📚 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。
Sigma notation provides a concise way to represent such sums. The capital Greek letter Σ means ‘sum’, and the expression ∑r=1n ar tells us to add a1 + a2 + … + an. 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 ar 表示将 a1 + a2 + … + an 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。
Sn = n/2 [2a + (n − 1)d]
This formula can also be written as Sn = n/2 (a + l), where l is the last term. The 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 Sn = 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。
For example, 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 arr−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 arr−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。
Sn = a(1 − rn) / (1 − r)
To prove it, subtract rSn from Sn; all intermediate terms cancel, leaving Sn(1 − r) = a(1 − rn). Then divide by (1 − r).
证明时，用 Sn 减去 rSn，中间各项全部抵消，剩下 Sn(1 − r) = a(1 − rn)，再除以 (1 − r) 即可。
For 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。
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
The third result is the 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
To sum a polynomial 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 (r² − 2r + 1) = ∑r² − 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
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