IB Math: Summation Notation & Geometric Series | IB数学：求和记号与等比级数

📚 IB Math: Summation Notation & Geometric Series | IB数学：求和记号与等比级数

Summation notation is a compact way to write the sum of many terms. In the IB Mathematics curriculum, it appears in arithmetic and geometric series, statistics, and calculus. A strong grasp of summation notation and geometric series is essential for both Analysis and Approaches (AA) and Applications and Interpretation (AI) exams.

求和记号是表示多项相加的简洁写法。在IB数学课程中，它出现在等差与等比级数、统计和微积分中。牢固掌握求和记号与等比级数，对分析与方法（AA）和应用与解释（AI）考试都至关重要。

1. The Summation Symbol | 求和符号

The capital Greek letter Σ (sigma) tells us to add a sequence of terms. If we write Σ (from k=1 to n) aₖ, we read it as “the sum of aₖ from k = 1 to n”.

大写希腊字母Σ（sigma）表示把一列项加起来。如果写“Σ（从 k=1 到 n）aₖ”，我们读作“从 k=1 到 n 对 aₖ 求和”。

The variable k is called the index of summation. The number below is the lower limit, and the number above is the upper limit. The expression aₖ is the general term.

变量 k 称为求和的指标（index）。下方的数字是下限，上方的数字是上限；表达式 aₖ 是通项。

Example: Σ (from k=1 to 5) k² = 1² + 2² + 3² + 4² + 5² = 55.

例如：Σ（从 k=1 到 5）k² = 1² + 2² + 3² + 4² + 5² = 55。

2. Term Index and Limits | 项下标与上下限

The index is a placeholder. Changing the letter from k to i or n does not change the sum.

指标只是一个占位符。把字母 k 换成 i 或 n 不会改变求和结果。

If a sequence starts at a₁, the first term is obtained by substituting the lower limit into the general term.

如果数列从 a₁ 开始，第一项就是把下限代入通项后得到的结果。

For example, Σ (from k=2 to 5) (3k + 1) = 7 + 10 + 13 + 16 = 46.

例如：Σ（从 k=2 到 5）(3k + 1) = 7 + 10 + 13 + 16 = 46。

3. Properties of Summation | 求和的运算性质

Summation is linear, which means constants can be factored out, and sums can be split term by term.

求和具有线性性质，即常数可以提出，和式可以按项拆分。

Σ (from k=1 to n) c aₖ = c Σ (from k=1 to n) aₖ

Σ（从 k=1 到 n）c aₖ = c Σ（从 k=1 到 n）aₖ
Σ (from k=1 to n) (aₖ + bₖ) = Σ aₖ + Σ bₖ

Σ（从 k=1 到 n）(aₖ + bₖ) = Σ aₖ + Σ bₖ
Σ (from k=1 to n) c = nc

Σ（从 k=1 到 n）c = nc

The following closed forms are often used in IB questions.

以下封闭形式在IB题目中经常出现。

Sum	Closed Form
Σ (k=1 to n) k	n(n + 1)/2
Σ (k=1 to n) k²	n(n + 1)(2n + 1)/6
Σ (k=1 to n) k³	[n(n + 1)/2]²

4. Geometric Sequences and Series | 等比数列与等比级数

A geometric sequence has a constant ratio r between consecutive terms. If the first term is a₁ and the ratio is r, the nth term is

等比数列相邻两项的比值 r 恒定。若首项为 a₁，公比为 r，则第 n 项为

aₙ = a₁ rⁿ⁻¹

A geometric series is the sum of the terms of a geometric sequence.

等比级数就是等比数列各项之和。

Example: 2, 6, 18, 54 is geometric with a₁ = 2, r = 3.

例如：2, 6, 18, 54 是等比数列，其中 a₁ = 2，r = 3。

5. Sum of a Finite Geometric Series | 有限等比级数的求和

For a geometric series with first term a, common ratio r, and n terms, the sum is

对于首项为 a、公比为 r、项数为 n 的等比级数，其和为

Sₙ = a(1 − rⁿ) / (1 − r), r ≠ 1

Or equivalently

等价形式为

Sₙ = a(rⁿ − 1) / (r − 1), r ≠ 1

If r = 1, every term is a, so Sₙ = na.

若 r = 1，则每一项都是 a，因此 Sₙ = na。

Example: For 3 + 6 + 12 + 24 + 48 + 96, a = 3, r = 2, n = 6.

例如：对 3 + 6 + 12 + 24 + 48 + 96，有 a = 3，r = 2，n = 6。

S₆ = 3(1 − 2⁶) / (1 − 2) = 189

6. Deriving the Formula | 公式推导

Start with Sₙ = a + ar + ar² + … + arⁿ⁻¹.

从 Sₙ = a + ar + ar² + … + arⁿ⁻¹ 出发。

Multiply both sides by r:

两边同时乘以 r：

rSₙ = ar + ar² + ar³ + … + arⁿ

Subtract the second equation from the first:

用第一式减去第二式：

Sₙ − rSₙ = a − arⁿ

Factor and solve:

提取公因式并求解：

Sₙ(1 − r) = a(1 − rⁿ), so Sₙ = a(1 − rⁿ)/(1 − r)

The same derivation works for the alternative form by multiplying the numerator and denominator by −1.

将分子分母同乘 −1，即可得到另一种等价形式。

7. Infinite Geometric Series | 无穷等比级数

An infinite series may converge to a finite limit if the terms approach zero quickly enough. For a geometric series, convergence happens exactly when |r| < 1.

如果各项足够快地趋近于零，无穷级数可能收敛到一个有限值。对于

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IB Math: Summation Notation & Geometric Series | IB数学：求和记号与等比级数