Unit A1: Sequences
& Series单元 A1：数列与级数
The first sub-unit of Topic 1 (Number & Algebra). Master arithmetic and geometric progressions, sigma notation, the convergence of an infinite GP, financial applications, and the patterns IB rewards on Papers 1 and 2.Topic 1（数与代数）的第一个子单元。掌握等差与等比数列（arithmetic & geometric progression）、求和记号（sigma notation）、无穷等比级数的收敛（convergence）、金融应用，以及 IB Paper 1 和 Paper 2 反复出现的考法。
How to use this guide本指南使用说明
This guide is built to serve two students at once:本指南同时为两类学生设计：
Read only the dashed-gold "Cram-Mode Cheat" box at the top of each section, plus the formula boxes. Skim one worked example per section. Skip the ▸ expandable details. Take the practice quiz. That's a 30-minute pass.
只看每节顶端的金色虚线 "Cram-Mode Cheat" 速记框和公式框（formula box）。每节挑一道 worked example（例题）扫一眼，跳过 ▸ 折叠的"深入"部分，做一次练习测验。约 30 分钟过完一遍。
Read straight through. Open every ▸ Going deeper details block — that's where derivations and proofs live. Do every worked example with the solution covered first. Re-do the quiz with no hints. Owning the why is what separates a 5 from a 7.
从头读到尾。打开每个 ▸ Going deeper 折叠块——推导和证明都在里面。每道 worked example 先遮住答案自己做，再对照解析。再做一遍测验不看提示。真正吃透 为什么，才是 5 分到 7 分的关键。
Arithmetic Sequences等差数列 SL 1.2
arithmetic sequence）每一步加一个固定的量。这个固定的量就是公差（common difference）$d = u_{n+1} - u_n$。第 $n$ 项为
$$ u_n = u_1 + (n-1)d. $$
给两个关于 $u_1$ 和 $d$ 的条件 → $2\times 2$ 方程组 → 唯一确定的数列。这一句话能解决约 80% 的 Paper 1 AP（arithmetic progression）题。
arithmetic）的，当且仅当相邻两项之差为常数：
$$ u_{n+1} - u_n = d \quad \text{for every } n \ge 1. $$
等价地，$u_n = u_1 + (n-1)d$。常数 $d$ 是公差（common difference），可以为正（递增）、为负（递减）或为零（常数列）。
(2) The arithmetic mean: $u_{n+1} = \tfrac{u_n + u_{n+2}}{2}$ — middle term is the average of its neighbours.
(3) Symmetry: $u_k + u_{n-k+1}$ is constant across $k$ for the first $n$ terms.
(2) 等差中项（
arithmetic mean）：$u_{n+1} = \tfrac{u_n + u_{n+2}}{2}$——中间项等于左右相邻两项的平均。(3) 对称性：在前 $n$ 项中，$u_k + u_{n-k+1}$ 对所有 $k$ 都相等。
Worked Example — Find a specific term例题——求指定项
Problem: An arithmetic sequence has $u_3 = 11$ and $u_8 = 26$. Find $u_{20}$.
Set up. Two facts in two unknowns. From identity (1):
$$ u_8 - u_3 = (8 - 3)d \;\Rightarrow\; 26 - 11 = 5d \;\Rightarrow\; d = 3. $$Find $u_1$. $u_3 = u_1 + 2d$, so $u_1 = 11 - 6 = 5$.
Apply. $u_{20} = 5 + 19(3) = 62$.
题目：已知等差数列中 $u_3 = 11$，$u_8 = 26$。求 $u_{20}$。
建立方程。两条信息、两个未知量。由恒等式 (1)：
$$ u_8 - u_3 = (8 - 3)d \;\Rightarrow\; 26 - 11 = 5d \;\Rightarrow\; d = 3. $$求 $u_1$。$u_3 = u_1 + 2d$，故 $u_1 = 11 - 6 = 5$。
代入。$u_{20} = 5 + 19(3) = 62$。
▸ Going deeper — Why "$2$ facts → unique sequence"▸ 深入——为何"两条信息就能唯一确定数列"
An AP has two parameters: $u_1$ and $d$. Each condition is one linear equation in them.
$$ \begin{aligned} u_p = a \;\;&\Longleftrightarrow\;\; u_1 + (p-1)\,d = a \\ S_q = b \;\;&\Longleftrightarrow\;\; q\,u_1 + \tfrac{q(q-1)}{2}\,d = b \end{aligned} $$Two independent equations $\Rightarrow$ unique $(u_1, d)$. Independence fails when one is a scalar multiple of the other — e.g.
$$ u_3 = 11 \;\;\text{and}\;\; u_3 + u_5 = 26 $$The second is $2u_1 + 6d = 26$, but $u_3 = 11$ gives $2u_1 + 4d = 22$, so subtracting yields $2d = 4$ alone — $u_1$ stays free. Recipe: write both as $\alpha u_1 + \beta d = \gamma$, then check $\det\!\begin{pmatrix}\alpha_1 & \beta_1 \\ \alpha_2 & \beta_2\end{pmatrix} \ne 0$.
一个 AP 只有两个参数：$u_1$ 和 $d$。每个条件就是关于它们的一条线性方程：
$$ \begin{aligned} u_p = a \;\;&\Longleftrightarrow\;\; u_1 + (p-1)\,d = a \\ S_q = b \;\;&\Longleftrightarrow\;\; q\,u_1 + \tfrac{q(q-1)}{2}\,d = b \end{aligned} $$两条独立方程 $\Rightarrow$ 唯一的 $(u_1, d)$。当一条是另一条的常数倍时独立性失败，例如：
$$ u_3 = 11 \;\;\text{and}\;\; u_3 + u_5 = 26 $$第二条化为 $2u_1 + 6d = 26$，而 $u_3 = 11$ 给出 $2u_1 + 4d = 22$，相减只得到 $2d = 4$——$u_1$ 仍是自由的。套路：把两条都写成 $\alpha u_1 + \beta d = \gamma$，然后验证行列式 $\det\!\begin{pmatrix}\alpha_1 & \beta_1 \\ \alpha_2 & \beta_2\end{pmatrix} \ne 0$。
Arithmetic Series等差级数 SL 1.2
arithmetic series）前 $n$ 项之和为 $\boxed{\,S_n = \tfrac{n}{2}(u_1 + u_n) = \tfrac{n}{2}\bigl(2u_1 + (n-1)d\bigr)\,}$。已知 $u_n$ 时用第一式；只有 $u_1$ 和 $d$ 时用第二式。两式恒等。
▸ Going deeper — The Gauss-pairing derivation▸ 深入——高斯配对推导
Write the sum forwards and backwards on top of each other:
$$ \begin{aligned} S_n &= u_1 + (u_1 + d) + (u_1 + 2d) + \cdots + (u_1 + (n-1)d) \\ S_n &= u_n + (u_n - d) + (u_n - 2d) + \cdots + (u_n - (n-1)d). \end{aligned} $$Add column-wise. Each column gives $u_1 + u_n$, and there are $n$ columns:
$$ 2 S_n = n(u_1 + u_n) \;\Rightarrow\; S_n = \tfrac{n}{2}(u_1 + u_n). $$Substituting $u_n = u_1 + (n-1)d$ gives the second form. The trick is named for Gauss who reportedly used it as a child to sum $1 + 2 + \cdots + 100$ in seconds — pair $1$ with $100$, $2$ with $99$, etc., to get $50$ pairs each summing to $101$.
把同一个和正序、逆序上下对齐写：
$$ \begin{aligned} S_n &= u_1 + (u_1 + d) + (u_1 + 2d) + \cdots + (u_1 + (n-1)d) \\ S_n &= u_n + (u_n - d) + (u_n - 2d) + \cdots + (u_n - (n-1)d). \end{aligned} $$按列相加，每一列都是 $u_1 + u_n$，共 $n$ 列：
$$ 2 S_n = n(u_1 + u_n) \;\Rightarrow\; S_n = \tfrac{n}{2}(u_1 + u_n). $$代入 $u_n = u_1 + (n-1)d$ 即得第二式。这一技巧得名于高斯——传说他童年时用 $50$ 对 $\,1\!+\!100,\,2\!+\!99,\ldots\,$ 每对都等于 $101$，瞬间算出 $1+2+\cdots+100$。
Worked Example — System for $u_1$ and $d$例题——求 $u_1$ 与 $d$ 的方程组
Problem: An arithmetic sequence has $u_4 = 11$ and $S_8 = 76$. Find $u_1$ and $d$.
Translate:
$$ u_4 = u_1 + 3d = 11 \tag{1} $$ $$ S_8 = \tfrac{8}{2}\bigl(2u_1 + 7d\bigr) = 4(2u_1 + 7d) = 76 \;\Rightarrow\; 2u_1 + 7d = 19. \tag{2} $$From (1), $u_1 = 11 - 3d$. Sub into (2):
$$ 2(11 - 3d) + 7d = 19 \;\Rightarrow\; 22 + d = 19 \;\Rightarrow\; d = -3. $$Then $u_1 = 11 - 3(-3) = 20$.
题目：等差数列满足 $u_4 = 11$，$S_8 = 76$。求 $u_1$ 和 $d$。
翻译为方程：
$$ u_4 = u_1 + 3d = 11 \tag{1} $$ $$ S_8 = \tfrac{8}{2}\bigl(2u_1 + 7d\bigr) = 4(2u_1 + 7d) = 76 \;\Rightarrow\; 2u_1 + 7d = 19. \tag{2} $$由 (1)：$u_1 = 11 - 3d$。代入 (2)：
$$ 2(11 - 3d) + 7d = 19 \;\Rightarrow\; 22 + d = 19 \;\Rightarrow\; d = -3. $$故 $u_1 = 11 - 3(-3) = 20$。
Worked Example — Sum of integers in a range例题——区间内整数求和
Problem: Find the sum of all multiples of $7$ between $100$ and $400$ inclusive.
Find $u_1$, $u_n$, $n$. First multiple $\ge 100$ is $105 = 7\cdot 15$; last $\le 400$ is $399 = 7 \cdot 57$. Number of terms: $57 - 15 + 1 = 43$.
$$ S_{43} = \tfrac{43}{2}(105 + 399) = \tfrac{43}{2}\cdot 504 = 43 \cdot 252 = 10\,836. $$Sanity check: the average term is $(105 + 399)/2 = 252$, times $43$ terms = $10\,836$. ✓
题目：求 $100$ 到 $400$（含端点）之间所有 $7$ 的倍数之和。
求 $u_1$、$u_n$、$n$。$\ge 100$ 的第一个倍数是 $105 = 7\cdot 15$；$\le 400$ 的最后一个是 $399 = 7 \cdot 57$。项数：$57 - 15 + 1 = 43$。
$$ S_{43} = \tfrac{43}{2}(105 + 399) = \tfrac{43}{2}\cdot 504 = 43 \cdot 252 = 10\,836. $$验算：平均项为 $(105 + 399)/2 = 252$，乘以 $43$ 项 $= 10\,836$。✓
Geometric Sequences等比数列 SL 1.3
geometric sequence）每一步乘以一个固定的量。这个固定的量就是公比（common ratio）$r = u_{n+1}/u_n$。第 $n$ 项为
$$ u_n = u_1 \, r^{\,n-1}. $$
复利（compound interest）、折旧（depreciation）、指数增长本质上都是 GP（geometric progression）问题。
geometric）的，当且仅当相邻两项之比为常数：
$$ \frac{u_{n+1}}{u_n} = r \quad \text{for every } n. $$
等价地，$u_n = u_1 \, r^{n-1}$，其中 $u_1 \ne 0$ 且 $r \ne 0$。
Geometric mean: if $a, b, c$ are consecutive terms of a GP with $a, c > 0$, then $b = \pm\sqrt{ac}$.
等比中项（
geometric mean）：若 $a, b, c$ 为某 GP 的连续三项且 $a, c > 0$，则 $b = \pm\sqrt{ac}$。
$|r| < 1$: terms shrink toward $0$.
$r < 0$: terms alternate sign. So $r = -\tfrac{1}{2}$ shrinks AND alternates.
$|r| < 1$：项趋近于 $0$。
$r < 0$：项的符号交替。所以 $r = -\tfrac{1}{2}$ 同时缩小且交替变号。
Worked Example — Finding $r$ and $u_1$ from two terms例题——由两项求 $r$ 与 $u_1$
Problem: A geometric sequence has $u_2 = 6$ and $u_5 = 48$. Find $u_1$ and $u_{10}$.
Eliminate $u_1$. Take ratios:
$$ \frac{u_5}{u_2} = r^{5-2} = r^3 = \frac{48}{6} = 8 \;\Rightarrow\; r = 2. $$Solve for $u_1$. $u_2 = u_1 r \Rightarrow u_1 = 6/2 = 3$.
Apply. $u_{10} = 3 \cdot 2^9 = 3 \cdot 512 = 1536$.
题目：等比数列满足 $u_2 = 6$，$u_5 = 48$。求 $u_1$ 与 $u_{10}$。
消去 $u_1$。作比：
$$ \frac{u_5}{u_2} = r^{5-2} = r^3 = \frac{48}{6} = 8 \;\Rightarrow\; r = 2. $$求 $u_1$。$u_2 = u_1 r \Rightarrow u_1 = 6/2 = 3$。
代入。$u_{10} = 3 \cdot 2^9 = 3 \cdot 512 = 1536$。
Worked Example — "Smallest $n$ such that"例题——"使……的最小 $n$"
Problem: A GP has $u_1 = 5$ and $r = 3$. Find the smallest $n$ such that $u_n > 10\,000$.
$$ u_n = 5 \cdot 3^{\,n-1} > 10\,000 \;\Rightarrow\; 3^{\,n-1} > 2000. $$Take log:
$$ (n-1)\log 3 > \log 2000 \;\Rightarrow\; n - 1 > \frac{\log 2000}{\log 3} \approx 6.92. $$So $n - 1 \ge 7 \Rightarrow n = 8$. Check: $u_8 = 5\cdot 3^7 = 10\,935 > 10\,000$. ✓
题目：GP 满足 $u_1 = 5$，$r = 3$。求使 $u_n > 10\,000$ 的最小 $n$。
$$ u_n = 5 \cdot 3^{\,n-1} > 10\,000 \;\Rightarrow\; 3^{\,n-1} > 2000. $$取对数：
$$ (n-1)\log 3 > \log 2000 \;\Rightarrow\; n - 1 > \frac{\log 2000}{\log 3} \approx 6.92. $$故 $n - 1 \ge 7 \Rightarrow n = 8$。验证：$u_8 = 5\cdot 3^7 = 10\,935 > 10\,000$。✓
▸ Going deeper — Negative $r$ makes "first term exceeding $M$" subtle▸ 深入——$r$ 为负时"首次超过 $M$"的微妙之处
If $r$ is negative, "exceeds $M$" usually means $|u_n| > M$, since the sequence alternates sign. Always read whether the question wants magnitude exceeded or the term itself to be greater than $M$ (a positive bound is unreachable from a negative term). Worked-out IB markschemes are explicit about this — match their convention by stating clearly whether you're working with $u_n$ or $|u_n|$.
$r$ 为负时，由于符号交替，"超过 $M$" 通常指 $|u_n| > M$。务必看清题目要求的是幅值超出，还是项本身大于 $M$（负数项永远达不到正的上界）。IB 标答（markscheme）一向写得很明确——你在解答中也要清楚指出处理的是 $u_n$ 还是 $|u_n|$。
Geometric Series等比级数 SL 1.3
geometric series）前 $n$ 项之和为 $\boxed{\,S_n = \dfrac{u_1(r^n - 1)}{r - 1} = \dfrac{u_1(1 - r^n)}{1 - r}\,}$，$r \ne 1$。两式代数等价——选分子和分母都为正的那个（更少算账）。$r = 1$ 时公式失效，此时 $S_n = nu_1$。
partial sum）If $r = 1$ then every term is $u_1$ and $S_n = nu_1$.
$r = 1$ 时每项都等于 $u_1$，故 $S_n = nu_1$。
▸ Going deeper — Derivation by $S_n - rS_n$▸ 深入——用 $S_n - rS_n$ 推导
Write the sum:
$$ S_n = u_1 + u_1 r + u_1 r^2 + \cdots + u_1 r^{n-1}. $$Multiply by $r$:
$$ r S_n = u_1 r + u_1 r^2 + \cdots + u_1 r^{n-1} + u_1 r^n. $$Subtract — most terms telescope away:
$$ S_n - r S_n = u_1 - u_1 r^n \;\Rightarrow\; (1 - r) S_n = u_1(1 - r^n). $$Divide by $(1 - r)$ (legal since $r \ne 1$):
$$ S_n = \frac{u_1(1 - r^n)}{1 - r}. $$The other form follows from multiplying numerator and denominator by $-1$.
写出和式：
$$ S_n = u_1 + u_1 r + u_1 r^2 + \cdots + u_1 r^{n-1}. $$两端乘 $r$：
$$ r S_n = u_1 r + u_1 r^2 + \cdots + u_1 r^{n-1} + u_1 r^n. $$相减，大部分项相互抵消：
$$ S_n - r S_n = u_1 - u_1 r^n \;\Rightarrow\; (1 - r) S_n = u_1(1 - r^n). $$两边除以 $(1 - r)$（因 $r \ne 1$ 合法）：
$$ S_n = \frac{u_1(1 - r^n)}{1 - r}. $$另一形式只需分子分母同乘 $-1$ 即可。
Worked Example — Direct sum例题——直接求和
Problem: Compute $2 + 6 + 18 + \cdots$ for the first $10$ terms.
$u_1 = 2$, $r = 3$, $n = 10$. Use the form with $r > 1$:
$$ S_{10} = \frac{2(3^{10} - 1)}{3 - 1} = \frac{2 \cdot 59\,048}{2} = 59\,048. $$题目：求 $2 + 6 + 18 + \cdots$ 前 $10$ 项之和。
$u_1 = 2$，$r = 3$，$n = 10$。$r > 1$，选第一式：
$$ S_{10} = \frac{2(3^{10} - 1)}{3 - 1} = \frac{2 \cdot 59\,048}{2} = 59\,048. $$Worked Example — Solving for $n$例题——反解 $n$
Problem: A GP has $u_1 = 4$ and $r = \tfrac{3}{2}$. Find the smallest $n$ such that $S_n > 200$.
$$ S_n = \frac{4\bigl((3/2)^n - 1\bigr)}{(3/2) - 1} = 8\bigl((3/2)^n - 1\bigr) > 200. $$ $$ (3/2)^n > 26 \;\Rightarrow\; n > \frac{\log 26}{\log(3/2)} \approx 8.04. $$So $n = 9$. Check: $S_9 = 8((3/2)^9 - 1) \approx 8(25.629) \approx 205.0 > 200$. ✓
题目：GP 满足 $u_1 = 4$，$r = \tfrac{3}{2}$。求使 $S_n > 200$ 的最小 $n$。
$$ S_n = \frac{4\bigl((3/2)^n - 1\bigr)}{(3/2) - 1} = 8\bigl((3/2)^n - 1\bigr) > 200. $$ $$ (3/2)^n > 26 \;\Rightarrow\; n > \frac{\log 26}{\log(3/2)} \approx 8.04. $$故 $n = 9$。验证：$S_9 = 8((3/2)^9 - 1) \approx 8(25.629) \approx 205.0 > 200$。✓
Sigma Notation求和记号 SL 1.2
sigma notation）就是求和的缩写。$\displaystyle\sum_{k=1}^{n} a_k = a_1 + a_2 + \cdots + a_n$。IB 所有 sigma 题的通用策略：识别底层是 AP 还是 GP，再套求和公式。不要逐项相加。
index）$k$ 从下界 $m$ 走到上界 $n$，两端含。$\Sigma$ 后的表达式就是每一项关于 $k$ 的公式。
Exponential in $k$: $\sum c \cdot r^{\,k-1}$ — geometric with $u_1 = c$, common ratio $r$.
Mixed: split with linearity, e.g. $\sum (3k + 2 \cdot 5^k) = 3\sum k + 2\sum 5^k$.
$k$ 的指数式：$\sum c \cdot r^{\,k-1}$——等比，$u_1 = c$，公比 $r$。
混合式：用线性性拆开，例如 $\sum (3k + 2 \cdot 5^k) = 3\sum k + 2\sum 5^k$。
Worked Example — Linear summand (AP)例题——一次式被加项（AP）
Evaluate: $\displaystyle\sum_{k=1}^{15}(4k - 3)$.
Arithmetic with $u_1 = 4(1) - 3 = 1$, $u_{15} = 4(15) - 3 = 57$, $n = 15$.
$$ S_{15} = \tfrac{15}{2}(1 + 57) = \tfrac{15}{2}\cdot 58 = 435. $$求值：$\displaystyle\sum_{k=1}^{15}(4k - 3)$。
等差，$u_1 = 4(1) - 3 = 1$，$u_{15} = 4(15) - 3 = 57$，$n = 15$。
$$ S_{15} = \tfrac{15}{2}(1 + 57) = \tfrac{15}{2}\cdot 58 = 435. $$Worked Example — Exponential summand (GP)例题——指数式被加项（GP）
Evaluate: $\displaystyle\sum_{k=1}^{8} 3 \cdot 2^{\,k-1}$.
Geometric with $u_1 = 3$, $r = 2$, $n = 8$.
$$ S_8 = \frac{3(2^8 - 1)}{2 - 1} = 3 \cdot 255 = 765. $$求值：$\displaystyle\sum_{k=1}^{8} 3 \cdot 2^{\,k-1}$。
等比，$u_1 = 3$，$r = 2$，$n = 8$。
$$ S_8 = \frac{3(2^8 - 1)}{2 - 1} = 3 \cdot 255 = 765. $$▸ Going deeper — Index shift▸ 深入——指标平移
Sometimes the lower bound isn't $1$. The trick is to substitute $j = k - m + 1$ so $j$ runs from $1$:
$$ \sum_{k=m}^{n} a_k = \sum_{j=1}^{n - m + 1} a_{j + m - 1}. $$For example, $\sum_{k=3}^{10} 2k = \sum_{j=1}^{8} 2(j+2) = \sum_{j=1}^{8}(2j + 4)$ which is now a clean AP on $j$.
Equivalently, use the "big sum minus small sum" identity:
$$ \sum_{k=m}^{n} a_k = \sum_{k=1}^{n} a_k - \sum_{k=1}^{m-1} a_k. $$Both work; pick whichever leaves cleaner arithmetic.
下界有时并非 $1$。诀窍是令 $j = k - m + 1$，让 $j$ 从 $1$ 开始：
$$ \sum_{k=m}^{n} a_k = \sum_{j=1}^{n - m + 1} a_{j + m - 1}. $$例如 $\sum_{k=3}^{10} 2k = \sum_{j=1}^{8} 2(j+2) = \sum_{j=1}^{8}(2j + 4)$，这是关于 $j$ 的干净 AP。
等价做法是用"大和减小和"恒等式：
$$ \sum_{k=m}^{n} a_k = \sum_{k=1}^{n} a_k - \sum_{k=1}^{m-1} a_k. $$两法皆可，挑算账更轻松的那个。
Infinite Geometric Series无穷等比级数 HL AHL 1.8
infinite geometric series）收敛（converge）当且仅当 $|r| < 1$，其和为
$$ S_\infty = \frac{u_1}{1 - r}. $$
先检验 $|r| < 1$，再用公式。若 $|r| \ge 1$ 仍硬套公式，会给发散（diverge）的级数算出一个有限的"伪和"——Paper 1 这步必扣分。
sum to infinity）▸ Going deeper — Why the condition is $|r| < 1$ and not just $r < 1$▸ 深入——为何条件是 $|r| < 1$ 而非仅 $r < 1$
Recall the partial sum:
$$ S_n = \frac{u_1(1 - r^n)}{1 - r}. $$For $S_n$ to settle to a finite limit as $n \to \infty$, the term $r^n$ must approach a finite value. Examine cases:
- $|r| < 1$: $r^n \to 0$, so $S_n \to \tfrac{u_1}{1 - r}$. ✓ Converges.
- $r = 1$: the formula breaks (denominator is $0$); each term equals $u_1$, so $S_n = nu_1$ which is unbounded if $u_1 \ne 0$.
- $r = -1$: $r^n$ alternates $\{-1, 1, -1, \ldots\}$, never settling. $S_n$ alternates between $u_1$ and $0$.
- $|r| > 1$: $|r^n| \to \infty$, so $|S_n| \to \infty$.
Only the first case yields a finite limit. The condition is therefore $|r| < 1$, strict on both sides.
回顾部分和：
$$ S_n = \frac{u_1(1 - r^n)}{1 - r}. $$$S_n$ 当 $n \to \infty$ 时要收敛到有限值，关键看 $r^n$ 的极限。分情况：
- $|r| < 1$：$r^n \to 0$，故 $S_n \to \tfrac{u_1}{1 - r}$。✓ 收敛。
- $r = 1$：公式失效（分母为 $0$）；每项都等于 $u_1$，$S_n = nu_1$，$u_1 \ne 0$ 时无界。
- $r = -1$：$r^n$ 在 $\{-1, 1, -1, \ldots\}$ 间交替，永不稳定。$S_n$ 在 $u_1$ 和 $0$ 间反复。
- $|r| > 1$：$|r^n| \to \infty$，故 $|S_n| \to \infty$。
只有第一种情形给出有限极限。所以条件就是 $|r| < 1$，两端严格不等。
Worked Example — Recurring decimal as a fraction例题——循环小数化分数
Problem: Express $0.\overline{63} = 0.636363\ldots$ as a fraction in lowest terms.
Write as a geometric series:
$$ 0.\overline{63} = 0.63 + 0.0063 + 0.000063 + \cdots $$$u_1 = 0.63 = \tfrac{63}{100}$, $r = \tfrac{1}{100}$, so $|r| < 1$.
$$ S_\infty = \frac{63/100}{1 - 1/100} = \frac{63/100}{99/100} = \frac{63}{99} = \frac{7}{11}. $$题目：把 $0.\overline{63} = 0.636363\ldots$ 写成最简分数。
写成等比级数：
$$ 0.\overline{63} = 0.63 + 0.0063 + 0.000063 + \cdots $$$u_1 = 0.63 = \tfrac{63}{100}$，$r = \tfrac{1}{100}$，故 $|r| < 1$。
$$ S_\infty = \frac{63/100}{1 - 1/100} = \frac{63/100}{99/100} = \frac{63}{99} = \frac{7}{11}. $$Worked Example — Geometric series with given infinite sum例题——已知无穷和求 GP
Problem: A convergent GP has $u_1 = 12$ and $S_\infty = 18$. Find $r$, and the sum of the first $5$ terms.
$$ S_\infty = \frac{12}{1 - r} = 18 \;\Rightarrow\; 1 - r = \tfrac{12}{18} = \tfrac{2}{3} \;\Rightarrow\; r = \tfrac{1}{3}. $$Check $|r| < 1$. ✓ Then
$$ S_5 = \frac{12\bigl(1 - (1/3)^5\bigr)}{1 - 1/3} = \frac{12(1 - 1/243)}{2/3} = 18 \cdot \tfrac{242}{243} = \tfrac{1452}{81} \approx 17.93. $$(Sanity: $S_5$ should be close to but less than $S_\infty = 18$. ✓)
题目：收敛的 GP 满足 $u_1 = 12$，$S_\infty = 18$。求 $r$ 和前 $5$ 项之和。
$$ S_\infty = \frac{12}{1 - r} = 18 \;\Rightarrow\; 1 - r = \tfrac{12}{18} = \tfrac{2}{3} \;\Rightarrow\; r = \tfrac{1}{3}. $$检验 $|r| < 1$ ✓。再算
$$ S_5 = \frac{12\bigl(1 - (1/3)^5\bigr)}{1 - 1/3} = \frac{12(1 - 1/243)}{2/3} = 18 \cdot \tfrac{242}{243} = \tfrac{1452}{81} \approx 17.93. $$（验算：$S_5$ 应接近但小于 $S_\infty = 18$。✓）
Worked Example — Bouncing ball (classic AP problem)例题——弹跳球（经典应用题）
Problem: A ball is dropped from $5$ m. Each bounce returns to $80\%$ of the previous height. Find the total vertical distance travelled before it comes to rest.
Down distances: $5, 4, 3.2, \ldots$ (a GP with $u_1 = 5$, $r = 0.8$).
Up distances: $4, 3.2, 2.56, \ldots$ (the same GP starting one term later: $u_1 = 4$, $r = 0.8$).
$$ D_{\text{down}} = \frac{5}{1 - 0.8} = 25 \text{ m}, \qquad D_{\text{up}} = \frac{4}{1 - 0.8} = 20 \text{ m}. $$ $$ \text{Total} = 25 + 20 = 45 \text{ m}. $$Common slip: double-counting the first drop or forgetting that the up-and-down distances are different series.
题目：从 $5$ 米高度释放一个球。每次弹回上一高度的 $80\%$。求球静止前在竖直方向上经过的总路程。
下落距离：$5, 4, 3.2, \ldots$（GP，$u_1 = 5$，$r = 0.8$）。
上升距离：$4, 3.2, 2.56, \ldots$（同一 GP 但晚一项起：$u_1 = 4$，$r = 0.8$）。
$$ D_{\text{down}} = \frac{5}{1 - 0.8} = 25 \text{ m}, \qquad D_{\text{up}} = \frac{4}{1 - 0.8} = 20 \text{ m}. $$ $$ \text{Total} = 25 + 20 = 45 \text{ m}. $$常见错误：把首次下落算两遍；或忘了上行与下行是两个不同的级数。
Financial Applications金融应用 SL 1.4
compound interest）本质上是 GP。以每期利率 $r$ 复利 $n$ 期后，$A = P(1 + r)^n$。年利率 $r$、每年复利 $k$ 次、共 $t$ 年时，$A = P\bigl(1 + \tfrac{r}{k}\bigr)^{kt}$。折旧（depreciation）公式相同，但用 $(1 - r)^n$。下笔前先写清楚"每期利率"是多少。
$P$ = principal, $r$ = annual rate (decimal), $k$ = compounds per year, $t$ = years, $A$ = future value.
$P$ = 本金（principal），$r$ = 年利率（小数），$k$ = 每年复利次数，$t$ = 年数，$A$ = 未来值（future value）。
$V_0$ = initial value, $r$ = depreciation rate per period, $n$ = number of periods, $V$ = final value.
$V_0$ = 初值，$r$ = 每期折旧率，$n$ = 期数，$V$ = 终值。
Worked Example — Compound interest, quarterly例题——按季度复利
Problem: $\$5000$ invested at $4\%$ p.a. compounded quarterly. Find the value after $7$ years.
Per-period rate $\tfrac{r}{k} = \tfrac{0.04}{4} = 0.01$; periods $kt = 28$.
$$ A = 5000(1.01)^{28} \approx \$6606.45. $$The balance is the GP $P, P(1.01), P(1.01)^2, \ldots$ ; the formula reads off the term after $28$ compoundings.
题目：$\$5000$ 投资，年利率 $4\%$，按季度（每年 4 次）复利。求 $7$ 年后的本利和。
每期利率 $\tfrac{r}{k} = \tfrac{0.04}{4} = 0.01$；总期数 $kt = 28$。
$$ A = 5000(1.01)^{28} \approx \$6606.45. $$余额本身就是 GP $P, P(1.01), P(1.01)^2, \ldots$；公式给出第 $28$ 期之后的那一项。
Worked Example — Depreciation例题——折旧
Problem: A car bought for $\$24\,000$ depreciates by $15\%$ each year. How long until its value drops below $\$10\,000$?
$$ 24\,000(0.85)^n < 10\,000 \;\Rightarrow\; (0.85)^n < \tfrac{5}{12}. $$Take log (note $\log 0.85 < 0$, so flip the inequality):
$$ n \log 0.85 < \log(5/12) \;\Rightarrow\; n > \frac{\log(5/12)}{\log(0.85)} \approx 5.39. $$So in the $6$th year. (Year $5$ value: $24000 \cdot 0.85^5 \approx \$10{,}649$; Year $6$: $\approx \$9051$.)
题目：$\$24\,000$ 购入的车每年折旧 $15\%$。多少年后价值首次低于 $\$10\,000$？
$$ 24\,000(0.85)^n < 10\,000 \;\Rightarrow\; (0.85)^n < \tfrac{5}{12}. $$取对数（注意 $\log 0.85 < 0$，除以负数要变号）：
$$ n \log 0.85 < \log(5/12) \;\Rightarrow\; n > \frac{\log(5/12)}{\log(0.85)} \approx 5.39. $$故在第 $6$ 年。（第 $5$ 年值 $24000 \cdot 0.85^5 \approx \$10{,}649$；第 $6$ 年值 $\approx \$9051$。）
▸ Going deeper — Annuity formulas (HL Paper 2 territory)▸ 深入——年金公式（HL Paper 2 范畴）
An annuity is a sequence of equal payments $P$ made each period at rate $r$. The future value of an annuity (ordinary, with payments at end of each period) is the GP-sum
$$ FV = P + P(1+r) + P(1+r)^2 + \cdots + P(1+r)^{n-1} = P \cdot \frac{(1+r)^n - 1}{r}. $$The present value is
$$ PV = \frac{P}{(1+r)} + \frac{P}{(1+r)^2} + \cdots + \frac{P}{(1+r)^n} = P \cdot \frac{1 - (1+r)^{-n}}{r}. $$Both follow directly from the GP partial-sum formula — no need to memorize them as separate facts. On Paper 2, IB usually wants you to set up the GP from scratch and then sum it; calculator handles the arithmetic.
Annuity-due variant: if payments are at the start of each period instead, multiply by $(1+r)$ — every payment earns one extra period of interest.
年金（annuity）是一系列等额支付：每期支付 $P$，利率为 $r$。普通年金（每期末支付）的未来值就是 GP 之和
现值（present value）为
两者直接由 GP 部分和公式推出——不必当成独立公式背。Paper 2 通常希望你从头建立 GP 再求和；算账交给计算器。
期初年金（annuity-due）：若每期支付改在期初，整体乘 $(1+r)$——每笔多享一期利息。
Mixed & Hybrid Patterns混合与综合题型
Pattern 1 — Both AP and GP题型 1——同时是 AP 又是 GP
▸ Going deeper — Quick proof▸ 深入——快速证明
AP: $u_{n+1} = u_n + d$. GP: $u_{n+1} = r\, u_n$. Equate for every $n$:
$$ u_n + d = r\, u_n \;\Longrightarrow\; d = (r - 1)\, u_n. $$RHS depends on $n$, LHS does not — only possible if $r = 1$ (so $d = 0$) or $u_n$ is constant. Either case gives a constant sequence. $\blacksquare$
AP：$u_{n+1} = u_n + d$。GP：$u_{n+1} = r\, u_n$。对每个 $n$ 取等：
$$ u_n + d = r\, u_n \;\Longrightarrow\; d = (r - 1)\, u_n. $$右边依赖 $n$，左边不依赖——只有当 $r = 1$（即 $d = 0$）或 $u_n$ 是常数时才可能成立。两种情形都给出常数列。$\blacksquare$
Pattern 2 — Inserting means between given terms题型 2——在已知两项间插入中项
For geometric means, replace differences with ratios: $r = \bigl(\tfrac{b}{a}\bigr)^{1/(k+1)}$, requiring $a, b$ same sign.
arithmetic means）：得到一个 AP，$u_1 = a$，$u_{k+2} = b$，故 $b - a = (k+1)d \Rightarrow d = \tfrac{b - a}{k + 1}$。中项为 $a + d, a + 2d, \ldots, a + kd$。插入等比中项（
geometric means）时，把差换成比：$r = \bigl(\tfrac{b}{a}\bigr)^{1/(k+1)}$，要求 $a, b$ 同号。
Worked Example — System with sum + term condition例题——同时含和与项的方程组
Problem: A GP has $u_1 + u_2 + u_3 = 14$ and $u_3 = 8$. Find all possible values of $r$.
Use $u_2 = u_1 r$, $u_3 = u_1 r^2 = 8$:
$$ u_1(1 + r + r^2) = 14 \quad\text{and}\quad u_1 r^2 = 8. $$Divide:
$$ \frac{1 + r + r^2}{r^2} = \frac{14}{8} = \tfrac{7}{4} \;\Rightarrow\; 4(1 + r + r^2) = 7 r^2 \;\Rightarrow\; 3 r^2 - 4 r - 4 = 0. $$Quadratic formula: $r = \tfrac{4 \pm \sqrt{16 + 48}}{6} = \tfrac{4 \pm 8}{6}$, giving $r = 2$ or $r = -\tfrac{2}{3}$.
Both are valid. Don't drop the negative root without checking — IB markschemes credit both.
题目：GP 满足 $u_1 + u_2 + u_3 = 14$ 且 $u_3 = 8$。求 $r$ 的所有可能值。
用 $u_2 = u_1 r$，$u_3 = u_1 r^2 = 8$：
$$ u_1(1 + r + r^2) = 14 \quad\text{and}\quad u_1 r^2 = 8. $$两式相除：
$$ \frac{1 + r + r^2}{r^2} = \frac{14}{8} = \tfrac{7}{4} \;\Rightarrow\; 4(1 + r + r^2) = 7 r^2 \;\Rightarrow\; 3 r^2 - 4 r - 4 = 0. $$求根公式：$r = \tfrac{4 \pm \sqrt{16 + 48}}{6} = \tfrac{4 \pm 8}{6}$，得 $r = 2$ 或 $r = -\tfrac{2}{3}$。
两根都有效。别不验证就丢掉负根——IB markscheme 两根都给分。
Pattern 3 — Telescoping (light intro)题型 3——裂项相消（简介）
telescoping）和式中相邻项互相抵消。经典例子：$\tfrac{1}{k(k+1)} = \tfrac{1}{k} - \tfrac{1}{k+1}$。于是
$$ \sum_{k=1}^{n} \frac{1}{k(k+1)} = \left(1 - \tfrac{1}{2}\right) + \left(\tfrac{1}{2} - \tfrac{1}{3}\right) + \cdots + \left(\tfrac{1}{n} - \tfrac{1}{n+1}\right) = 1 - \tfrac{1}{n+1}. $$
中间项几乎全部抵消，只剩首末两项。出现部分分式（partial fractions）时最常用。
Exam Strategy & Common Pitfalls考试策略与常见陷阱
- $u_n = u_1 + (n-1)d$ & both forms of $S_n$ for an AP
- $u_n = u_1 r^{n-1}$ & both forms of $S_n$ for a GP (and the $r=1$ exception)
- $S_\infty = \tfrac{u_1}{1-r}$, valid only when $|r| < 1$
- Compound interest $A = P(1 + r/k)^{kt}$
- Identity: $u_m - u_n = (m-n)d$ (AP), $u_m / u_n = r^{m-n}$ (GP)
- Sigma linearity properties
- AP 的 $u_n = u_1 + (n-1)d$ 和两式 $S_n$
- GP 的 $u_n = u_1 r^{n-1}$ 和两式 $S_n$（以及 $r=1$ 的例外）
- $S_\infty = \tfrac{u_1}{1-r}$，仅在 $|r| < 1$ 时有效
- 复利公式 $A = P(1 + r/k)^{kt}$
- 恒等式：$u_m - u_n = (m-n)d$（AP），$u_m / u_n = r^{m-n}$（GP）
- $\Sigma$ 的线性性质
- Why $|r| < 1$ is the exact convergence condition
- Gauss-pairing derivation of $S_n^{\text{AP}}$
- The $S_n - r S_n$ trick for $S_n^{\text{GP}}$
- Why both $r = 1$ and $r = -1$ are excluded from $S_\infty$
- Recognizing AP / GP under sigma — picking the right sum formula
- Why the per-period rate is $r/k$, not $r$, in compound interest
- 为何 $|r| < 1$ 是精确的收敛条件
- $S_n^{\text{AP}}$ 的高斯配对推导
- $S_n^{\text{GP}}$ 的 $S_n - r S_n$ 技巧
- 为何 $r = 1$ 与 $r = -1$ 同被 $S_\infty$ 排除
- 识别 sigma 后面的 AP / GP——选对求和公式
- 为何复利里每期利率是 $r/k$ 而不是 $r$
Common Pitfalls常见陷阱
2. Off-by-one on the index — confusing $u_5$ ("$5$th term") with $u_4 = u_1 + 3d$ ("$4$ steps from $u_1$").
3. Forgetting the $r = 1$ case in geometric sums (formula divides by zero).
4. Mis-identifying a sequence: $\{1, 2, 4, 7, 11, \ldots\}$ has differences $1, 2, 3, 4, \ldots$ — that's quadratic, neither AP nor GP.
5. In compound interest, plugging the annual rate without dividing by the compounding frequency.
6. Bouncing-ball: counting the initial drop twice or forgetting that "up" and "down" distances form different series.
7. On a sigma question, expanding term-by-term instead of recognizing the AP/GP — works but eats clock.
2. 指标差一——把 $u_5$（"第 5 项"）和 $u_4 = u_1 + 3d$（"从 $u_1$ 走 4 步"）搞混。
3. 忘记 GP 求和的 $r = 1$ 情形（公式分母为零）。
4. 误判数列性质：$\{1, 2, 4, 7, 11, \ldots\}$ 的差是 $1, 2, 3, 4, \ldots$——这是二次，既非 AP 也非 GP。
5. 复利题里没把年利率除以复利频率，直接套年利率。
6. 弹跳球：把首次下落算两遍，或忘了上行与下行是两个不同的级数。
7. sigma 题逐项展开而不识别底层 AP/GP——也能做对，但浪费考试时间。
Paper 2 (calc): long compound-interest problems, "smallest $n$ such that" with non-trivial bases, multi-stage financial problems. Don't over-use the calculator on the AP/GP set-up — they're testing whether you can identify the structure first.
Paper 2（可用计算器）：长复利题、底数非平凡的"最小 $n$"题、多阶段金融题。AP/GP 建模阶段别过度依赖计算器——他们要先考你识别结构的能力。
Flashcards闪卡
GP: $u_m / u_n = r^{m-n}$
$\sum c\, a_k = c \sum a_k$
$\sum_{k=1}^{n} c = nc$
so dividing flips the inequality.$|r| < 1$ 时 $\log r < 0$，
除以负数不等号变号。
Unit A1 — Practice Quiz单元 A1——练习测验
Ten mixed-difficulty items. Your score updates in real time at the top of the page. Aim for 8/10 before exam day.十道难度不一的题。得分实时显示在页面顶部。考前目标 8/10。
Readiness Checklist备考清单
Click each item you've mastered. Aim for 100% before exam day. Items marked HL are HL-only.点击你已经掌握的条目。考前目标 100%。标有 HL 的为 HL 专属内容。
- State and use $u_n = u_1 + (n-1)d$ for an AP写出并应用等差数列（
AP）的通项 $u_n = u_1 + (n-1)d$ - Solve $2\times 2$ systems in $u_1$ and $d$ from two AP conditions由两条 AP 条件解关于 $u_1$ 与 $d$ 的 $2\times 2$ 方程组
- Use both forms of $S_n$ for an AP and choose the cleaner one熟练 AP 求和 $S_n$ 的两种形式，按题目条件选用更简洁的一种
- Reproduce the Gauss-pairing derivation of $S_n^{\text{AP}}$能复现 $S_n^{\text{AP}}$ 的高斯配对推导
- State and use $u_n = u_1 r^{n-1}$ for a GP写出并应用等比数列（
GP）的通项 $u_n = u_1 r^{n-1}$ - Use $u_m/u_n = r^{m-n}$ to recover $r$ from two terms利用 $u_m/u_n = r^{m-n}$ 由任意两项反推公比 $r$
- State and use $S_n^{\text{GP}}$ in both forms; handle the $r = 1$ case写出并应用 GP 求和 $S_n^{\text{GP}}$ 的两种形式，能单独处理 $r = 1$ 的退化情形
- Reproduce the $S_n - r S_n$ derivation of the GP sum formula能复现 $S_n - r S_n$ 推导出 GP 求和公式的过程
- Translate sigma notation into AP or GP form before summing求和前先把求和记号（
sigma notation）化为 AP 或 GP 形式 - Apply linearity and index-shift identities for sigma sums熟练 $\Sigma$ 的线性性和换元（index shift）恒等式
- HL Recognize when an infinite GP converges (exact condition $|r| < 1$)辨认无穷 GP 收敛（
convergence）的精确条件 $|r| < 1$ - HL Compute $S_\infty$ and convert recurring decimals into fractions能计算 $S_\infty$ 并将循环小数化为分数
- Set up and solve compound-interest / depreciation problems with the right per-period rate能正确设定每期利率（per-period rate）解复利与折旧问题
- Recognize and resolve mixed AP+GP and "both AP and GP" patterns能识别并求解 AP 与 GP 混合题型及"同时为 AP 和 GP"的特例
IB Paper-Style PracticeIB 试卷风格练习
IB exam-style questions across Paper 1A (short response, no calc), Paper 1B (extended response, no calc), and Paper 2 (calculator). EMH difficulty mix with fully worked solutions. Use this after the in-page quiz and flashcards.
IB 考试风格题，涵盖 Paper 1A（短答，无计算器）、Paper 1B（长答，无计算器）、Paper 2（可用计算器）。难度按 EMH 分级，配完整解题过程。建议在做完本页测验与闪卡后再来。