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。
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。
How to use this guide本指南使用说明
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。