Sum of an Arithmetic Sequence | 等差数列的和

📚 Sum of an Arithmetic Sequence | 等差数列的和

An arithmetic sequence is a list of numbers in which each term after the first is obtained by adding a fixed number. This fixed number is called the common difference. When you add the terms of an arithmetic sequence, you form an arithmetic series. In the Edexcel IGCSE Mathematics course, you need to be able to find the sum of the first n terms quickly and accurately, without adding every single term by hand.

等差数列是指从第二项起，每一项与它的前一项的差都等于同一个常数的数列，这个常数叫做公差。把等差数列的项依次相加，就得到一个等差级数。在 Edexcel IGCSE 数学考试中，你需要能够快速、准确地求出前 n 项的和，而不是一个一个地把项手动相加。

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1. Recognising an Arithmetic Sequence | 认识等差数列

Consider the sequence 3, 7, 11, 15, 19, … The first term is a = 3 and the common difference is d = 4, because 7 − 3 = 4, 11 − 7 = 4, and so on. The difference between consecutive terms must be constant for the sequence to be arithmetic.

例如数列 3, 7, 11, 15, 19, … 中，首项 a = 3，公差 d = 4，因为 7 − 3 = 4，11 − 7 = 4，依此类推。只有在任意相邻两项的差都相等时，这个数列才是等差数列。

The nth term of an arithmetic sequence is given by the formula

aₙ = a + (n−1)d

where a is the first term and d is the common difference. For the sequence above, the 10th term is a₁₀ = 3 + 9 × 4 = 39.

其中 a 是首项，d 是公差。对于上面的数列，第 10 项是 a₁₀ = 3 + 9 × 4 = 39。

The common difference can also be negative. If d is negative, the sequence decreases. For example, 20, 17, 14, 11, … is arithmetic with a = 20 and d = −3.

公差也可以是负数。如果 d 是负数，数列就会递减。例如 20, 17, 14, 11, … 是等差数列，其中 a = 20，d = −3。

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2. The Sum Formulas | 求和公式的两种形式

Let Sₙ denote the sum of the first n terms of an arithmetic sequence. There are two equivalent formulas that you can use.

设 Sₙ 表示等差数列前 n 项的和。你需要掌握以下两个等价公式。

Sₙ = n/2 [2a + (n−1)d]

Sₙ = n/2 (a + l)

In the second formula, l is the last term of the sum. Since the last term is exactly the nth term, l = a + (n−1)d. The second formula is especially useful when you already know the first term and the last term.

在第二个公式中，l 是求和中的末项。因为末项正好是第 n 项，所以 l = a + (n−1)d。当你已经知道首项和末项时，第二个公式尤其方便。

Do not confuse the nth term aₙ with the sum Sₙ. The expression aₙ gives only one term, while Sₙ gives the total of all the first n terms.

不要把第 n 项 aₙ 与和 Sₙ 混淆。aₙ 表示单独的一项，而 Sₙ 表示前 n 项的总和。

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3. Deriving the Formula | 推导求和公式

To understand why Sₙ = n/2 (a + l), write the sum forwards and then backwards:

为了理解为什么 Sₙ = n/2 (a + l)，先把和由前向后写一遍，再由后向前写一遍：

S = a + (a+d) + (a+2d) + … + (l−d) + l

S = l + (l−d) + (l−2d) + … + (a+d) + a

Add the two equations column by column. The first pair gives a + l, the second pair also gives a + l, and every pair gives the same result a + l. There are n pairs, so 2S = n(a + l). Dividing by 2 gives the formula.

将这两个式子逐列相加。第一组得到 a + l，第二组仍然得到 a + l，每一组都得到相同的 a + l。一共有 n 组，因此 2S = n(a + l)。两边除以 2 就得到公式。

If you replace l with a + (n−1)d, you obtain the first formula:

如果把 l 替换为 a + (n−1)d，就得到第一个公式：

Sₙ = n/2 [2a + (n−1)d]

You are not required to reproduce this derivation in the exam, but understanding it helps you remember the formula and explains why the factor n/2 appears.

考试不要求你默写这个推导过程，但理解推导能够帮助你记住公式，并明白为什么会出现因子 n/2。

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4. Worked Example 1 | 例题 1

Find the sum of the first 20 terms of the arithmetic sequence 5, 8, 11, 14, …

求等差数列 5, 8, 11, 14, … 的前 20 项之和。

Here a = 5, d = 3 and n = 20. Substitute into the formula:

这里 a = 5，d = 3，n = 20。代入公式：

S₂₀ = 20/2 [2 × 5 + (20−1) × 3]

S₂₀ = 10 [10 + 57] = 670

So the sum of the first 20 terms is 670.

所以前 20 项的和是 670。

A useful check: when n is even, the average of each pair of terms from the two ends is (first term + last term)/2. The 20th term is 5 + 19 × 3 = 62, so the average of all terms is (5 + 62)/2 = 33.5. Multiply by 20, and you get 670.

一个有用的验算方法：当 n 是偶数时，首尾配对后的平均数是 (首项 + 末项)/2。第 20 项是 5 + 19 × 3 = 62，所以所有项的平均数是 (5 + 62)/2 = 33.5。再乘以 20，得到 670。

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5. Sum When the Last Term Is Known | 已知末项求和

Sometimes a question gives you the last term directly. For example, find the sum 2 + 6 + 10 + … + 50.

有时题目会直接给出末项。例如，求 2 + 6 + 10 + … + 50 的和。

The first term is a = 2 and the common difference is d = 4. The last term is l = 50. To use Sₙ = n/2(a + l), first find n.

首项 a = 2，公差 d = 4，末项 l = 50。要使用 Sₙ = n/2(a + l)，需要先求出 n。

50 = 2 + (n−1) × 4

48 = (n−1) × 4

n − 1 = 12, so n = 13

There are 13 terms in the sum. Then use the second sum formula:

这个和一共有 13 项。然后使用第二个求和公式：

S₁₃ = 13/2 (2 + 50) = 13 × 26 = 338

The sum is 338.

所求的和是 338。

When the last term is known, the formula Sₙ = n/2(a + l) saves you from expanding the longer expression inside the first formula.

当末项已知时，使用 Sₙ = n/2(a + l) 可以避免展开第一个公式中较长的表达式。

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6. Sum of the First n Positive Integers | 前 n 个正整数的和

A very common special case is the sum of the first n positive integers: 1 + 2 + 3 + … + n. This is arithmetic with a = 1, d = 1 and l = n.

一个非常常见的特例是前 n 个正整数的和：1 + 2 + 3 + … + n。这是一个等差数列，其中 a = 1，d = 1，l = n。

Sₙ = n/2 (1 + n) = n(n+1)/2

For example, the sum of the integers from 1 to 100 is

例如，从 1 到 100 的所有整数之和为

1 + 2 + … + 100 = 100 × 101 / 2 = 5050

This result is useful in many non-calculator questions and is worth memorising.

这个结论在许多不使用计算器的题目中非常有用，值得记住。

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7. Using Two Sums to Solve Simultaneous Equations | 利用两组和建立联立方程

An exam-style problem may give you two different sums and ask you to find the first term and the common difference. For example:

一种常见的考试题型是给出两组不同的和，要求你求首项和公差。例如：

The sum of the first 10 terms of an arithmetic sequence is 100. The sum of the first 20 terms is 400. Find a and d.

一个等差数列的前 10 项和是 100，前 20 项和是 400。求 a 和 d。

Use Sₙ = n/2 [2a + (n−1)d]. For n = 10:

使用 Sₙ = n/2 [2a + (n−1)d]。当 n = 10 时：

10/2 [2a + 9d] = 100

2a + 9d = 20

For n = 20:

当 n = 20 时：

20/2 [2a + 19d] = 400

2a + 19d = 40

Subtract the first equation from the second:

用第二个方程减去第一个方程：

10d = 20, so d = 2

Then substitute d = 2 into 2a + 9d = 20:

再把 d = 2 代入 2a + 9d = 20：

2a + 18 = 20, so a = 1

Therefore

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