Geometric Series Sum Formulas and Typical Problems | 等比级数求和公式与典型题型

📚 Geometric Series Sum Formulas and Typical Problems | 等比级数求和公式与典型题型

A geometric series is one of the most frequently tested topics in A-Level Mathematics. It appears in questions on sequences, financial mathematics, binomial expansions and even integration by series. Mastering its sum formulas and recognising the conditions for convergence is essential for higher grades.

等比级数是A-Level数学中考查频率最高的内容之一。无论是在数列题、金融数学、二项式展开，还是在利用级数求积分的问题中都会出现。熟练掌握其求和公式并理解收敛条件，是冲刺高分的关键。

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

A geometric sequence is a sequence in which each term after the first is obtained by multiplying the previous term by a fixed constant r, called the common ratio. If the first term is a, then the nth term is uₙ = a rⁿ⁻¹.

等比数列是指从第二项起，每一项都与前一项的比值等于同一个常数 r 的数列，这个常数 r 称为公比。若首项为 a，则第 n 项为 uₙ = a rⁿ⁻¹。

The corresponding geometric series is the sum of the terms. We write Sₙ = u₁ + u₂ + … + uₙ.

相应的等比级数就是这些项的和，记作 Sₙ = u₁ + u₂ + … + uₙ。

For example, 2, 6, 18, 54, … is geometric with a = 2 and r = 3.

例如，2, 6, 18, 54, … 就是一个等比数列，其中 a = 2，r = 3。

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

For a geometric series with first term a and common ratio r ≠ 1, the sum of the first n terms is

对于首项为 a、公比 r ≠ 1 的等比级数，前 n 项和为

Sₙ = a(1 – rⁿ) / (1 – r)

or equivalently

或者等价地写成

Sₙ = a(rⁿ – 1) / (r – 1)

If r = 1, every term equals a, so Sₙ = n a.

当 r = 1 时，所有项都等于 a，因此 Sₙ = n a。

The formula can be proved by multiplying Sₙ by r and subtracting. For example, Sₙ = a + ar + … + arⁿ⁻¹ and rSₙ = ar + … + arⁿ, so Sₙ(1-r) = a(1-rⁿ).

这个公式可通过错位相减法证明：将 Sₙ 乘以 r 后与原式相减，即 Sₙ = a + ar + … + arⁿ⁻¹，rSₙ = ar + … + arⁿ，两式相减得 Sₙ(1 – r) = a(1 – rⁿ)。

Be careful when using the formula: the index in (1 – rⁿ) is exactly the number of terms being added.

使用公式时要注意：(1 – rⁿ) 中的指数 n 必须与所加项数完全一致。

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3. Infinite Geometric Series and Convergence | 无穷等比级数与收敛条件

An infinite geometric series is S∞ = a + ar + ar² + … . When |r| < 1, the terms approach zero and the partial sums approach a limit. The sum to infinity is

无穷等比级数为 S∞ = a + ar + ar² + …。当 |r| < 1 时，项趋近于零，部分和趋于一个极限，其和（极限值）为

S∞ = a / (1 – r)

If |r| ≥ 1, the series does not converge. In particular, when r = 1, all terms are equal to a and the partial sums grow without bound; when r = -1, the partial sums oscillate between a and 0.

如果 |r| ≥ 1，级数不收敛。特别地，当 r = 1 时，所有项都等于 a，部分和无限增大；当 r = -1 时，部分和在 a 和 0 之间交替振荡。

The condition |r| < 1 is also written as -1 < r < 1. Note that r must be strictly less than 1 in absolute value; equality is not allowed.

条件 |r| < 1 也可写成 -1 < r < 1。注意绝对值必须严格小于 1，等于 1 是不允许的。

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4. Typical Problem 1: Finding the Number of Terms | 典型题型一：求项数

Given a, r and the sum Sₙ, you may be asked to find n. This requires solving an equation of the form rⁿ = constant, usually by taking logarithms.

已知 a、r 和和 Sₙ，要求项数 n。这类问题需要解形如 rⁿ = 常数的方程，通常借助对数求解。

Example: A geometric series has first term 5 and common ratio 2. Find the least number of terms needed for the sum to exceed 5000.

例：一个等比级数的首项为 5，公比为 2。求使和超过 5000 所需的最少项数。

The solution uses Sₙ = 5(2ⁿ – 1)/(2 – 1) = 5(2ⁿ – 1). We need 5(2ⁿ – 1) > 5000, so 2ⁿ > 1001.

解法：Sₙ = 5(2ⁿ – 1)/(2 – 1) = 5(2ⁿ – 1)。要使 5(2ⁿ – 1) > 5000，即 2ⁿ > 1001。

Taking logarithms gives n log 2 > log 1001, so n > log 1001 / log 2 ≈ 9.97. Hence the least integer n is 10.

两边取对数得 n log 2 > log 1001，所以 n > log 1001 / log 2 ≈ 9.97，因此最小整数 n 为 10。

Always check whether the inequality requires rounding up or down, and remember that n must be a positive integer.

要注意根据不等号方向选择向上取整或向下取整，并且 n 必须是正整数。

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5. Typical Problem 2: Convergence of an Infinite Series | 典型题型二：判断无穷级数是否收敛

For an infinite geometric series, you only need to examine the common ratio r. The series converges if and only if |r| < 1.

对于无穷等比级数，只需考察公比 r。级数收敛当且仅当 |r| < 1。

Example 1: Does the series 100 − 80 + 64 − 51.2 + … have a sum to infinity? Here r = -80/100 = -0.8, so |r| = 0.8 < 1. The sum is S∞ = 100 / (1 - (-0.8)) = 100 / 1.8 = 500/9 ≈ 55.56.

例1：级数 100 − 80 + 64 − 51.2 + … 是否有无穷和？这里 r = -80/100 = -0.8，|r| = 0.8 < 1，所以存在。S∞ = 100 / (1 - (-0.8)) = 100 / 1.8 = 500/9 ≈ 55.56。

Example 2: The series 3 + 6 + 12 + 24 + … has r = 2, so |r| ≥ 1 and it diverges; no finite sum to infinity exists.

例2：级数 3 + 6

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