Sigma Notation | 求和符号

📚 Sigma Notation | 求和符号

Sigma notation is a concise way to write the sum of a sequence. The Greek capital letter Σ (sigma) tells you to add up all terms of the form uᵣ from the lower limit r = a to the upper limit r = b. It appears throughout Edexcel A-Level Mathematics, especially in sequences, series, binomial expansions and further pure topics.

求和符号是一种简洁表示数列求和的方法。希腊大写字母 Σ（西格玛）表示将 uᵣ 从下限 r = a 到上限 r = b 的所有项相加。它在 Edexcel A-Level 数学中广泛出现，尤其在数列、级数、二项式展开和进阶纯数学专题中。

1. Definition and Notation | 定义与符号

The expression ∑ᵣ₌₁ⁿ uᵣ means u₁ + u₂ + u₃ + … + uₙ. The variable r is called the index of summation, 1 is the lower limit, n is the upper limit, and uᵣ is the general term.

表达式 ∑ᵣ₌₁ⁿ uᵣ 表示 u₁ + u₂ + u₃ + … + uₙ。变量 r 称为求和指标，1 是下限，n 是上限，uᵣ 是通项。

∑ᵣ₌₁⁵ r = 1 + 2 + 3 + 4 + 5 = 15
∑ᵣ₌₁⁴ r² = 1² + 2² + 3² + 4² = 1 + 4 + 9 + 16 = 30
∑ᵣ₌₃⁶ (2r − 1) = (2×3 − 1) + (2×4 − 1) + (2×5 − 1) + (2×6 − 1) = 5 + 7 + 9 + 11 = 32

∑ᵣ₌₁⁵ r = 1 + 2 + 3 + 4 + 5 = 15

∑ᵣ₌₁⁴ r² = 1² + 2² + 3² + 4² = 30

∑ᵣ₌₃⁶ (2r − 1) = 5 + 7 + 9 + 11 = 32

2. Expanding a Sigma Expression | 展开求和式

To expand a sigma expression, substitute each integer value of r from the lower limit to the upper limit into the general term, then connect the results with plus signs. This is often the first step before evaluating a sum.

要展开求和式，将 r 的每一个整数值从下限到上限代入通项，然后用加号连接结果。这通常是求值前的第一步。

∑ₖ₌₁⁴ (3k + 1) = 4 + 7 + 10 + 13
∑ᵢ₌₀³ (−1)ⁱ = 1 − 1 + 1 − 1
∑ₙ₌₂⁵ n(n − 1) = 2×1 + 3×2 + 4×3 + 5×4 = 2 + 6 + 12 + 20

∑ₖ₌₁⁴ (3k + 1) = 4 + 7 + 10 + 13 = 34

∑ᵢ₌₀³ (−1)ⁱ = 1 − 1 + 1 − 1 = 0

∑ₙ₌₂⁵ n(n − 1) = 2 + 6 + 12 + 20 = 40

3. Linearity Properties | 线性性质

Sigma notation has two key linearity rules. A constant factor can be taken outside the sum, and the sum of two terms can be split into separate sums. These rules allow complicated expressions to be simplified before using standard results.

求和符号有两个关键的线性规则。常数因子可以提到求和号外，两项之和可以拆分为多个求和式。这些规则允许在使用标准结果之前简化复杂表达式。

∑ₖ₌₁ⁿ c aₖ = c ∑ₖ₌₁ⁿ aₖ

∑ₖ₌₁ⁿ (aₖ + bₖ) = ∑ₖ₌₁ⁿ aₖ + ∑ₖ₌₁ⁿ bₖ

∑ᵣ₌₁ⁿ 5r = 5∑ᵣ₌₁ⁿ r
∑ᵣ₌₁ⁿ (r² + 3r) = ∑ᵣ₌₁ⁿ r² + 3∑ᵣ₌₁ⁿ r
∑ᵣ₌₁ⁿ (4r − 7) = 4∑ᵣ₌₁ⁿ r − ∑ᵣ₌₁ⁿ 7 = 4∑ᵣ₌₁ⁿ r − 7n

4. Standard Summation Results | 标准求和公式

Edexcel A-Level expects you to know and use three standard results for sums of positive integers, squares and cubes. They are given in the formulae booklet and are essential for evaluating polynomial sums.

Edexcel A-Level 要求你掌握并使用三个正整数、平方数和立方数和的标准公式。这些公式在公式手册中给出，是计算多项式求和的基础。

∑ᵣ₌₁ⁿ r = n(n + 1) / 2

∑ᵣ₌₁ⁿ r² = n(n + 1)(2n + 1) / 6

∑ᵣ₌₁ⁿ r³ = [n(n + 1) / 2]²

∑ᵣ₌₁¹⁰ r = 10×11 / 2 = 55
∑ᵣ₌₁¹⁰ r² = 10×11×21 / 6 = 385
∑ᵣ₌₁¹⁰ r³ = (10×11 / 2)² = 55² = 3025

5. Using the Standard Results with Constants | 带常数的标准公式运用

When the general term contains a constant multiple of r, r² or r³, apply the linearity rule first. Then substitute one of the standard results. Always be careful with constants that do not depend on r.

当通项含有 r、r² 或 r³ 的常数倍时，首先使用线性规则。然后代入标准公式之一。对于不依赖 r 的常数项要特别小心。

∑ᵣ₌₁ⁿ 3r = 3 × n(n + 1) / 2
∑ᵣ₌₁ⁿ 2r² = 2 × n(n + 1)(2n + 1) / 6 = n(n + 1)(2n + 1) / 3
∑ᵣ₌₁ⁿ (6r + 4) = 6 × n(n + 1) / 2 + 4n = 3n(n + 1) + 4n = 3n² + 7n

∑ᵣ₌₁ⁿ 3r = 3n(n + 1) / 2

∑ᵣ₌₁ⁿ (6r + 4) = 3n(n + 1) + 4n = 3n² + 7n

6. Splitting Combined Sums | 拆分组合求和

A common exam question asks you to evaluate ∑(ar² + br + c). Split the expression into three separate sums, factor out constants, then apply the standard results for r², r and the constant term. This method avoids expanding the whole sum term by term.

常见的考试题要求计算 ∑(ar² + br + c)。将表达式拆分为三个独立的求和式，提出常数，然后应用 r²、r 和常数项的标准公式。这种方法避免逐项展开整个求和。

∑ᵣ₌₁ⁿ (ar² + br + c) = a∑ᵣ₌₁ⁿ r² + b∑ᵣ₌₁ⁿ r + cn

∑ᵣ₌₁ⁿ (3r² + 2r + 1) = 3 × n(n+1)(2n+1)/6 + 2 × n(n+1)/2 + n
= n(n+1)(2n+1)/2 + n(n+1) + n
= n[(n+1)(2n+1)/2 + (n+1) + 1]

Always simplify your final answer as a single fraction where possible. If n = 10, the sum is 3×385 + 2×55 + 10 = 1155 + 110 + 10 = 1275.

尽可能将最终答案化简为单个分式。如果 n = 10，和为 3×385 + 2×55 + 10 = 1155 + 110 + 10 = 1275。

7. Changing the Lower Limit | 改变下限

Sometimes the lower limit is not 1, or the standard results are easiest to use from r = 1. You can rewrite a sum from r = k to n as the sum from 1 to n minus the sum from 1 to k − 1.

有时下限不是 1，或者标准公式从 r = 1 开始使用最方便。你可以将从 r = k 到 n 的和改写为从 1 到 n 的和减去从 1 到 k − 1 的和。

∑ᵣ₌ₖⁿ f(r) = ∑ᵣ₌₁ⁿ f(r) − ∑ᵣ₌₁ᵏ⁻¹ f(r)

∑ᵣ₌₆²⁰ r = ∑ᵣ₌₁²⁰ r − ∑ᵣ₌₁⁵ r
= 20×21/2 − 5×6/2 = 210 − 15 = 195
∑ᵣ₌₄¹² r² = ∑ᵣ₌₁¹² r² − ∑ᵣ₌₁³ r²
= 12×13×25/6 − 3×4×7/6 = 650 − 14 = 636

8. Sigma Notation for Arithmetic Series | 等差数列的求和符号

An arithmetic series has the general term uᵣ = a + (r − 1)d, where a is the first term and d is the common difference. Sigma notation can be used to write the sum of the first n terms compactly, and the standard arithmetic series formula gives the value.

等差数列的通项为 uᵣ = a + (r − 1)d，其中 a 是首项，d 是公差。求和符号可以简洁地表示前 n 项和，标准等差级数公式给出其值。

Sₙ = ∑ᵣ₌₁ⁿ [a + (r − 1)d] = n/2 [2a + (n − 1)d]

∑ᵣ₌₁²⁰ (3 + 5(r − 1)) = 20/2 [2×3 + 19×5] = 10 × (6 + 95) = 1010
∑ᵣ₌₁¹⁵ (2r + 1) = 2∑ᵣ₌₁¹⁵ r + 15 = 2×15×16/2 + 15 = 240 + 15 = 255

9. Sigma Notation for Geometric Series | 等比数列的求和符号

A geometric series has the general term uᵣ = arʳ⁻¹, where a is the first term and r is the common ratio. The sum to n terms can be written with sigma notation and evaluated using the geometric series formula.

等比数列的通项为 uᵣ = arʳ⁻¹，其中 a 是首项，r 是公比。前 n 项和可以用求和符号表示，并使用等比级数公式计算。

Sₙ = ∑ₖ₌₁ⁿ arᵏ⁻¹ = a(1 − rⁿ) / (1 − r), r ≠ 1

∑ₖ₌₁¹⁰ 3×2ᵏ⁻¹ = 3(1 − 2¹⁰)/(1 − 2) = 3(2¹⁰ − 1) = 3×1023 = 3069
∑ₖ₌₀⁹ 5×(1/2)ᵏ = 5 × 10 terms from k=0 to 9, but first term a = 5 and ratio r = 1/2.
Here the sum from k=0 to 9 has 10 terms: 5 + 5/2 + … = 5(1 − (1/2)¹⁰)/(1 − 1/2) = 10(1 − 1/1024) = 10230/1024.

10. Method of Differences | 差分法

Sigma notation is especially useful in the method of differences. If the general term can be written as f(r) − f(r + 1), then most terms cancel when the sum is expanded. This technique appears frequently in Edexcel A-Level exam questions.

求和符号在差分法中尤其有用。如果通项可以写成 f(r) − f(r + 1)，那么展开后大多数项会相互抵消。这一技巧在 Edexcel A-Level 考试题中频繁出现。

∑ᵣ₌₁ⁿ [1/r − 1/(r + 1)] = (1/1 − 1/2) + (1/2 − 1/3) + … + (1/n − 1/(n+1))
= 1 − 1/(n + 1) = n/(n + 1)
∑ᵣ₌₁ⁿ [1/(r+1) − 1/(r+2)] = 1/2 − 1/(n+2)

To use this method, express the given algebraic fraction as partial fractions first. Then write out the first few and last few terms to identify the cancellation pattern.

使用这种方法时，首先将给定的代数分式表示为部分分式。然后写出前几项和后几项，以识别抵消模式。

11. Common Mistakes and Exam Tips | 常见错误与考试提示

Students often forget that ∑ₖ₌₁ⁿ c = cn, not c, because there are n identical constant terms. Another common error is applying the standard square or cube formula to sums like ∑(2r)² without expanding the bracket first.

学生经常忘记 ∑ₖ₌₁ⁿ c = cn，而不是 c，因为有 n 个相同的常数项。另一个常见错误是在没有先展开括号的情况下，将平方或立方公式用于 ∑(2r)² 这样的求和。

∑ᵣ₌₁ⁿ 5 = 5n, not 5
∑ᵣ₌₁ⁿ (2r)² = ∑ᵣ₌₁ⁿ 4r² = 4∑ᵣ₌₁ⁿ r², not 2∑ᵣ₌₁ⁿ r²
∑ᵣ₌₁ⁿ (r + 1)² = ∑ᵣ₌₁ⁿ (r² + 2r + 1), then split

Always check the lower limit carefully. If the lower limit is zero or another value, adjust the standard results accordingly. Show clear substitution lines in exam answers to gain method marks.

始终仔细检查下限。如果下限是零或其他值，请相应调整标准公式。在考试答案中展示清晰的代入步骤，以获得方法分。

12. Worked Exam-Style Example | 考试型例题解析

Question: Evaluate ∑ᵣ₌₁ⁿ (2r + 3)², giving your answer as a fully simplified polynomial in n.

问题：计算 ∑ᵣ₌₁ⁿ (2r + 3)²，并将答案表示为关于 n 的完全化简多项式。

First expand the bracket: (2r + 3)² = 4r² + 12r + 9. Then split the sum and use standard results.

首先展开括号：(2r + 3)² = 4r² + 12r + 9。然后拆分求和并使用标准公式。

∑ᵣ₌₁ⁿ (4r² + 12r + 9) = 4∑ᵣ₌₁ⁿ r² + 12∑ᵣ₌₁ⁿ r + 9n

Substitute the standard results: ∑ r² = n(n+1)(2n+1)/6 and ∑ r = n(n+1)/2.

代入标准公式：∑ r² = n(n+1)(2n+1)/6，∑ r = n(n+1)/2。

= 4 × n(n+1)(2n+1)/6 + 12 × n(n+1)/2 + 9n

= 2n(n+1)(2n+1)/3 + 6n(n+1) + 9n

Write all terms over a common denominator of 3, then expand and simplify.

将所有项写成公分母 3，然后展开并化简。

= [2n(n+1)(2n+1) + 18n(n+1) + 27n] / 3

= n[2(2n² + 3n + 1) + 18n + 18 + 27] / 3

= n(4n² + 6n + 2 + 18n + 45) / 3 = n(4n² + 24n + 47) / 3

Therefore, ∑ᵣ₌₁ⁿ (2r + 3)² = n(4n² + 24n + 47) / 3. This is the fully simplified polynomial form.

因此，∑ᵣ₌₁ⁿ (2r + 3)² = n(4n² + 24n + 47) / 3。这是完全化简的多项式形式。

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Sigma Notation | 求和符号