📚 Arithmetic Series: Sum Formula and Applications | 等差级数求和公式及应用
An arithmetic sequence is a sequence of numbers in which the difference between consecutive terms is constant. An arithmetic series is the sum of the terms of an arithmetic sequence. In this article, we will derive the sum formula, explore its alternative forms, and apply it to typical A-Level problems.
等差数列是指从第二项起，每一项与它的前一项的差等于同一个常数的数列。等差级数就是这个数列各项之和。本文将推导等差级数求和公式，介绍其等价形式，并应用于典型的 A-Level 题目中。
1. Arithmetic Sequences and Series | 等差数列与级数
An arithmetic sequence is often written as a, a+d, a+2d, … where a is the first term and d is the common difference. For example, 3, 7, 11, 15, … is an arithmetic sequence with a = 3 and d = 4.
等差数列通常写作 a, a+d, a+2d, …，其中 a 为首项，d 为公差。例如 3, 7, 11, 15, … 就是一个首项 a = 3，公差 d = 4 的等差数列。
The corresponding arithmetic series is the sum a + (a+d) + (a+2d) + … + [a+(n-1)d]. We denote this sum by Sₙ.
相应的等差级数为 a + (a+d) + (a+2d) + … + [a+(n-1)d]，我们用 Sₙ 表示这个和。
2. The nth Term Formula | 通项公式
Before studying the sum, we need the formula for the nth term of an arithmetic sequence:
在研究求和之前，我们需要等差数列的第 n 项公式：
uₙ = a + (n − 1)d
Here uₙ is the nth term, a is the first term, n is the number of terms, and d is the common difference.
这里 uₙ 是第 n 项，a 是首项，n 是项数，d 是公差。
For the sequence 5, 9, 13, 17, …, the 10th term is u₁₀ = 5 + (10−1)×4 = 41.
对于数列 5, 9, 13, 17, …，第 10 项为 u₁₀ = 5 + (10−1)×4 = 41。
3. Deriving the Sum Formula | 求和公式的推导
To find the sum of the first n terms, Sₙ, write the series in two ways:
为了求前 n 项和 Sₙ，我们把级数正写和倒写各列一遍：
Sₙ = a + (a+d) + … + [a+(n−1)d]
Sₙ = [a+(n−1)d] + [a+(n−2)d] + … + a
Adding these two expressions vertically gives 2Sₙ = n[2a + (n−1)d], because each of the n columns sums to 2a + (n−1)d.
将这两个式子逐项相加，得到 2Sₙ = n[2a + (n−1)d]，因为每一列之和都是 2a + (n−1)d，共有 n 列。
Dividing by 2 gives the standard sum formula:
两边除以 2，就得到标准求和公式：
Sₙ = n/2 [2a + (n − 1)d]
4. Alternative Formula Using the Last Term | 用末项表示的公式
If the last term l = uₙ = a + (n−1)d is known, the sum can be written more compactly:
如果已知末项 l = uₙ = a + (n−1)d，求和公式可以写成更紧凑的形式：
Sₙ = n/2 (a + l)
This formula is especially useful when the last term is given directly rather than the common difference.
当题目直接给出末项而不是公差时，这个公式特别方便。
For example, the sum of the series 2 + 5 + 8 + 11 + 14 can be found with a=2, l=14, n=5:
例如，求级数 2 + 5 + 8 + 11 + 14 的和，可用 a=2, l=14, n=5：
S₅ = 5/2 × (2 + 14) = 40
5. Finding the Number of Terms | 求项数
Given the first term, common difference and the last term, we can find n using the nth-term formula.
已知首项、公差和末项时，我们可以利用通项公式求出项数 n。
For example, how many terms are there in 7, 10, 13, …, 61?
例如，数列 7, 10, 13, …, 61 共有多少项？
61 = 7 + (n−1)×3 ⇒ n = 19
Then the sum is S₁₉ = 19/2 × (7 + 61) = 646.
因此和为 S₁₉ = 19/2 × (7 + 61) = 646。
6. Finding the First Term or Common Difference | 求首项或公差
The sum formula can also be rearranged to find an unknown a or d when the sum is given.
当给出和时，也可以由求和公式反求未知的首项 a 或公差 d。
For instance, the sum of the first 20 terms of an arithmetic series is 650 and the first term is 5. Find d.
例如，某等差级数的前 20 项和为 650，首项为 5，求公差 d。
650 = 20/2 [2×5 + (20−1)d]
650 = 10(10 + 19d) ⇒ d = 55/19
Notice that d need not be an integer; it is perfectly valid to have a fractional common difference.
注意公差不一定是整数，分数公差完全合理。
7. Real-World Applications | 实际应用
Arithmetic series appear in many practical contexts, such as rows of seats, stacks of objects, depreciation, and salary increases.
等差级数在许多实际问题中都有应用，例如座位排数、物体堆叠、折旧计算和工资增长等。
-
Seating in a theatre: each row has 2 more seats than the previous row, so total capacity forms an arithmetic series.
剧场座位：每一排比前一排多 2
Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导