Summation of Finite Series | 有限级数求和

📚 Summation of Finite Series | 有限级数求和

In A-Level mathematics, the summation of finite series is a core skill. A finite series is the sum of a finite number of terms of a sequence. Mastering summation formulae and techniques allows you to evaluate expressions quickly without adding term by term, which is essential for exams and further calculus.

在 A-Level 数学中，有限级数求和是一项核心技能。有限级数是指一个数列中有限项的和。掌握求和公式与技巧，可以让你不必逐项相加而快速求值，这在考试和后续微积分学习中至关重要。

---

1. Summation Notation | 求和符号

The Greek capital letter sigma (Σ) is used to denote summation. For a sequence with general term uᵣ, the sum from r = 1 to r = n is written as:

希腊大写字母 Σ 用于表示求和。对于通项为 uᵣ 的数列，从 r = 1 到 r = n 的和写作：

Σ₍ᵣ₌₁₎ⁿ uᵣ = u₁ + u₂ + u₃ + … + uₙ

The variable r is called the index of summation; it is a dummy variable and can be replaced by any other letter, such as k or i, without changing the sum.

变量 r 称为求和指标；它是一个“哑变量”，可以用任何其他字母（如 k 或 i）替换，而不改变总和的值。

• Linear property: Σ(aᵣ ± bᵣ) = Σaᵣ ± Σbᵣ.

  线性性质：Σ(aᵣ ± bᵣ) = Σaᵣ ± Σbᵣ。

• Constant multiple: Σ(caᵣ) = cΣaᵣ, where c is a constant.

  常数倍性质：Σ(caᵣ) = cΣaᵣ，其中 c 为常数。

• Constant term: Σ₍ᵣ₌₁₎ⁿ c = nc.

  常数项：Σ₍ᵣ₌₁₎ⁿ c = nc。

---

2. Arithmetic Series | 等差级数

An arithmetic sequence has a common difference d between consecutive terms. The first term is a, and the n-th term is a + (n−1)d. The sum of the first n terms, Sₙ, can be expressed in two equivalent ways:

等差数列的相邻两项之差为常数 d。设首项为 a，第 n 项为 a + (n−1)d。前 n 项和 Sₙ 有两种等价形式：

Sₙ = ⁿ⁄₂ [2a + (n−1)d] = ⁿ⁄₂ (first term + last term)

The second form is especially useful when the last term is known. For example, the sum of the first 50 positive integers is 50⁄2(1 + 50) = 1275.

第二种形式在已知末项时尤为方便。例如，前 50 个正整数的和为 50⁄2(1 + 50) = 1275。

• Proof strategy: Write Sₙ forwards and backwards, then add corresponding terms.

  证明策略：将 Sₙ 正写和倒写，然后对应项相加。

• Exam tip: Always check whether the series is arithmetic by verifying a constant difference.

  考试提示：通过检验相邻项之差是否恒定来判断是否为等差级数。

---

3. Geometric Series | 等比级数

A geometric sequence has a common ratio r between consecutive terms. With first term a, the sum of the first n terms is:

等比数列的相邻两项之比为常数 r。设首项为 a，则前 n 项和为：

Sₙ = a(1 − rⁿ) ⁄ (1 − r) (r ≠ 1)

This formula is derived by multiplying Sₙ by r and subtracting it from Sₙ. If r = 1, then Sₙ = na.

该公式通过将 Sₙ 乘以 r 后与原式相减得到。若 r = 1，则 Sₙ = na。

• Valid for all real n as long as r ≠ 1; no restriction on |r| for finite series.

  对于有限级数，只要 r ≠ 1，该公式对所有实数 r 都适用；不要求 |r| < 1。

• Common mistake: sign errors when r is negative. For example, 1 − (−2)ⁿ must be handled carefully.

  常见错误：当 r 为负数时符号错误。例如 1 − (−2)ⁿ 需要仔细处理。

---

4. Standard Power Sums | 常用幂和公式

Three standard results are essential for summing series involving powers of integers. They are often given in formula booklets, but you must know how to apply them:

三个标准结果对于求整数幂的级数和至关重要。它们通常出现在公式册中，但你必须知道如何应用：

Σr = ⁿ(n+1)⁄₂
Σr² = ⁿ(n+1)(2n+1)⁄₆
Σr³ = [ⁿ(n+1)⁄₂]²

Here Σr means Σ₍ᵣ₌₁₎ⁿ r, and similarly for higher powers. These formulas are linear in the sense that sums of polynomial expressions can be decomposed into these basic sums.

这里的 Σr 表示 Σ₍ᵣ₌₁₎ⁿ r，高次幂类似。由于求和是线性的，任何关于 r 的多项式表达式都可分解为这些基本和来计算。

• Example: Σ(2r − 1)² = 4Σr² − 4Σr + Σ1.

  例：Σ(2r − 1)² = 4Σr² − 4Σr + Σ1。

• Always expand carefully and combine like terms before substituting limits.

  务必先仔细展开并合并同类项，再代入求和限。

---

5. Splitting and Grouping | 拆项分组法

Many finite series can be evaluated by splitting the general term into a sum of simpler parts, each of which fits a standard formula. For example, a quadratic expression like ar² + br + c can be summed term by term.

许多有限级数可以通过将通项拆成若干个更简单的部分来求和，每一部分都可用标准公式。例如，形如 ar² + br + c 的二次表达式可以逐项求和。

Σ(3r² − 2r + 1) = 3Σr² − 2Σr + n

After substituting the standard formulae, simplify the resulting polynomial in n. This method is reliable and avoids the need to identify special patterns.

代入标准公式后，将得到的关于 n 的多项式化简。这种方法可靠，且无需识别特殊规律。

• Group terms that share the same power of r.

  将 r 的相同次幂的项合并。

• Check your answer with a small value of n, such as n = 1 or 2.

  用 n 的小值（如 n = 1 或 2）检验答案。

---

6. Method of Differences | 差分法

The method of differences is powerful for series whose terms can be written as the difference of two consecutive terms of another sequence. Suppose uᵣ = f(r) − f(r−1). Then the sum telescopes:

差分法对于通项可以写成某个数列连续两项之差的级数非常有效。设 uᵣ = f(r) − f(r−1)，则求和会相互抵消：

Σ₍ᵣ₌₁₎ⁿ uᵣ = f(n) − f(0)

This is also called the telescoping method. A classic example is summing 1/[r(r+1)] because 1/[r(r+1)] = 1/r − 1/(r+1).

这也称为“裂项相消法”。经典例子是求 1/[r(r+1)] 的和，因为 1/[r(r+1)] = 1/r − 1/(r+1)。

• Write out the first few terms and the last few terms explicitly to see which terms cancel.

  写出前几项和后几项，观察哪些项相消。

• Be cautious with index shifts: f(r) − f(r−1) vs f(r+1) − f(r).

  注意指标平移：f(r) − f(r−1) 与 f(r+1) − f(r) 的区别。

---

7. Summing Rational Functions | 有理函数求和

Sometimes the general term is a rational function of r. Partial fractions often convert it into a telescoping form. For instance, 1/[(2r−1)(2r+1)] can be split into ½[1/(2r−1) − 1/(2r+1)].

有时通项是 r 的有理函数。部分分式常常能将其转化为裂项相消的形式。例如，1/[(2r−1)(2r+1)] 可拆为 ½[1/(2r−1) − 1/(2r+1)]。

Σ₍ᵣ₌₁₎ⁿ 1/[(2r−1)(2r+1)] = ½[1 − 1/(2n+1)]

When applying partial fractions, always solve for the unknown constants by substituting convenient values of r or equating coefficients.

在使用部分分式时，通过代入 r 的方便值或比较系数来求解未知常数。

• If the denominator has repeated linear factors, the partial fraction form includes terms with increasing powers.

  若分母有重因式，部分分式形式需包含递增次幂的项。

• Check whether the remainder after cancellation is a constant plus a simple fraction.

  检查相消后的剩余部分是否为常数加简单分式。

---

8. Change of Limits and Index Shift | 求和限变换与指标平移

Sometimes a sum does not start at r = 1. You can shift the index or split the sum to use standard formulae. For example, Σ₍ᵣ₌₅₎ⁿ r = Σ₍ᵣ₌₁₎ⁿ r − Σ₍ᵣ₌₁₎₄ r.

有时求和并非从 r = 1 开始。你可以平移指标或将和拆开以使用标准公式。例如，Σ₍ᵣ₌₅₎ⁿ r = Σ₍ᵣ₌₁₎ⁿ r − Σ₍ᵣ₌₁₎₄ r。

Σ₍ᵣ₌ₘ₎ⁿ uᵣ = Σ₍ᵣ₌₁₎ⁿ uᵣ − Σ₍ᵣ₌₁₎ᵐ₋₁ uᵣ

Alternatively, substitute s = r − m + 1 to make the lower limit 1, then compute the general term in terms of s.

另一种方法是令 s = r − m + 1 将下限变为 1，然后用 s 表示通项。

• When shifting, do not forget to adjust the upper limit and the expression inside the term.

  平移时，不要忘记调整上限和通项中的表达式。

• This technique is often needed when summing from r = 0 or from r = 3, etc.

  当求和的起点为 r = 0 或 r = 3 等时，常常需要此技巧。

---

9. Common Pitfalls and Exam Tips | 常见失误与考试技巧

Students often lose marks in finite series questions due to small algebraic errors. Here are the most common pitfalls and how to avoid them.

学生在有限级数题目中常因细小的代数错误而失分。以下是最常见的失误及避免方法。

• Misreading Σr as Σ₁ᵣ or forgetting that Σr means the sum over r, not the term r.

  误读 Σr，或忘记 Σr 表示对 r 求和，而不是仅有 r 这一项。

• Using the infinite geometric series formula for a finite series; the finite formula requires the factor (1 − rⁿ).

  将无穷等比级数公式用于有限级数；有限公式必须包含 (1 − rⁿ) 因子。

• Arithmetic series: confusing n (number of terms) with the value of the last term.

  等差级数：混淆项数 n 与末项的值。

• Expanding (n+1)³ incorrectly; always write out multiplication carefully.

  错误展开 (n+1)³；务必仔细写出乘法过程。

• Not verifying the result with a numerical example for small n.

  没有用小 n 的数值例子验证结果。

---

10. Worked Example: Mixed Sum | 综合例题

Find Σ₍ᵣ₌₁₎ⁿ (r+1)(r+2) and simplify your answer.

求 Σ₍ᵣ₌₁₎ⁿ (r+1)(r+2) 并化简答案。

First expand the general term: (r+1)(r+2) = r² + 3r + 2. Therefore,

首先展开通项：(r+1)(r+2) = r² + 3r + 2。因此，

Σ = Σr² + 3Σr + 2Σ1

Substitute the standard formulae:

代入标准公式：

Σ = ⁿ(n+1)(2n+1)⁄₆ + 3·ⁿ(n+1)⁄₂ + 2n

Writing over a common denominator 6:

通分，以 6 为公分母：

Σ = [n(n+1)(2n+1) + 9n(n+1) + 12n] ⁄ 6 = [n(n+1)(2n+1+9) + 12n] ⁄ 6 = [n(n+1)(2n+10) + 12n] ⁄ 6

Simplify further: n(n+1)(2n+10) = 2n(n+1)(n+5). Hence

进一步化简：n(n+1)(2n+10) = 2n(n+1)(n+5)。因此

Σ = [2n(n+1)(n+5) + 12n] ⁄ 6 = ⁿ⁄₃[(n+1)(n+5) + 6] = ⁿ⁄₃(n² + 6n + 11)

Check with n = 1: left side (1+1)(1+2) = 6; right side 1⁄3(1+6+11) = 6. Correct.

用 n = 1 检验：左边 (1+1)(1+2) = 6；右边 1⁄3(1+6+11) = 6。正确。

---

11. Applications to Recurrence and Estimation | 递推与估计中的应用

Finite series appear in many contexts, such as compound interest calculations, area approximations, and recurrence relations. Understanding summation helps you compute total amounts over a fixed number of periods.

有限级数出现在许多情境中，例如复利计算、面积近似和递推关系。理解求和有助于计算固定期数内的总量。

For an arithmetic sequence, the sum gives the total of equally spaced values. For a geometric sequence, the sum gives the total of a quantity growing or decaying by a fixed factor.

等差级数的和给出等间隔值的总量；等比级数的和给出按固定因子增长或衰减的量在若干期内的总和。

• In exam questions, often you are given the sum and need to find n or r; use the appropriate formula and solve the resulting equation.

  考试中，常给出和的值要求 n 或 r；选用合适的公式并解方程。

• When solving for n in a geometric series, you may need logarithms after isolating (1 − rⁿ).

  在等比级数中解 n 时，隔离 (1 − rⁿ) 后可能需要使用对数。

---

12. Summary | 总结

Mastering finite series requires fluency with sigma notation, arithmetic and geometric formulas, standard power sums, and telescoping techniques. Always expand expressions fully, use linearity to split sums, and verify results with small n values.

掌握有限级数需要熟练使用求和符号、等差与等比公式、常用幂和公式以及裂项技巧。始终完整展开表达式，利用线性性质拆分求和，并用小的 n 值验证结果。

Remember the key formulas:

记住关键公式：

Sₙ(arithmetic) = ⁿ⁄₂[2a + (n−1)d]
Sₙ(geometric) = a(1 − rⁿ)⁄(1 − r)
Σr = ⁿ(n+1)⁄₂, Σr² = ⁿ(n+1)(2n+1)⁄₆, Σr³ = [ⁿ(n+1)⁄₂]²

With practice, these techniques become intuitive and save considerable time in the exam.

通过练习，这些技巧会变得直观，并在考试中节省大量时间。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）