Arithmetic Series | 等差数列求和

📚 Arithmetic Series | 等差数列求和

An arithmetic series is the sum of the terms of an arithmetic sequence, where the difference between consecutive terms is constant. In A-Level Edexcel Mathematics, you need to use and derive the sum formulae, apply them to unknown quantities, and solve modelling problems.

等差数列是等差数列各项的和，其中相邻两项的差为常数。在 A-Level Edexcel 数学中，你需要使用并推导求和公式、将其应用于未知量，并解决建模问题。

1. Arithmetic Sequence and Common Difference | 等差数列与公差

An arithmetic sequence has the form a, a+d, a+2d, … where a is the first term and d is the common difference. The difference d = uₙ₊₁ − uₙ is constant for all n.

等差数列的形式为 a、a+d、a+2d……其中 a 是首项，d 是公差。对任意 n，差值 d = uₙ₊₁ − uₙ 保持不变。

Example: 3, 7, 11, 15, … has a = 3 and d = 4. The sequence is increasing because d > 0; if d < 0 it is decreasing.

例如：3、7、11、15……的首项 a = 3，公差 d = 4。因为 d > 0，该数列递增；若 d < 0，则递减。

2. The nth Term Formula | 第 n 项公式

The nth term of an arithmetic sequence is uₙ = a + (n − 1)d. This formula allows you to find any term directly or to set up equations when terms are unknown.

等差数列的第 n 项为 uₙ = a + (n − 1)d。该公式可用于直接求任意项，或在项未知时建立方程。

For example, the 20th term of 3, 7, 11, … is u₂₀ = 3 + 19 × 4 = 79.

例如，3、7、11……的第 20 项为 u₂₀ = 3 + 19 × 4 = 79。

3. Defining an Arithmetic Series | 等差数列的级数定义

A series is the sum of terms of a sequence. For an arithmetic sequence, the sum of the first n terms is written Sₙ = a + (a+d) + (a+2d) + … + (a+(n−1)d).

级数是数列各项之和。对于等差数列，前 n 项和写作 Sₙ = a + (a+d) + (a+2d) + … + (a+(n−1)d)。

You may meet sigma notation: Sₙ = Σᵣ₌₁ⁿ (a + (r−1)d). This compact form is common in exam questions.

你可能会遇到求和符号：Sₙ = Σᵣ₌₁ⁿ (a + (r−1)d)。这种紧凑写法在考试题中很常见。

4. Deriving the Sum Formula | 求和公式推导

To derive Sₙ, write the sum forwards and backwards:

为推导 Sₙ，将和式正写与倒写：

Sₙ = a + (a+d) + … + (a+(n−1)d)

Sₙ = (a+(n−1)d) + (a+(n−2)d) + … + a

Adding term by term gives 2Sₙ = n × [2a + (n−1)d], because each column sums to the same value 2a + (n−1)d.

逐项相加得到 2Sₙ = n × [2a + (n−1)d]，因为每一列的和都等于相同的值 2a + (n−1)d。

Therefore Sₙ = n/2 [2a + (n−1)d]. This is the standard formula on the Edexcel formula booklet.

因此 Sₙ = n/2 [2a + (n−1)d]。这是 Edexcel 公式手册中的标准公式。

5. Alternative Formula Using the Last Term | 使用末项的另一公式

If the last term l is known, Sₙ = n/2 (a + l). Since l = a + (n−1)d, this is equivalent to the previous formula and is quicker in many questions.

如果已知末项 l，则 Sₙ = n/2 (a + l)。由于 l = a + (n−1)d，它与前面的公式等价，在许多题目中更快捷。

For example, find the sum of the first 50 positive odd numbers. Here a = 1, d = 2, l = u₅₀ = 1 + 49 × 2 = 99, so S₅₀ = 50/2 × (1 + 99) = 2500.

例如，求前 50 个正奇数之和。这里 a = 1，d = 2，l = u₅₀ = 1 + 49 × 2 = 99，因此 S₅₀ = 50/2 × (1 + 99) = 2500。

6. Solving for Unknowns | 求解未知量

Exam questions often give Sₙ, n, and one other quantity, asking you to find a or d. Substitute the known values into Sₙ = n/2 [2a + (n−1)d] and solve the resulting linear or quadratic equation.

考试题常给出 Sₙ、n 和另一个量，要求你求 a 或 d。将已知值代入 Sₙ = n/2 [2a + (n−1)d]，并求解所得的一次或二次方程。

Always check that your solution makes sense in context: the first term can be negative or fractional, but n must be a positive integer.

务必检查解是否符合题意：首项可以为负或分数，但 n 必须是正整数。

7. Proof Questions | 证明题

Edexcel may ask you to prove the sum formula. The backward-forward addition method above is the expected approach, and you should explain why each pair sums to 2a+(n−1)d.

Edexcel 可能会要求你证明求和公式。上述倒序相加法是预期方法，你应解释为什么每一对的和都是 2a+(n−1)d。

A common alternative is to use the formula for the sum of the first n natural numbers: 1+2+…+n = n(n+1)/2, but this is not a direct proof for the general arithmetic series.

常见的替代方法是使用前 n 个自然数之和公式：1+2+…+n = n(n+1)/2，但这并非一般等差数列求和公式的直接证明。

8. Sigma Notation and Arithmetic Series | 求和符号与等差数列

You may see expressions such as Σᵣ₌₁ⁿ (3r + 2). This represents an arithmetic series with uᵣ = 3r + 2. To find the sum, identify a = u₁ and d = the coefficient of r.

你可能会看到诸如 Σᵣ₌₁ⁿ (3r + 2) 的表达式。它表示 uᵣ = 3r + 2 的等差数列。要求其和，先确定 a = u₁，d = r 的系数。

Example: Σᵣ₌₁⁴⁰ (3r + 2) has a = 5, d = 3, n = 40, so S₄₀ = 40/2 [2 × 5 + 39 × 3] = 20 × 127 = 2540.

例如：Σᵣ₌₁⁴⁰ (3r + 2) 中 a = 5，d = 3，n = 40，因此 S₄₀ = 40/2 [2 × 5 + 39 × 3] = 20 × 127 = 2540。

9. Finding the Number of Terms | 求项数

If you know a, l, and d, you can find n using l = a+(n−1)d. Rearranging gives n = (l − a)/d + 1.

如果已知 a、l 和 d，你可以用 l = a+(n−1)d 求 n。整理得 n = (l − a)/d + 1。

This is useful when the series is given in expanded form, such as 5 + 9 + 13 + … + 101. Here n = (101 − 5)/4 + 1 = 25.

当级数以展开形式给出时这很有用，例如 5 + 9 + 13 + … + 101。这里 n = (101 − 5)/4 + 1 = 25。

10. Word Problems and Modelling | 应用题与建模

Arithmetic series questions often describe savings, seating, production, or simple interest-free instalments. Identify a, d, and n from the wording, then apply Sₙ.

等差数列题目常描述储蓄、座位、产量或不含利息的分期付款。从文字中识别 a、d 和 n，然后应用 Sₙ。

Example: A student saves £5 in week 1, £8 in week 2, £11 in week 3, and so on. The total after 20 weeks is S₂₀ = 20/2 [2 × 5 + 19 × 3] = £670.

例如：一名学生第 1 周存 £5，第 2 周存 £8，第 3 周存 £11，以此类推。20 周后的总储蓄为 S₂₀ = 20/2 [2 × 5 + 19 × 3] = £670。

11. Common Mistakes | 常见错误

Using n in the last term instead of n−1: the final term is a+(n−1)d, not a+nd.

在末项中使用 n 而不是 n−1：最后一项是 a+(n−1)d，而不是 a+nd。
Forgetting to divide by 2 in the sum formula.

忘记在求和公式中除以 2。
Confusing Sₙ with the nth term uₙ.

将 Sₙ 与第 n 项 uₙ 混淆。
Using d incorrectly when the sequence is decreasing, such as 20, 17, 14, … where d = −3.

当数列递减时误用 d，如 20、17、14……中 d = −3。

12. Summary and Key Formulae | 总结与核心公式

Key formulae for arithmetic series:

等差数列的核心公式：

uₙ = a + (n − 1)d

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

Always write down the values of a,

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

更多咨询请联系16621398022（同微信）

Arithmetic Series | 等差数列求和