Sigma Notation and Summation Representation | 西格玛记号与求和表示

📚 Sigma Notation and Summation Representation | 西格玛记号与求和表示

The sigma notation is a compact way to represent the sum of a sequence of terms. It is one of the most important tools in A-level mathematics, appearing in sequences, series, integration, and statistics.

西格玛记号是表示数列求和的简洁方式。它是 A-level 数学中最重要工具之一，出现在数列、级数、积分和统计等领域。

---

1. The Basic Form of Sigma Notation | 西格玛记号的基本形式

The Greek capital letter Σ (sigma) is used to denote summation. A general sum is written as:

希腊大写字母 Σ（西格玛）用于表示求和。一个一般的求和式写作：

Σ_i=mⁿ a_i = a_m + a_m+1 + a_m+2 + ... + a_n

Here, i is called the index of summation, m is the lower limit, and n is the upper limit. The expression a_i is the general term of the sequence being summed.

其中 i 称为求和指标，m 是下限，n 是上限。表达式 a_i 是被求和数列的通项。

For example, Σ_i=1⁴ i² means 1² + 2² + 3² + 4² = 1 + 4 + 9 + 16 = 30.

例如，Σ_i=1⁴ i² 表示 1² + 2² + 3² + 4² = 1 + 4 + 9 + 16 = 30。

---

2. The Index and the Limits | 求和指标与上下限

The index of summation is a dummy variable: it can be replaced by any other letter without changing the value of the sum. Thus Σ_i=1ⁿ a_i = Σ_k=1ⁿ a_k.

求和指标是一个哑变量：它可以用任何其他字母替代而不改变求和的值。因此 Σ_i=1ⁿ a_i = Σ_k=1ⁿ a_k。

The lower limit does not have to be 1. For instance, Σ_i=3⁷ (2i - 1) means the odd numbers from 5 to 13: 5 + 7 + 9 + 11 + 13 = 45.

下限不一定是 1。例如，Σ_i=3⁷ (2i - 1) 表示从 5 到 13 的奇数：5 + 7 + 9 + 11 + 13 = 45。

It is essential to read the limits carefully: the index starts at the lower limit and increases by 1 until it reaches the upper limit, inclusive.

必须仔细读取上下限：指标从下限开始，每次增加 1，直到达到上限（包含上限）。

---

3. Linear Properties of Sigma | 西格玛的线性性质

Sigma notation respects addition and scalar multiplication. If c is a constant, then:

西格玛记号遵循加法和标量乘法的线性性质。如果 c 是常数，则：

Σ_i=1ⁿ (c a_i + b_i) = c Σ_i=1ⁿ a_i + Σ_i=1ⁿ b_i

This means a constant factor can be pulled out of the sum, and a sum of two expressions can be split into two separate sums. These properties are widely used to evaluate complicated series.

这意味着常数因子可以提到求和号之外，并且两个表达式之和可以拆成两个独立的求和。这些性质在求解复杂级数时被广泛使用。

For example, Σ_i=1¹⁰ (3i + 2) = 3Σ_i=1¹⁰ i + Σ_i=1¹⁰ 2 = 3 × 55 + 20 = 185.

例如，Σ_i=1¹⁰ (3i + 2) = 3Σ_i=1¹⁰ i + Σ_i=1¹⁰ 2 = 3 × 55 + 20 = 185。

---

4. Summing a Constant Term | 常数项求和

When the general term is a constant c, the sum simply counts how many terms there are:

当通项是常数 c 时，求和结果就是项数乘以该常数：

Σ_i=1ⁿ c = c + c + ... + c = c n

Here there are exactly n terms. This is a special case of the linear property and is often used together with other formulas.

这里恰好有 n 项。这是线性性质的一个特例，常与其他公式联合使用。

For example, Σ_i=1²⁰ 7 = 7 × 20 = 140. Also Σ_i=5¹⁰ 3 = 3 × 6 = 18 because there are 6 integers from 5 to 10 inclusive.

例如，Σ_i=1²⁰ 7 = 7 × 20 = 140。同样，Σ_i=5¹⁰ 3 = 3 × 6 = 18，因为从 5 到 10 共有 6 个整数。

---

5. Standard Formulas for Powers of i | 幂的常用求和公式

The following standard results are essential for A-level mathematics:

以下标准结果对于 A-level 数学至关重要：

• Σ_i=1ⁿ 1 = n
• Σ_i=1ⁿ i = n(n+1)/2
• Σ_i=1ⁿ i² = n(n+1)(2n+1)/6
• Σ_i=1ⁿ i³ = [n(n+1)/2]²

These formulas allow us to evaluate polynomial-like sums without adding each term individually. The first formula is the sum of the first n positive integers, and the second is the sum of their squares. The third gives the sum of cubes, which is the square of the sum of the integers.

这些公式使我们能够在不逐项相加的情况下求多项式型求和。第一个是前 n 个正整数的和，第二个是它们平方的和，第三个是它们立方和，后者等于整数和的平方。

For example, Σ_i=1¹⁰⁰ i = 100 × 101 / 2 = 5050, and Σ_i=1⁵ i³ = 1³ + 2³ + 3³ + 4³ + 5³ = (5×6/2)² = 225.

例如，Σ_i=1¹⁰⁰ i = 100 × 101 / 2 = 5050，而 Σ_i=1⁵ i³ = 1³ + 2³ + 3³ + 4³ + 5³ = (5×6/2)² = 225。

---

6. Deriving the Square Sum Formula | 平方和公式的推导

Σi² is by using the identity (i+1)³ - i³ = 3i² + 3i + 1, and then summing from i = 1 to n.

推导 Σi² 的一种常用方法是利用恒等式 (i+1)³ - i³ = 3i² + 3i + 1，然后从 i=1 到 n 求和。

Σ_i=1ⁿ [(i+1)³ - i³] = (n+1)³ - 1

Expanding the right side gives:

展开右边得到：

3Σi² + 3Σi + Σ1 = (n+1)³ - 1

Using Σi = n(n+1)/2 and Σ1 = n, we solve for Σi²:

利用 Σi = n(n+1)/2 和 Σ1 = n，我们解出 Σi²：

Σi² = n(n+1)(2n+1)/6

This technique is an example of a telescoping sum, which is extremely useful in many contexts.

这种技巧是裂项求和（telescoping sum）的一个例子，在许多情境下非常有用。

---

广告 / Sponsored

7. Geometric Series in Sigma Form | 几何级数的西格玛形式

For a geometric sequence with first term a and common ratio r, the sum of the first n+1 terms can be written as:

对于首项为 a、公比为 r 的等比数列，前 n+1 项的和可以写成：

Σ_i=0ⁿ a rⁱ = a(1 - rⁿ⁺¹) / (1 - r), r ≠ 1

This formula is fundamental for both finite and infinite geometric series. If |r| < 1, the infinite sum is:

该公式对有限和无限等比级数都是基础性的。若 |r| < 1，无穷和为：

Σ_i=0^∞ a rⁱ = a / (1 - r)

For example, Σ_i=1¹⁰ 2 × 3^i-1 = 2(3¹⁰ - 1)/(3 - 1) = 59048.

例如，Σ_i=1¹⁰ 2 × 3^i-1 = 2(3¹⁰ - 1)/(3 - 1) = 59048。

---

8. Changing the Limits of a Sum | 改变求和上下限

that the lower limit becomes 0 or 1. A change of variable j = i - c rewrites the same sum.

有时需要通过平移指标使下限变为 0 或 1。变量替换 j = i - c 可以改写同一个求和。

Σ_i=mⁿ a_i = Σ_j=m-c^n-c a_j+c

For instance, Σ_i=2⁵ (i-1)² can be rewritten by letting j = i - 1, giving Σ_j=1⁴ j² = 1 + 4 + 9 + 16 = 30.

例如，Σ_i=2⁵ (i-1)² 可通过令 j = i - 1 改写为 Σ_j=1⁴ j² = 1 + 4 + 9 + 16 = 30。

Another common shift is from i = 0 to n to i = 1 to n+1: Σ_i=0ⁿ a_i = Σ_i=1ⁿ⁺¹ a_i-1.

另一个常见平移是从 i=0 到 n 变为 i=1 到 n+1：Σ_i=0ⁿ a_i = Σ_i=1ⁿ⁺¹ a_i-1。

Correctly shifting limits is vital when applying standard formulas that require the index to start at 1.

正确平移上下限在应用需要指标从 1 开始的标准公式时至关重要。

---

9. Splitting and Combining Sums | 拆项与合并求和

The linear property allows a complicated sum to be broken into simpler parts. For example, Σ_i=1ⁿ (i+1)(i+2) can be expanded as Σ_i=1ⁿ (i² + 3i + 2).

线性性质允许将复杂的求和拆成更简单的部分。例如，Σ_i=1ⁿ (i+1)(i+2) 可以展开为 Σ_i=1ⁿ (i² + 3i + 2)。

Then we evaluate each part separately:

然后分别求每一部分：

Σ(i² + 3i + 2) = Σi² + 3Σi + 2Σ1 = n(n+1)(2n+1)/6 + 3n(n+1)/2 + 2n

Combining sums in reverse is equally useful: if two sums share the same limits, they can be written as one sum of the combined general terms.

反向合并求和同样有用：如果两个求和具有相同的上下限，它们可以写成一个通项合并后的求和。

---

10. Using Sigma to Evaluate a Series | 用西格玛求级数值

Consider the sum S = Σ_i=1ⁿ (2i - 1). This is the sum of the first n odd numbers. Using linearity:

考虑和式 S = Σ_i=1ⁿ (2i - 1)。这是前 n 个奇数的和。利用线性性质：

S = 2Σi - Σ1 = 2 × n(n+1)/2 - n = n²

Hence the sum of the first n odd numbers is exactly n². This is a beautiful result and a direct application of sigma notation.

因此前 n 个奇数之和恰好是 n²。这是一个优美的结果，也是西格玛记号的直接应用。

Another example: Σ_i=1ⁿ i(i+1) = Σ(i² + i) = n(n+1)(2n+1)/6 + n(n+1)/2 = n(n+1)(n+2)/3.

另一个例子：Σ_i=1ⁿ i(i+1) = Σ(i² + i) = n(n+1)(2n+1)/6 + n(n+1)/2 = n(n+1)(n+2)/3。

---

11. Common Pitfalls and Tips | 常见误区与技巧

• The index is a dummy variable: it does not appear in the final answer. Do not confuse it with a fixed number.
• 指标是哑变量：它不会出现在最终答案中。不要将其与固定数字混淆。
• Always check the lower and upper limits. A sum from i = 0 to n has n+1 terms, not n terms.
• 始终检查上下限。从 i=0 到 n 的求和有 n+1 项，而不是 n 项。
• Do not pull out a factor that depends on the index. Only constants can be moved outside Σ.
• 不要提出依赖于指标的因子。只有常数才能移到 Σ 外面。
• When using standard formulas, make sure the index starts at 1 and the upper limit is n. If not, shift the index first.
• 使用标准公式时，确保指标从 1 开始且上限为 n。如果不是，先平移指标。

Sigma notation is not just a notational convenience; it is a mental tool that organises the evaluation of series. Mastering it will greatly improve your ability to handle sequences, binomial expansions, and numerical methods.

西格玛记号不仅仅是一种书写上的便利，更是一种组织级数求解的思维工具。掌握它将极大提升你处理数列、二项式展开和数值方法的能力。

---

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）