Arithmetic Series | 等差数列

📚 Arithmetic Series | 等差数列

In A-Level Mathematics, an arithmetic series is the sum of the terms of an arithmetic sequence. Understanding how to sum these sequences efficiently is a fundamental skill that appears in pure mathematics and applied contexts. This article will guide you through the key formulas, derivations, and problem-solving techniques for Edexcel A-Level success.

在 A-Level 数学中，等差数列的和是指等差数列各项之和。熟练掌握如何高效地求出这些数列的和是纯数及应用背景中的基础技能。本文将带你深入探讨针对 Edexcel A-Level 考试的关键公式、推导过程和解题技巧。

---

1. What is an Arithmetic Series? | 什么是等差数列的和？

An arithmetic series results from adding the terms of an arithmetic progression (AP). An AP is a sequence where the difference between consecutive terms is constant, known as the common difference d.

等差数列的和是由等差数列 (AP) 的各项相加得到的。等差数列是指相邻两项之差恒定的数列，这个差值称为公差 d。

fference d = 4. The series for the first four terms is 3 + 7 + 11 + 15, and its sum is 36.

例如，数列 3, 7, 11, 15, ... 的首项 a = 3，公差 d = 4。前四项构成的级数为 3 + 7 + 11 + 15，其和为 36。

The sum of the first n terms is denoted by S_n. If the last term is known, it is often denoted by l.

前 n 项的和用 S_n 表示。如果末项已知，通常用 l 表示。

---

2. Key Notation | 关键符号

The first term of an arithmetic sequence is denoted by a.

等差数列的首项用 a 表示。

The common difference, d, is the constant amount added to each term to obtain the next term.

公差 d 是加到每一项以得到下一项的固定数值。

The number of terms is always represented by n.

项数始终用 n 表示。

nformation, we use l. The relationship l = a + (n−1)d connects the last term with a and d.

如果已知末项，我们用 l 表示。关系式 l = a + (n−1)d 将末项与 a 和 d 联系起来。

The sum of the first n terms is S_n. You will see this symbol in questions asking for totals.

前 n 项和记为 S_n。在求和的问题中你会频繁看到这个符号。

---

3. The Sum Formula (Given Last Term) | 使用末项的求和公式

When both the first term a and the last term l are known, the sum can be found directly without using d.

当首项 a 和末项 l 都已知时，可以直接求和而不需要用到 d。

S_n = ½ n (a + l)

This formula is simply the average of the first and last terms multiplied by the number of terms. It is extremely efficient when the endpoint of the series is given.

该公式实际上就是首末项的平均值乘以项数。当给出级数的终点时，该公式非常高效。

For example, to find the sum of the first 25 terms of the arithmetic series that starts with 4 and ends with 100, compute S₂₅ = ½ × 25 × (4 + 100) = 12.5 × 104 = 1300.

例如，求首项为 4、末项为 100 的等差数列的前 25 项和，计算 S₂₅ = ½ × 25 × (4 + 100) = 12.5 × 104 = 1300。

---

4. The Sum Formula (Using Common Difference) | 使用公差的求和公式

term. By substituting l = a + (n−1)d into the formula above, we obtain the standard form.

更多情况下，你知道首项 a、公差 d 和项数 n，但不知道末项。将 l = a + (n−1)d 代入上述公式即可得到标准形式。

S_n = ½ n [2a + (n−1)d]

This is the most frequently used formula in Edexcel examinations. Always write it down at the start of a question to avoid sign errors with the (n−1) factor.

这是 Edexcel 考试中最常用的公式。解题时一定要先写下该公式，避免 (n−1) 部分的符号出错。

For instance, an arithmetic series has a = 10, d = −3, and n = 12. The sum is S₁₂ = ½ × 12 × [2×10 + (12−1)(−3)] = 6 × [20 − 33] = 6 × (−13) = −78.

例如，一个等差数列首项 a = 10，公差 d = −3，项数 n = 12。求和 S₁₂ = ½ × 12 × [2×10 + (12−1)(−3)] = 6 × [20 − 33] = 6 × (−13) = −78。

---

5. Deriving the Formula by Reversing | 通过倒序相加推导公式

A classic and elegant derivation involves writing the series forwards and backwards, then adding the two expressions.

一种经典且巧妙的推导是将级数正向和反向书写，然后将两个表达式相加。

Write S_n = a + (a+d) + (a+2d) + ... + l.

写出 S_n = a + (a+d) + (a+2d) + ... + l。

Now write the same sum in reverse order: S_n = l + (l−d) + (l−2d) + ... + a.

现在将同一个和逆序写出：S_n = l + (l−d) + (l−2d) + ... + a。

Adding these two equations gives 2S_n = (a+l) + (a+l) + ... + (a+l), where there are n identical brackets. Thus 2S_n = n(a+l).

将这两个等式相加得到 2S_n = (a+l) + (a+l) + ... + (a+l)，共有 n 个相同的括号。因此 2S_n = n(a+l)。

Dividing both sides by 2 yields S_n l). This derivation explains why the formula works and is a common exam proof requirement.

两边除以 2 即得 S_n = ½ n (a+l)。这个推导解释了公式的原理，也是考试中常见的证明要求。

---

广告 / Sponsored

6. Using Sigma Notation | 使用 Σ 符号

Arithmetic series can be expressed compactly using the sigma notation. For an AP with first term a and common difference d, the general term is u_r = a + (r−1)d.

等差数列可以用 sigma 符号简洁地表示。对于首项为 a、公差为 d 的等差数列，其通项为 u_r = a + (r−1)d。

The sum of the first n terms is written as ∑_r=1ⁿ [a + (r−1)d].

前 n 项和可写作 ∑_r=1ⁿ [a + (r−1)d]。

In practice, you may see expressions like ∑_r=1³⁰ ms: when r=1, term = 7; when r=2, term = 9; r=3 gives 11. This is an AP with a = 7, d = 2.

实际应用中，你可能会看到形如 ∑_r=1³⁰ (2r + 5) 的表达式。要识别这个级数，计算前几项：当 r=1 时，项为 7；r=2 时为 9；r=3 时为 11。这是一个首项 a = 7、公差 d = 2 的等差数列。

the constant term plus the coefficient gives the first term. Always check by plugging r=1.

此处，r 的系数就是公差，常数项加上系数即得首项。务必代入 r=1 进行检验。

Using the formula, S₃₀ = 1080. Alternatively, you can treat the sigma as 2∑r + ∑5, but the AP formula is faster.

代入公式，S₃₀ = ½ × 30 × [2×7 + (30−1)×2] = 15 × [14 + 58] = 15 × 72 = 1080。当然，你也可以将 sigma 分解为 2∑r + ∑5，但等差数列公式更加快捷。

---

7. Finding the Sum of the First n Terms | 求前 n 项和

Let us apply the formula to a typical exam-style question: Find the sum of the first 20 terms of the series 6 + 13 + 20 + 27 + ...

让我们将公式应用到一个典型考题：求级数 6 + 13 + 20 + 27 + ... 的前 20 项和。

Here a = 6, d = 7, and n = 20. Substitute into S_n = ½ n [2a + (n−1)d].

这里 a = 6，d = 7，n = 20。代入 S_n = ½ n [2a + (n−1)d]。

S₂₀ = ½ × 20 × [2×6 + (20−1)×7] = 10 × [12 + 133] = 10 × 145 = 1450

Always double-check by computing a few terms manually to confirm a and d, especially if the series is given in words.

务必手动计算前几项以确认 a 和 d，尤其是当题目以文字描述给出时。

---

8. Finding the Number of Terms | 求项数

A common problem gives the sum, first term, and common difference, and asks for n. You will usually need to solve a quadratic equation.

常见问题是已知总和、首项和公差，要求 n。此时通常需要解一个二次方程。

Example: The sum of the first n terms of an arithmetic series is 210. The first term is 3 and the common difference is 4. Find n.

例题：一个等差数列的前 n 项和为 210，首项为 3，公差为 4。求 n。

Set up the formula: S_n = ½ n [2×3 + (n−1)×4] = ½ n (6 + 4n − 4) = ½ n (4n + 2) = n(2n + 1).

建立等式：S_n = ½ n [2×3 + (n−1)×4] = ½ n (6 + 4n − 4) = ½ n (4n + 2) = n(2n + 1)。

+ n − 210 = 0. Factorising gives (2n + 21)(n − 10) = 0. Since n must be positive, n = 10.

令其等于 210：n(2n + 1) = 210 → 2n² + n − 210 = 0。因式分解得 (2n + 21)(n − 10) = 0。由于 n 必须为正，故 n = 10。

Always reject negative solutions. If the quadratic does not factorise, use the quadratic formula and round to the nearest whole number only if the context allows non-integer n; but in APs, n is a positive integer.

负根舍去。如果二次式不能因式分解，使用求根公式，并仅在题目允许非整数的前提下取近似整数；但在等差数列中，n 必须是正整数。

---

9. Finding the First Term or Common Difference | 求首项或公差

Sometimes you are given the sum, the number of terms, and either a or d, and you must find the missing parameter.

有时题目给出总和、项数以及 a 或 d 中的一个，需要你求出缺失的参数。

Example: The sum of the first 15 terms of an AP is 345, and the first term is 2. Find the common difference.

例题：某等差数列前 15 项和为 345，首项为 2。求公差。

345 = ½ × 15 × [2×2 + (15−1)d] = 7.5 × [4 + 14d] = 30 + 105d. So 105d = 315, giving d = 3.

代入：345 = ½ × 15 × [2×2 + (15−1)d] = 7.5 × [4 + 14d]

Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）