Summation of Finite Series | 有限级数求和

📚 Summation of Finite Series | 有限级数求和

Finite series summation is a central A-level Pure Mathematics topic. You need to handle sigma notation fluently, apply standard results accurately, and choose the right method for arithmetic, geometric, polynomial, or difference-type sums.

有限级数求和是 A-level 纯数学的核心主题。你需要熟练处理 sigma 记号、准确应用标准公式，并针对等差、等比、多项式或差分型求和选择正确方法。

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1. Sigma Notation and Key Terms | Sigma 记号与基本术语

e Mathematics, a finite series is written compactly using sigma notation. The expression ∑_r=1ⁿ u_r means u₁ + u₂ + ... + u_n, where r is the dummy index and u_r is the general term.

在 AQA 纯数学中，有限级数用 sigma 记号紧凑表示。表达式 ∑_r=1ⁿ u_r 表示 u₁ + u₂ + ... + u_n，其中 r 是哑指标，u_r 是通项。

∑_r=1ⁿ u_r = u₁ + u₂ + ... + u_n

The index r is a dummy variable, so ∑_r=1ⁿ r² and ∑_k=1ⁿ k² are exactly the same sum. Always check the lower and upper limits before applying any formula.

指标 r 是哑变量，因此 ∑_r=1ⁿ r² 与 ∑_k=1ⁿ k² 是完全相同的求和。在应用任何公式之前，始终要检查下限和上限。

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2. Standard Results for ∑r, ∑r², ∑r³ | ∑r、∑r²、∑r³ 的标准结果

Three standard results must be memorised for AQA. They are used repeatedly for polynomial sums and can be quoted directly unless a question asks for a proof.

三个标准结果必须牢记。它们反复用于多项式求和，除非题目要求证明，否则可以直接引用。

Sum | 求和	Result | 结果
∑r=1n 1	n
∑r=1n r	n(n+1)/2
∑r=1n r²	n(n+1)(2n+1)/6
∑r=1n r³	n²(n+1)²/4 = [n(n+1)/2]²

The result ∑_r=1ⁿ 1 = n simply adds the number 1 n times.

结果 ∑_r=1ⁿ 1 = n 只是将数字 1 加 n 次。

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3. Linearity: Splitting and Factoring | 线性性质：拆分与提取

Summation is linear, meaning that a constant multiple can be taken outside and a sum can be split term by term. This is the key tool for handling polynomial sums.

求和运算是线性的，这意味着常数倍数可以提到外面，求和可以逐项拆分。这是处理多项式求和的关键工具。

∑_r=1ⁿ [a u_r + b v_r] = a∑_r=1ⁿ u_r + b∑_r=1ⁿ v_r

, ∑(5r² - 2r) = 5∑r² - 2∑r. After splitting, substitute the standard results and simplify.

例如，∑(5r² - 2r) = 5∑r² - 2∑r。拆分后代入标准结果并化简。

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4. Arithmetic Series Sum | 等差级数求和

An arithmetic sequence has a constant difference d between consecutive terms. If the first term is a, then u_r = a + (r - 1)d. The sum of the first n terms is given by either form below.

等差数列的相邻项之差为常数 d。如果首项为 a，则 u_r = a + (r - 1)d。前 n 项和由以下任一形式给出。

S_n = n/2 [2a + (n - 1)d] = n/2 (a + l)

rm. The second form is especially useful when the first and last terms are known directly.

其中 l = a + (n - 1)d 是末项。当首项和末项直接已知时，第二种形式尤其有用。

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5. Geometric Series Sum | 等比级数求和

A geometric sequence has a constant ratio r between consecutive terms. If the first term is a, then the nth term is arⁿ⁻¹. The finite geometric sum is below, provided r ≠ 1.

等比数列的相邻项之比为常数 r。如果首项为 a，则第 n 项为 arⁿ⁻¹。当 r ≠ 1 时，有限等比和为：

S_n = a(1 - rⁿ)/(1 - r) = a(rⁿ - 1)/(r - 1)

If r = 1, every term is equal to a, so S_n = an. Be careful: there are n terms, so the power is rⁿ, not rⁿ⁻¹.Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com

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