Summation of Finite Series | 有限级数的求和

📚 Summation of Finite Series | 有限级数的求和

A sequence is a list of numbers written in a definite order. When we add the terms of a sequence, we obtain a series; if the number of terms is fixed, we call it a finite series. For example, 1 + 2 + 3 + ... + 100 is a finite series.

数列是按照确定顺序排列的一列数。当我们把数列中的项依次相加时，就得到级数；如果项数是固定的，就称为有限级数。例如，1 + 2 + 3 + ... + 100 就是一个有限级数。

a notation provides a concise way to represent such sums. The capital Greek letter Σ means 'sum', and the expression ∑_r=1ⁿ a_r tells us to add a₁ + a₂ + ... + a_n. The variable r is called the index of summation, and it changes from the lower limit 1 to the upper limit n.

西格玛记号为这类求和提供了一种简洁的表达方式。大写希腊字母 Σ 表示“求和”，表达式 ∑_r=1ⁿ a_r 表示将 a₁ + a₂ + ... + a_n 相加。变量 r 称为求和指标，它会从下限 1 变化到上限 n。

To evaluate such a series, we can either expand and add directly, or use a known formula. In this article we review the standard formulas and methods that AQA A-Level students are expected to apply for finite sums, including arithmetic and geometric series, standard power sums, and telescoping methods.

要计算这样的级数，我们可以直接展开相加，也可以使用已知公式。本文将复习 AQA A-Level 考试预期学生掌握的有限求和标准公式和方法，包括等差与等比级数、标准幂和以及相消求和法。

1. Arithmetic Series | 等差级数

An arithmetic sequence has a common difference d between consecutive terms. The r-th term is a + (r − 1)d, where a is the first term. The sum of the first n terms is given by a simple formula.

等差数列相邻两项之差为公差 d。若首项为 a，则第 r 项为 a + (r − 1)d。前 n 项和由一条简单公式给出。

S_n = n/2 [2a + (n − 1)d]

This formula can also be written as S_n he proof is simple: write the sum forward, then backward, add the two lines. Each pair sums to a + l, and there are n pairs.

这个公式也可写作 S_n = n/2 (a + l)，其中 l 是末项。证明很简单：把和式正写一遍、倒写一遍再相加，每一对都等于 a + l，共有 n 对。

the sum of the first 50 even positive integers is S = 50/2 × (2 + 100) = 25 × 102 = 2550.

例如，前 50 个正偶数的和为 S = 50/2 × (2 + 100) = 25 × 102 = 2550。

2. Geometric Series | 等比级数

A geometric sequence has a common ratio r between consecutive terms. The r-th term is ar^r−1. For a finite geometric series with n terms, the sum is given by the following formula, provided r ≠ 1.

等比数列相邻两项之比为公比 r。第 r 项为 ar^r−1。对于有 n 项的有限等比级数，其和由以下公式给出，要求 r ≠ 1。

S_n = a(1 − rⁿ) / (1 − r)

To prove it, subtract rS_n from S_n; all intermediate terms cancel, leaving S_n(1 − r) = a(1 − rⁿ). Then divide by (1 − r).

证明时，用 S_n 减去 rS_n，中间各项全部抵消，剩下 S_n(1 − r) = a(1 − rⁿ)，再除以 (1 − r) 即可。

or example, the sum of the first 5 terms of 3, 6, 12, ... is S = 3(1 − 2⁵) / (1 − 2) = 93.

例如，数列 3, 6, 12, ... 的前 5 项和为 S = 3(1 − 2⁵) / (1 − 2) = 93。

广告 / Sponsored

3. Standard Results for ∑r, ∑r² and ∑r³ | r、r²、r³ 求和的标准结果

For powers of the first n positive integers, the following standard results are essential. These are given in the AQA formula booklet and can be used without proof in examinations.

对于前 n 个正整数各次幂的求和，以下标准结果至关重要。这些公式在 AQA 公式书中提供，考试时可直接使用而无需证明。

• ∑_r=1ⁿ r = n(n + 1) / 2
• ∑_r=1ⁿ r² = n(n + 1)(2n + 1) / 6
• ∑_r=1ⁿ r³ = [n(n + 1) / 2]²

he square of the first result. It is sometimes stated as (∑r)² = ∑r³ for sums from 1 to n.

第三个结果是第一个结果的平方。有时也表述为：从 1 到 n 的求和满足 (∑r)² = ∑r³。

4. Using Standard Results | 使用标准结果求和

omial in r, split the sum term by term using linearity. For example, ∑(2r + 3) = 2∑r + 3n.

要计算 r 的多项式求和，可利用线性性质逐项拆分。例如，∑(2r + 3) = 2∑r + 3n。

Worked example: find ∑_r=1ⁿ (2r + 3). We have 2 × n(n + 1)/2 + 3n = n(n + 1) + 3n = n² + 4n = n(n + 4).

例：求 ∑_r=1ⁿ (2r + 3)。可得 2 × n(n + 1)/2 + 3n = n(n + 1) + 3n = n² + 4n = n(n + 4)。

Another example: ∑_r=1ⁿ 2∑r + n. Substituting the standard results gives n(2n − 1)(n − 1)/6 after simplification.

另一个例子：∑_r=1ⁿ (r² − 2r + 1) = ∑r² − 2∑r + n。代入标准结果并化简后可得 n(2n − 1)(n − 1)/6。

5. Method of Differences | 差分法（裂项相消）

The method of differences uses

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

更多咨询请联系16621398022（同微信）

备考资讯每日更新，关注不错过

考纲变化 · 科目分析 · 学习方法

在微信里长按二维码识别关注
或截图后用「扫一扫 → 相册」选择本图

公众号：tutorhao