1
等差数列进阶 Arithmetic Sequences
必考
📝 等差数列核心公式
通项公式：an = a1 + (n−1)d
前 n 项和：Sn = n(a1 + an) / 2 = na1 + n(n−1)d / 2
d 为公差（common difference），a₁ 为首项
等差数列是 AMC 10 中最基础的数列类型。常见考点包括：
- 性质：等距三项 a, b, c 成等差数列 ⟺ 2b = a + c Three equally spaced terms form an arithmetic sequence if and only if 2b = a + c.
- 等差中项：a 和 c 的等差中项为 b = (a+c)/2 The arithmetic mean of a and c is b = (a+c)/2.
- 子数列：等差数列中每隔 k 项取一个，仍为等差数列，新公差为 kd Every k-th term of an arithmetic sequence also forms an arithmetic sequence with common difference kd.
💡 求和方法：等差数列前 n 项和 Sn = n × 中位数（当 n 为奇数时特别好用）
When n is odd, Sₙ = n × median, which is often the quickest approach.
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等比数列进阶 Geometric Sequences
高频
📝 等比数列核心公式
通项公式：an = a1 · rn−1
前 n 项和：Sn = a1(1 − rn) / (1 − r)，其中 r ≠ 1
无穷递缩等比级数：S∞ = a1 / (1 − r)，其中 |r| < 1
r 为公比（common ratio），a₁ 为首项
等比数列在 AMC 10 中常以以下形式出现：
- 性质：等距三项 a, b, c 成等比数列 ⟺ b² = ac（b ≠ 0） Three equally spaced terms form a geometric sequence if and only if b² = ac (b ≠ 0).
- 等比中项：a 和 c 的等比中项为 b = ±√(ac)（需注意正负号） The geometric mean of a and c is b = ±√(ac). Watch for the ± sign!
- 无穷级数：当 |r| < 1 时，S∞ = a₁/(1−r) 是 AMC 10 高频考点 When |r| < 1, the infinite sum converges: S∞ = a₁/(1−r).
⚠️ 注意区分"前 n 项和"与"无穷级数和"。只有当 |r| < 1 时无穷级数才有意义。
Distinguish finite sum Sₙ from infinite sum S∞. The infinite sum only converges when |r| < 1.
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递推数列 Recursive Sequences
进阶
递推数列通过前项定义后项，在 AMC 10 中需要掌握以下策略：
- 列举法：直接写出前若干项，观察规律或周期性 List the first several terms and look for patterns or periodicity.
- 周期性检测：如果递推涉及取模（mod），数列往往呈现周期性 Recurrence relations involving modular arithmetic often produce periodic sequences.
- 差分法：对数列求差，判断一阶差分或二阶差分是否为常数（等差数列/二次数列） Take differences to check if first or second differences are constant.
- 转化法：将递推关系转化为等差或等比数列求解 Transform the recurrence into an arithmetic or geometric sequence.
📝 常见递推类型
an = an−1 + d → 等差数列
an = an−1 · r → 等比数列
an = pan−1 + q → 可转化为等比数列
第三种类型的通解为 aₙ + q/(p−1) = p(aₙ₋₁ + q/(p−1))，即等比数列
💡 周期性数列技巧：若递推涉及 mod m，则周期不超过 m²。列出前 m² 项即可找到周期。
For modular recurrences, the period is at most m². Listing the first m² terms suffices.
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特殊求和技巧 Summation Techniques
竞赛必会
📝 裂项求和（Telescoping Series）
Σ [1/(k(k+1))] = Σ [1/k − 1/(k+1)] = 1 − 1/(n+1) = n/(n+1)
关键是找到通项的裂项形式 1/(k(k+1)) = 1/k − 1/(k+1)
📝 常见裂项模式
1/(k(k+2)) = ½[1/k − 1/(k+2)]
1/(k(k+1)(k+2)) = ½[1/(k(k+1)) − 1/((k+1)(k+2))]
(2k+1)/(k(k+1)) = 1/k + 1/(k+1)
📝 错位相减法
用于求 {aₙ · bₙ} 的和，其中 {aₙ} 为等差、{bₙ} 为等比
设 S = a₁b₁ + a₂b₂ + ... + aₙbₙ，则 rS = a₁b₂ + a₂b₃ + ... + aₙbₙ₊₁
两式相减，中间项大量抵消，剩余部分即可求和
📝 常用求和公式
Σ k = n(n+1)/2
Σ k² = n(n+1)(2n+1)/6
Σ k³ = [n(n+1)/2]²
以上求和范围均为 k = 1 到 n
💡 裂项求和的关键在于"拆"：把每一项拆成两项之差，使得相邻项正负抵消，最终只剩首尾两项。
The key to telescoping: decompose each term into a difference so adjacent terms cancel, leaving only the first and last.
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例题精讲 Worked Examples
5 题
例题 1 等差数列
一个等差数列的第 3 项为 7，第 7 项为 19。求前 20 项的和。
In an arithmetic sequence, the 3rd term is 7 and the 7th term is 19. Find the sum of the first 20 terms.
📋 解题过程
由 a₃ = a₁ + 2d = 7，a₇ = a₁ + 6d = 19。
两式相减：4d = 12，所以 d = 3。代入得 a₁ = 7 − 2×3 = 1。
a₂₀ = a₁ + 19d = 1 + 19×3 = 58。
S₂₀ = 20 × (a₁ + a₂₀) / 2 = 10 × (1 + 58) = 590。
a₃ = a₁ + 2d = 7, a₇ = a₁ + 6d = 19 → 4d = 12 → d = 3, a₁ = 1. a₂₀ = 58. S₂₀ = 10(1+58) = 590.
例题 2 等比数列
一个等比数列的首项为 3，公比为 1/2。求其无穷级数的和。
A geometric sequence has first term 3 and common ratio 1/2. Find the sum of its infinite series.
📋 解题过程
|r| = 1/2 < 1，所以无穷级数收敛。
S∞ = a₁ / (1 − r) = 3 / (1 − 1/2) = 3 / (1/2) = 6。
Since |r| = 1/2 < 1, the infinite sum converges. S∞ = 3/(1 − 1/2) = 6.
例题 3 裂项求和
求 1/(1×2) + 1/(2×3) + 1/(3×4) + … + 1/(99×100) 的值。
Find the value of 1/(1×2) + 1/(2×3) + 1/(3×4) + … + 1/(99×100).
📋 解题过程
每一项可裂项：1/(k(k+1)) = 1/k − 1/(k+1)。
所以原式 = (1 − 1/2) + (1/2 − 1/3) + (1/3 − 1/4) + … + (1/99 − 1/100)。
中间项全部抵消，只剩 1 − 1/100 = 99/100。
Each term telescopes: 1/(k(k+1)) = 1/k − 1/(k+1). All intermediate terms cancel, leaving 1 − 1/100 = 99/100.
例题 4 递推数列
数列 {aₙ} 满足 a₁ = 1，aₙ₊₁ = 2aₙ + 1。求 a₅ 的值。
A sequence satisfies a₁ = 1 and aₙ₊₁ = 2aₙ + 1. Find a₅.
📋 解题过程
方法一（列举法）：
a₁ = 1, a₂ = 2×1+1 = 3, a₃ = 2×3+1 = 7, a₄ = 2×7+1 = 15, a₅ = 2×15+1 = 31。
方法二（公式法）：aₙ₊₁ + 1 = 2(aₙ + 1)，所以 {aₙ+1} 是公比为 2 的等比数列。
aₙ + 1 = 2ⁿ，即 aₙ = 2ⁿ − 1。a₅ = 2⁵ − 1 = 31。
Method 1: a₁=1, a₂=3, a₃=7, a₄=15, a₅=31. Method 2: aₙ₊₁+1=2(aₙ+1), so aₙ=2ⁿ−1, giving a₅=31.
例题 5 综合应用
三个数成等比数列，它们的和为 14，积为 64。求这三个数中最大的数。
Three numbers form a geometric sequence. Their sum is 14 and their product is 64. Find the largest number.
📋 解题过程
设三个数为 a/r, a, ar（对称设项）。
积：(a/r) · a · ar = a³ = 64，所以 a = 4。
和：a/r + a + ar = 4/r + 4 + 4r = 14。
化简：4r + 4 + 4r = 14（乘以 r），即 4r² − 10r + 4 = 0。
整理：2r² − 5r + 2 = 0，(2r−1)(r−2) = 0，所以 r = 2 或 r = 1/2。
当 r = 2 时，三数为 2, 4, 8，最大为 8。
当 r = 1/2 时，三数为 8, 4, 2，最大仍为 8。
Let the three numbers be a/r, a, ar. Product: a³=64 → a=4. Sum: 4/r+4+4r=14 → 2r²−5r+2=0 → r=2 or 1/2. Either way, the largest number is 8.
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巩固练习 Practice Problems
8 题
巩固练习
第 1 题：等差数列 {aₙ} 中，a₅ = 13，a₁₅ = 33，求公差 d。
In an arithmetic sequence, a₅ = 13 and a₁₅ = 33. Find the common difference d.
第 2 题：等比数列 {aₙ} 中，a₁ = 2，a₄ = 54，求公比 r。
In a geometric sequence, a₁ = 2 and a₄ = 54. Find the common ratio r.
第 3 题：求 1 + 2 + 3 + … + 100 的值。
Find the value of 1 + 2 + 3 + … + 100.
第 4 题：数列 {aₙ} 满足 a₁ = 2，aₙ₊₁ = aₙ + 3，求 a₁₀。
A sequence satisfies a₁ = 2 and aₙ₊₁ = aₙ + 3. Find a₁₀.
第 5 题：等比数列 {aₙ} 的首项为 5，公比为 1/3，求无穷级数 S∞。
A geometric sequence has first term 5 and common ratio 1/3. Find S∞.
第 6 题：求 1/(1×3) + 1/(3×5) + 1/(5×7) + … + 1/(99×101) 的值。
Find the value of 1/(1×3) + 1/(3×5) + 1/(5×7) + … + 1/(99×101).
第 7 题：三个数成等比数列，它们的积为 64，和为 14。求这三个数中最大的数。
Three numbers form a geometric sequence. Their product is 64 and their sum is 14. Find the largest number.
第 8 题：一个等差数列的前 n 项和为 Sn = 2n² + 3n。求该数列的第 10 项 a₁₀。
The sum of the first n terms of an arithmetic sequence is Sₙ = 2n² + 3n. Find the 10th term a₁₀.