Sequences and Series | 数列与级数考点精讲

📚 Sequences and Series | 数列与级数考点精讲

Sequences and series form the backbone of many A-Level OCR Mathematics problems, linking algebra, modelling, and limits. A strong command of arithmetic and geometric progressions, sigma notation, and convergence tests is essential for top marks. This revision guide breaks down every key concept, from nth terms to infinite sums, with paired English–Chinese explanations to clarify theory and technique.

数列与级数是 A-Level OCR 数学中许多问题的基础，连接着代数、建模与极限。熟练掌握等差数列与等比数列、∑ 符号以及收敛性检验是取得高分的关键。本复习指南逐一剖析每一个重要概念，从通项到无穷和，配以中英对照讲解，帮助理清理论与解题技巧。

1. Arithmetic Sequences | 等差数列

An arithmetic sequence has a constant difference between consecutive terms. If the first term is a and the common difference is d, then the nth term is given by the formula:

等差数列的相邻两项之差为常数。若首项为 a，公差为 d，则第 n 项的通项公式为：

u_n = a + (n – 1)d

For example, the sequence 5, 9, 13, 17, … has a = 5 and d = 4, so its 20th term is u₂₀ = 5 + 19 × 4 = 81.

例如，数列 5, 9, 13, 17, … 的首项 a = 5，公差 d = 4，因此第 20 项为 u₂₀ = 5 + 19 × 4 = 81。

To check if a number belongs to an arithmetic sequence, set u_n equal to that number, solve for n, and confirm that n is a positive integer.

要判断某个数是否属于某等差数列，可令 u_n 等于该数，解出 n，并验证 n 是否为正整数。

2. Sum of an Arithmetic Series | 等差数列求和

The sum of the first n terms of an arithmetic series is found using the formula:

等差数列前 n 项和的公式为：

S_n = n/2 [2a + (n – 1)d]

Alternatively, if the last term l is known, S_n = n/2 (a + l). This formula comes from pairing terms that add to a constant sum.

若已知末项 l，也可使用 S_n = n/2 (a + l)。该公式源于将首尾配对相加得到常数的思想。

For the sequence 3, 7, 11, …, find the sum of the first 12 terms: S₁₂ = 12/2 [2×3 + (12 – 1)×4] = 6 × (6 + 44) = 300.

求数列 3, 7, 11, … 的前 12 项和：S₁₂ = 12/2 [2×3 + (12 – 1)×4] = 6 × (6 + 44) = 300。

When the number of terms is unknown, use the given S_n and solve the resulting quadratic equation for n.

当项数未知时，可利用已知的 S_n 建立二次方程求解 n。

3. Geometric Sequences | 等比数列

A geometric sequence has a constant ratio between consecutive terms. With first term a and common ratio r, the nth term is:

等比数列的相邻两项之比为常数。设首项为 a，公比为 r，则第 n 项为：

u_n = ar^n-1

For instance, 2, 6, 18, 54, … has a = 2, r = 3, giving u₅ = 2 × 3⁴ = 162.

例如，2, 6, 18, 54, … 中 a = 2，r = 3，则第 5 项 u₅ = 2 × 3⁴ = 162。

If the terms alternate in sign, r is negative. The sequence 4, -8, 16, -32, … has r = -2.

若数列正负交替，公比 r 为负数。如 4, -8, 16, -32, … 的 r = -2。

To find missing terms, use the ratio of known terms: if u₃ = 12 and u₅ = 48, then ar² = 12 and ar⁴ = 48, so r² = 4, giving r = ±2.

求解缺失项时，可利用已知项的比值：若 u₃ = 12 且 u₅ = 48，则 ar² = 12，ar⁴ = 48，相除得 r² = 4，故 r = ±2。

4. Sum of a Geometric Series | 等比数列求和

The sum of the first n terms of a geometric series is:

等比数列前 n 项和的公式为：

S_n = a(1 – rⁿ) / (1 – r)

for r ≠ 1. This is derived by multiplying the series by r and subtracting. For r = 1, the sum is simply na.

适用于 r ≠ 1。该公式通过将数列乘以 r 后相减得出。当 r = 1 时，和为 na。

If you prefer a numerator with rⁿ – 1, use S_n = a(rⁿ – 1) / (r – 1); both forms are equivalent.

若你偏好分子为 rⁿ – 1，可使用 S_n = a(rⁿ – 1) / (r – 1)，两者等价。

Example: sum the first 6 terms of 3, 6, 12, … Here a = 3, r = 2, so S₆ = 3(1 – 2⁶) / (1 – 2) = 3(1 – 64) / (-1) = 189.

示例：求 3, 6, 12, … 的前 6 项和。a = 3，r = 2，S₆ = 3(1 – 2⁶) / (1 – 2) = 3(1 – 64) / (-1) = 189。

Always check that the correct number of terms is used; n appears as the exponent of r, not n−1.

务必核实项数是否正确；n 是 r 的指数，切勿误用 n−1。

5. Infinite Geometric Series | 无穷等比级数

An infinite geometric series converges if |r| < 1. In that case, the sum to infinity is:

当 |r| < 1 时，无穷等比级数收敛。此时无穷和为：

S_∞ = a / (1 – r)

If |r| ≥ 1, the series diverges, meaning the sum grows without bound or oscillates indefinitely.

若 |r| ≥ 1，级数发散，意味着和无限增大或无限振荡。

For the series 8 + 4 + 2 + 1 + …, a = 8, r = 1/2, so S_∞ = 8 / (1 – 0.5) = 16.

对于级数 8 + 4 + 2 + 1 + …，a = 8，r = 1/2，因此 S_∞ = 8 / (1 – 0.5) = 16。

This formula is often used in recurring decimal-to-fraction conversions: 0.2̇3̇ = 23/100 + 23/10000 + … with a = 23/100, r = 1/100.

该公式常用于将循环小数化为分数：0.2̇3̇ = 23/100 + 23/10000 + …，其中 a = 23/100，r = 1/100。

6. Sigma Notation | ∑ 符号

Sigma (∑) provides a compact way to write the sum of terms following a rule. The expression ∑_k=1ⁿ (2k + 1) means sum the values of 2k + 1 for k = 1 to n.

∑ 符号可简洁地表示按规则求和。表达式 ∑_k=1ⁿ (2k + 1) 表示对 k = 1 到 n 计算 2k + 1 的总和。

You can split sums: ∑ (ar^k-1) from k=1 to n is exactly the geometric series a + ar + ar² + … + ar^n-1.

可以拆分求和：从 k=1 到 n 的 ∑ (ar^k-1) 正是等比级数 a + ar + ar² + … + ar^n-1。

Be careful with the lower limit: if it starts at r = 0, the first term is ar⁰ = a, which affects the number of terms.

注意下标的起始值：若从 r = 0 开始，首项为 ar⁰ = a，这将影响项数的计算。

English: ∑_r=1ⁿ r = n(n+1)/2
中文： ∑_r=1ⁿ r = n(n+1)/2
English: ∑_r=1ⁿ r² = n(n+1)(2n+1)/6
中文： ∑_r=1ⁿ r² = n(n+1)(2n+1)/6

7. Recurrence Relations | 递推关系

A recurrence relation defines each term using the previous one, e.g. u_n+1 = 2u_n + 3 with a starting value. You generate terms step by step.

递推关系通过前一项来定义每一项，例如 u_n+1 = 2u_n + 3，并给定初始值。你需要逐步生成各项。

If a sequence defined this way approaches a limit L, then L satisfies L = 2L + 3, giving L = -3. However, this only holds if the sequence actually converges.

若按此方式定义的数列趋近于极限 L，则 L 满足 L = 2L + 3，解得 L = -3。但这只有在数列确实收敛时才成立。

Some recurrence relations lead to periodic sequences. For instance, u_n+1 = -u_n with u₁ = 5 gives 5, -5, 5, -5, … (period 2).

某些递推关系会产生周期数列。例如 u_n+1 = -u_n 且 u₁ = 5，得到 5, -5, 5, -5, …（周期为 2）。

When finding a limit, replace both u_n+1 and u_n by L and solve, but always verify convergence first.

求极限时，将 u_n+1 与 u_n 都替换为 L 并求解，但切记先验证是否收敛。

8. Convergence and Divergence | 收敛与发散

A sequence converges if its terms approach a fixed finite value as n → ∞. For geometric sequences rⁿ converges to 0 when |r| < 1 and diverges otherwise.

若数列的项随着 n → ∞ 趋近于某个固定的有限值，则称该数列收敛。对于等比数列 rⁿ，当 |r| < 1 时收敛于 0；否则发散。

An arithmetic sequence always diverges unless d = 0, because terms march to ±∞.

等差数列除非 d = 0，否则总是发散，因为项会趋向正无穷或负无穷。

For series, an infinite geometric series converges if |r| < 1; its sum is a/(1 - r). The harmonic series ∑ 1/n diverges, but you are not required to prove it at OCR – just know that not all series converge.

对于级数，无穷等比级数在 |r| < 1 时收敛，和为 a/(1 - r)。调和级数 ∑ 1/n 发散，但 OCR 不要求证明——只需知道并非所有级数都收敛。

When modelling with recurrence relations, convergence behaviour helps interpret long-term trends, such as a population stabilising.

在使用递推关系建模时，收敛性有助于解释长期趋势，例如种群数量趋于稳定。

9. Modelling with Sequences | 数列模型应用

Arithmetic models suit linear growth, such as a fixed annual salary increase. If your salary rises by £1000 each year, the total earnings over 20 years form an arithmetic series.

等差数列模型适用于线性增长，例如固定的年薪增长。若每年加薪 1000 英镑，则 20 年的总收入构成一个等差数列。

Geometric models describe multiplicative change, like compound interest. A £500 investment at 4% annual interest becomes a geometric sequence with r = 1.04.

等比数列模型描述倍增式变化，如复利计息。一笔 500 英镑的投资，年利率 4%，转化为公比 r = 1.04 的等比数列。

For depreciation, a car losing 15% of its value each year produces a geometric sequence with r = 0.85. After n years, its value is arⁿ.

对于折旧问题，一辆车每年贬值 15%，产生公比 r = 0.85 的等比数列。n 年后的价值为 arⁿ。

Always define your terms and state clearly whether you are using u_n or S_n. In modelling, S_n often represents total accumulated quantity.

务必定义各项，并清楚说明使用的是 u_n 还是 S_n。在建模中，S_n 常表示累计总量。

10. Exam Tips and Common Pitfalls | 考试技巧与常见错误

Many students confuse the number of terms in a sum. In sigma notation ∑_k=3²⁰, there are 20 – 3 + 1 = 18 terms, not 17.

许多学生容易算错求和的项数。在 ∑_k=3²⁰ 中，项数为 20 – 3 + 1 = 18 项，而非 17 项。

Using u_n = a + (n-1)d correctly means the first term corresponds to n = 1. If a problem gives terms like ‘the third term is 7’, set u₃ = 7, not u₂.

正确使用 u_n = a + (n-1)d 意味着首项对应 n = 1。若题目给出“第三项为 7”，应设 u₃ = 7，而非 u₂。

With infinite geometric series, always state the condition |r| < 1 before quoting S_∞. Missing this condition loses marks.

在处理无穷等比级数时，引用 S_∞ 前务必先说明条件 |r| < 1，遗漏该条件会导致失分。

For recurrence limits, check that the sequence is indeed convergent by evaluating a few terms; a simple equality L = f(L) may give a spurious limit if the sequence diverges.

对于递推极限，需通过计算若干项来检验数列是否确实收敛；若数列发散，直接由 L = f(L) 解出的“极限”可能是伪解。

Finally, always show substitution steps clearly. Examiners award method marks even if the final answer is slightly off.

最后，务必清晰展示代入步骤。即使最终答案略有偏差，考官仍会给步骤分。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Sequences and Series | 数列与级数 考点精讲