚 Modelling with Series | 用级数建模

At A-Level, modelling with sequences and series means translating everyday situations – like savings, loans or population change – into arithmetic or geometric patterns so that we can calculate totals, long‑term values or required payments. Edexcel exam questions often ask you to set up the correct series, write a sum and then apply the relevant formula. This article guides you through the most common models, linking each to the key techniques you will need in your exam.

在 A-Level 考试中，用数列和级数建模是指将储蓄、贷款或人口变化等日常情景转化为等差或等比模式，以便计算总量、长期值或所需付款额。爱德思考试题经常要求你建立正确的级数、写出总和，然后应用相应公式。本文将带你梳理最常见的模型，将每个模型与你需要的核心技巧联系起来。

1. Understanding Modelling with Sequences and Series | 理解用数列和级数建模

When we model with sequences, each term of a sequence represents a quantity that occurs at a specific time – for example, the amount saved in year k, the number of bacteria on day k, or the payment made at the end of month k. The series is simply the sum of these terms, giving the accumulated total over a period. Recognising whether a model produces an arithmetic progression (constant difference) or a geometric progression (constant ratio) is the first and most important step.

当我们用数列建模时，数列的每一项代表某个特定时间发生的量——例如第 k 年储蓄的金额、第 k 天的细菌数量或第 k 月末的付款额。级数就是这些项的总和，给出一个时间段内的累计总量。判断一个模型生成的是等差数列（公差恒定）还是等比数列（公比恒定），是第一步也是最关键的一步。

Almost all series modelling questions in Edexcel Pure Maths require you to do four things: define the first term and the common difference or ratio, write an expression for the nth term, form the sum of the first n terms, and then answer a contextual question such as "after how many years will the total exceed £10000?" or "what is the present value?" This framework keeps your working clear and earns method marks.

爱德思纯数学中几乎所有的级数建模题都要求你做四件事：定义首项与公差或公比，写出第 n 项的表达式，列出前 n 项的和，然后回答一个情景问题，例如“经过多少年后总额会超过 10000 英镑？”或“现值是多少？”。这个框架能让你的解题过程清晰，并获得方法分。

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2. Arithmetic Series and Linear Models | 等差数列与线性模型

An arithmetic series models situations where a quantity increases or decreases by the same fixed amount each period. Classic examples include the total cost of hiring a car for n days if the daily charge rises by a constant amount, the total number of seats in an arc of a theatre where each row has a fixed number more than the previous row, or a savings plan where the amount deposited rises by £d each term.

等差数列适用于某个量每个时期以相同的固定数量增加或减少的情景。经典例子包括：如果租车每日费用按固定金额增加，n 天的总费用；剧院的弧形座位，每一排比前一排多固定数量的座位；或者每次存款额比上一次增加 d 英镑的储蓄计划。

and common difference d, the nth term is uₙ = a + (n−1)d. The sum of the first n terms is

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

n modelling, a is the initial value and d is the constant step. If d > 0 the model shows linear growth.

对于首项为 a、公差为 d 的等差数列，第 n 项为 uₙ = a + (n−1)d。前 n 项的和为

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

，其中 l = a + (n−1)d 为末项。在建模中，a 是初始值，d 是固定步长。若 d > 0，模型呈线性增长。

Example: A company plans to increase its annual training budget by £500 each year. In the first year it allocates £ The total spent over the first 8 years is S₈ = 8/2 [2×3000 + 7×500] = 4 × (6000 + 3500) = £38000. Always check that the problem is truly arithmetic – hmetic model when the quantity changes by a percentage, which requires a geometric series.

示例：一家公司计划每年增加培训预算 500 英镑，第一年拨款 3000 英镑。第 n 年的金额为 uₙ = 3000 + 500(n−1)。前 8 年的总支出为 S₈ = 8/2 [2×3000 + 7×500] = 4 × (6000 + 3500) = 38000 英镑。务必检查问题是否真的是等差——许多学生错误地对百分比变化的情况使用等差模型，而这种情况需要用等比级数。

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3. Geometric Series and Exponential Models | 几何级数与指数模型

Geometric series describe situations where a quantity changes by a constant factor each period. This factor is often (1 + r) for growth or (1 − r) for decay, where r is the percentage change written as a decimal. Examples include compound interest, depreciation of machinery, population growth and the elimination of a drug from the body.

几何级数描述的是某个量每个时期以恒定倍数变化的情景。该倍数通常是 (1 + r) 表示增长，或 (1 − r) 表示衰减，其中 r 是以小数表示的百分比变化。例子包括复利、机器折旧、人口增长和药物从体内的清除。

mmon ratio is r, the nth term is uₙ = a rⁿ⁻¹. The sum of the first n terms (when r ≠ 1) is

Sₙ = a(1 − rⁿ)/(1 − r)

for |r| < 1, or

Sₙ = a(rⁿ − 1)/(r − 1)

when r > 1. If |r| < 1, the sum to infinity exists:

S∞ = a/(1 − r)

. This is extremely useful for modelling long‑term steady states, such as the eventual level of a drug in the bloodstream or the maximum possible sum saved with a perpetuity.

若首项为 a，公比为 r，则第 n 项为 uₙ = a rⁿ⁻¹。当 r ≠ 1 时，前 n 项的和为

Sₙ = a(1 − rⁿ)/(1 − r)

（当 |r| < 1），或当 r > 1 时用

Sₙ = a(rⁿ − 1)/(r − 1)

。若 |r| < 1，无穷和存在：

S∞ = a/(1 − r)

。这对于模拟长期稳态非常有用，例如血液中药物的最终水平或永续年金的最大可能储蓄总额。

Always identify a clearly; it is the starting quantity at the point from which the constant ratio applies. If a process starts at time zero with a different value, make sure your first term is the first value that follows the geometric pattern.

务必清晰识别 a；它是恒定比值开始作用时的起始量。如果过程从零时刻开始且具有不同值，请确保首项是遵循几何规律的首个值。

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4. Savings Plans with Compound Interest | 储蓄计划与复利

One of the most important modelling applications is a regular savings plan where deposits earn compound interest. Suppose you deposit £P at the end of each year into an account paying a fixed annual interest rate of i% (expressed as a decimal). Each deposit grows for a different number of years. The first deposit gains interest for (n−1) years, the second for (n−2) years, and the last deposit earns no interest in that final year.

最重要的建模应用之一是定期储蓄计划，存款以复利增值。假设你在每年末存入 P 英镑到一个支付固定年利率 i（以小数表示）的账户。每笔存款增值的年数不同：第一笔存款获得 (n−1) 年的利息，第二笔获得 (n−2) 年的利息，最后一笔存款在最后一年没有利息。

The total amount just after the nth deposit is therefore the geometric series

S = P + P(1+i) + P(1+i)² + ... + P(1+i)ⁿ⁻¹

. Notice the first term is P (the last deposit) and the common ratio is (1+i). Using the sum formula with r > 1:

S = P[(1+i)ⁿ − 1] / i

. This formula can be used to find the future value of an annuity. If deposits are made at the start of each year, the series shifts and the common ratio still multiplies the whole sum.

因此，第 n 次存款刚结束时的总额为几何级数

S = P + P(1+i) + P(1+i)² + ... + P(1+i)ⁿ⁻¹

。注意首项是 P（最后一笔存款），公比为 (1+i)。使用 r > 1 时的和公式：

S = P[(1+i)ⁿ − 1] / i

。此公式可用于求年金的终值。如果存款在每年初进行，级数会移位，但公比仍乘以整个总和。

Example: You save £2000 at the end of each year in an account earning 5% interest. After 10 years the total is S = 2000[(1.05)¹⁰ − 1] / 0.05 ≈ £ nterest. In an exam always show the series explicitly before substituting into the formula – this demonstrates understanding and prevents errors with the exponent count.

示例：你每年末存入 2000 英镑，账户年利率 5%。10 年后的总额为 S = 2000[(1.05)¹⁰ − 1] / 0.05 ≈ 25155.80 英镑。这种计算显示了复利的威力。考试中始终应先明确列出级数再代入公式——这既能展示理解，又能避免指数计数错误。

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5. Loan Repayments and Amortisation | 贷款偿还与分期摊还

elate the borrowed amount, the regular repayment and the interest rate. Imagine you borrow £L at an annual interest rate of i, and you repay £ s L(1+i) − A, after 2 years it is [L(1+i) − A](1+i) − A = L(1+i)² − A(1+i) − A, and so on.

贷款偿还模型利用几何级数将贷款额、定期还款额和利率联系起来。设想你以年利率 i 借入 L 英镑，在每年末偿还 A 英镑，共 n 年。每次还款后的未清余额可以追踪：第 1 年后为 L(1+i) − A，第 2 年后为 L(1+i)² − A(1+i) − A，以此类推。

After n years the balance must be zero, giving

L(1+i)ⁿ = A[ (1+i)ⁿ⁻¹ + (1+i)ⁿ⁻² + ... + 1 ]

. The right‑hand side is a geometric series with first term 1 (or we can re‑index) and common ratio (1+i). Its sum is [ (1+i)ⁿ − 1 ] / i . Therefore the relationship becomes

L(1+i)ⁿ = A × [(1+i)ⁿ − 1] / i

. This formula can be rearranged to find A, L, i or n depending on the question.

n 年后余额必须为零，因此得到

L(1+i)ⁿ = A[ (1+i)ⁿ⁻¹ + (1+i)ⁿ⁻² + ... + 1 ]

。右侧为等比级数，首项为 1，公比为 (1+i)，其和为 [ (1+i)ⁿ − 1 ] / i。因此关系式变为

L(1+i)ⁿ = A × [(1+i)ⁿ − 1] / i

。该公式可变形求出 A、L、i 或 n，取决于题目要求。

Edexcel questions often ask you to prove this formula or to use it in a slightly different context, such as an overdraft or a car lease. The key skill is building the balance equation step by step and recognising the hidden geometric series. Never shortcut the derivation unless the formula is given; method marks are awarded for setting up the series.

爱德思考题常要求你证明该公式，或在稍微不同的情景中使用它，如透支或汽车租赁。核心技能是逐步建立余额方程并识别出隐藏的几何级数。除非题目直接给出公式，否则不要走捷径；设置级数的过程会获得方法分。

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6. Depreciation of Assets | 资产折旧

Depreciation can be modelled in two distinct ways. Straight‑line depreciation reduces the asset’s value by a fixed amount each year, producing an arithmetic sequence. Reducing‑balance depreciation reduces the value by a fixed percentage each year, which generates a geometric sequence. In A‑Level exams, the geometric version appears far more frequently because it reflects the way many assets lose value and leads to series calculations.

折旧可以两种截然不同的方式建模。直线折旧每年减少资产固定金额，形成等差数列。余额递减折旧每年按固定百分比减少价值，生成等比数列。在 A‑Level 考试中，等比版本出现的频率高得多，因为它反映了许多资产贬值的方式，并需用级数计算。

For an asset costing £V₀ that depreciates at r% s is V₀ − Vₙ, but questions often ask for the sum of the annual depreciation charges. The depreciation in year k is the difference between the values at the start and end of the year: Dₖ = V₀(1 − r)ᵏ⁻¹ × r . These annual charges form a geometric series with first term rV₀ and common ratio (1 − r).

对于成本为 V₀ 的资产，每年按 r% 折旧（余额递减），n 年后的价值为 Vₙ = V₀(1 − r)ⁿ，其中 r 以小数表示。n 年的总折旧额为 V₀ − Vₙ，但题目常要求求每年折旧费用的总和。第 k 年的折旧费为年初与年末价值之差：Dₖ = V₀(1 − r)ᵏ⁻¹ × r。这些年折旧费构成等比级数，首项为 rV₀，公比为 (1 − r)。

preciation equals the original cost minus the book value after n years. You can practise by picking V₀ = 20000, r = 0.2, n = 5 and verifying both the term‑by‑term sum and the formula.

因此，前 n 年折旧费用的总和为 Sₙ = rV₀[1 − (1 − r)ⁿ] / [1 − (1 − r)] = V₀[1 − (1 − r)ⁿ]。这在直观上合理：总折旧等于原始成本减去 n 年后的账面价值。可以通过取 V₀ = 20000, r = 0.2, n = 5 练习，逐项求和并与公式结果对比验证。

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