📚 Modelling with Series | 数列建模
In A-Level Mathematics, modelling with series means using arithmetic and geometric sequences to describe real-world situations such as savings, loans, depreciation and population growth. Understanding when to use each type of series is essential for Edexcel exam questions.
在 A-Level 数学中，数列建模是指用等差数列和等比数列来描述储蓄、贷款、折旧和人口增长等现实情境。掌握何时使用哪种数列是 Edexcel 考试的关键。
1. What is Modelling with Series? | 什么是数列建模
A sequence is an ordered list of numbers. A series is the sum of the terms of a sequence. In modelling, we usually split time into equal steps, such as years, months or days, and use a sequence to represent a quantity at each step.
数列是一个有序的数字列表，级数是数列各项的和。在建模中，我们通常把时间分成相等的步长，例如年、月或日，并用数列表示每一步的数量。
Arithmetic: Uₙ = a + (n - 1)d, Sₙ = n/2 [2a + (n - 1)d]
中文：等差数列第 n 项为 Uₙ = a + (n-1)d，前 n 项和为 Sₙ = n/2 [2a + (n-1)d]，其中 a 为首项，d 为公差。
Geometric: Uₙ = arⁿ⁻¹, Sₙ = a(1 - rⁿ)/(1 - r) for r ≠ 1
中文：等比数列第 n 项为 Uₙ = arⁿ⁻¹，前 n 项和为 Sₙ = a(1 - rⁿ)/(1 - r)，其中 r ≠ 1，a 为首项，r 为公比。
For a geometric series with |r| infinity is S∞ = a/(1 - r). This is useful for long-term modelling such as repeated decay.
当 |r| < 1 时，等比级数可以求和到无穷：S∞ = a/(1 - r)。这在长期衰减等建模中很有用。
2. Arithmetic Series in Financial Modelling | 金融建模中的等差数列
Arithmetic sequences model situations where a quantity changes by a constant amount in each period. Simple interest and fixed annual salary increases are common examples.
等差数列用于模拟每个周期数量变化固定值的情况。单利和固定年薪增长是常见例子。
(1 + nr). The balance increases by the same amount Pr each period, so it follows an arithmetic sequence.
对于单利，如果本金 P 每期利率为 r，那么 n 期后的余额为 P + nPr = P(1 + nr)。余额每期增加相同的金额 Pr，因此它遵循等差数列。
Example: A salary starts at £20,000 and increases by £1,500 each year. Find the salary in year 10 and the total earnings over the first 10 years.
示例：起薪为 £20,000，每年增加 £1,500。求第 10 年的薪水以及前 10 年的总收入。
a = 20000, d = 1500, n = 10
U₁₀ = 20000 + 9 × 1500 = 33500
S₁₀ = 10/2 [2 × 20000 + 9 × 1500] = 5 (40000 + 13500) = £267,500
So the salary in year 10 is £33,500 and the total earned over 10 years is £267,500.
因此，第 10 年的薪水为 £33,500，10 年总收入为 £267,500。
3. Geometric Series in Financial Modelling | 金融建模中的等比数列
Geometric sequences model situations where a quantity grows or decays by a constant percentage each period. Compound interest, inflation and population growth are typical examples.
等比数列用于模拟每个周期数量按固定百分比增长或衰减的情况。复利、通货膨胀和人口增长是典型例子。
The balances form a geometric sequence with first term P(1 + r) and common ratio (1 + r).
如果金额 P 按每期复利率 r 投资，则 n 期后的价值为 Vₙ = P(1 + r)ⁿ。余额构成等比数列，首项为 P(1 + r)，公比为 (1 + r)。
The sum of the first n terms of a geometric series can be used to calculate the total of several growing deposits or cumulative amounts.
等比数列前 n 项和公式可用于计算若干笔增长存款的总和或累计金额。
4. Savings and Regular Deposits | 储蓄与定期存款
When a fixed amount is deposited at the end of each period into an account that pays compound interest, the future value is the sum of a geometric series.
当在每个周期末将固定金额存入支付复利的账户时，未来价值是等比级数的和。
Let A be the regular deposit, i the interest rate per period, and n the number of deposits. The first deposit at the end of period 1 grows for (n - 1) periods, the second for (n - 2) periods, and the last deposit grows for 0 periods.
设 A 为每期固定存款，i 为每期利率，n 为存款次数。第 1 期末的存款增长 (n - 1) 期，第 2 笔增长 (n - 2) 期，最后一笔增长 0 期。
FV = A + A(1 + i) + A(1 + i)² + ... + A(1 + i)ⁿ⁻¹ = A[(1 + i)ⁿ - 1]/i
中文：未来价值 FV = A[(1 + i)ⁿ - 1]/i，其中 A 是每期存款，i 是每期利率，n 是存款次数。
Example: You deposit £200 at the end of each month for 24 months. The monthly interest rate is 0.5%. Calculate the total value at the end of 24 months.
示例：每月末存入 £200，存 24 个月。月利率为 0.5%。计算 24 个月末的总价值。
FV = 200[(1.005)²⁴ - 1]/0.005
(1.005)²⁴ ≈ 1.12716, so FV ≈ 200 × 0.12716 / 0.005 ≈ £5,086.36
The total value is approximately £5,086.36. Notice that the interest earned is £5,086.36 - 24 × £200 = £286.36.
总价值约为 £5,086.36。注意所得利息为 £5,086.36 - 24 × £200 = £286.36。
5. Loan Repayments and Amortisation | 贷款偿还与摊销
Loans can be modelled with a geometric series if repayments are made at regular intervals. The outstanding balance after n repayments can be found by combining compound interest on the loan with the accumulated value of repayments.
如果定期还款，贷款可以用等比数列建模。n 次还款后的未偿还余额可以通过贷款复利增长减去还款累计价值来计算。
If a loan L is taken at interest rate i per period, and a repayment R is made at the end of each period, then after n payments the outstanding balance is:
如果贷款 L 每期利率为 i，每期末还款 R，那么 n 次还款后的未偿还余额为：
Bₙ = L(1 + i)ⁿ - R[(1 + i)ⁿ - 1]/i
中文：Bₙ = L(1 + i)ⁿ - R[(1 + i)ⁿ - 1]/i，其中 L 是贷款额，i 是每期利率，R 是每期还款额。
, so L(1 + i)ⁿ = R[(1 + i)ⁿ - 1]/i. This can be rearranged to find the required repayment.
如果贷款恰好经过 N 次还款还清，则 Bₙ = 0，因此 L(1 + i)ⁿ = R[(1 + i)ⁿ - 1]/i。可以重新整理来求所需还款额。
Example: A loan of £10,000 is charged at 6% per year compounded monthly. It is repaid over 24 equal monthly instalments. Find the monthly repayment.
示例：贷款 £10,000，年利率 6% 按月复利，分 24 个月等额偿还。求每月还款额。
i = 0.06/12 = 0.005, L = 10000, N = 24
R = L i (1 + i)ⁿ / [(1 + i)ⁿ - 1] = 10000 × 0.005 × 1.005²⁴ / (1.005²⁴ - 1) ≈ £443.21
The monthly repayment is approximately £443.21.
每月还款额约为 £443.21。
6. Depreciation of Assets | 资产折旧
Reducing balance depreciation means an asset loses a fixed percentage of its current value each year. This produces a geometric sequence with common ratio (1 - d), where d is the depreciation rate.
余额递减折旧是指资产每年按其当前价值损失一个固定百分比。这产生公比为 (1 - d) 的等比数列，其中 d 为折旧率。
Vₙ = V₀(1 - d)ⁿ
中文：Vₙ = V₀(1 - d)ⁿ，其中 V₀ 为初始价值，d 为年折旧率，n 为年数。
Example: A machine costs £20,000 and depreciates at 15% per year using the reducing balance method. Find its value after 5 years and the total depreciation over those 5 years.
示例：一台机器成本为 £20,000，按余额递减法每年折旧 15%。求 5 年后的价值以及这 5 年的总折旧。
V₅ = 20000 × 0.85⁵ ≈ 20000 × 0.44371 = £8,874.10
Total depreciation = 20000 - 8874.10 = £11,125.90
The value after 5 years is about £8,874.10 and the total depreciation is about £11,125.90.
5 年后
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