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Sequences and Series · Solutions数列与级数 · 详解

Companion to the Practice Set · Mark-by-mark walkthroughs · SAT / AP-Feeder / ON / BC styles练习题配套答案 · 逐分讲解 · SAT / AP 衔接 / 安 / 卑诗省考风格

EASY易 MEDIUM中 HARD难 🇺🇸 US美 🇨🇦 ON安 🇨🇦 BC卑 SAT-style MCQSAT 风格选择题 AP-feeder FRQAP 衔接简答题 ON Provincial-style安大略省考风格 BC Provincial-style卑诗省考风格 Honors荣誉级
PART I  ·  SHORT RESPONSE · SOLUTIONS第一部分  ·  短答题 · 详解SAT MCQ + ON/BC short answer · 22 marksSAT 选择题 + 安/卑省考短答 · 共 22 分

Section A · Worked SolutionsA 部分 · 详解

Q1EASY易 🇺🇸 US美 SAT-style MCQSAT 风格选择题 §1 Recursive vs Explicit递推公式与通项公式 · HSF-IF.A.3 [3 marks][3 分]

Recursion $a_1 = 4$, $a_{n+1} = a_n + 6$, convert to explicit.递推 $a_1 = 4$、$a_{n+1} = a_n + 6$，化为通项公式。

Answer:答案：  (C)  $a_n = 6n - 2$

(a) Read the recursion as arithmetic将递推识别为等差 M1·A1·A1

$a_{n+1} - a_n = 6$ for every $n$, so the sequence is arithmetic with first term $a_1 = 4$ and common difference $d = 6$. The explicit form is对每个 $n$ 都有 $a_{n+1} - a_n = 6$，故该数列为等差数列，首项 $a_1 = 4$，公差 $d = 6$。通项公式为 $$ a_n \;=\; a_1 + (n - 1) d \;=\; 4 + 6(n - 1) \;=\; 6n - 2. $$ Spot-check at $n = 1$: $6(1) - 2 = 4$. $\checkmark$  At $n = 2$: $6(2) - 2 = 10 = a_1 + 6$. $\checkmark$ Matches option (C).$n = 1$ 时核对：$6(1) - 2 = 4$。 $\checkmark$  $n = 2$ 时：$6(2) - 2 = 10 = a_1 + 6$。 $\checkmark$ 对应选项 (C)。
Why the wrong choices fail.错误选项剖析。
  • (A) $a_n = 6n + 4$, uses $a_0 = 4$ instead of $a_1 = 4$; gives $a_1 = 10$, off by one term (the classic indexing slip).：把 $4$ 当成 $a_0$ 而不是 $a_1$，得 $a_1 = 10$，整体错一项（典型的下标偏移错误）。
  • (B) $a_n = 4n + 6$, swaps the roles of $a_1$ and $d$: writes "first term times $n$ plus the step" instead of "step times $(n-1)$ plus the first term".：把首项与公差的角色对调，写成"首项乘 $n$ 再加步长"，而不是"步长乘 $(n-1)$ 再加首项"。
  • (D) $a_n = 4 \cdot 6^{n-1}$, reads the $+6$ as a multiplicative ratio (geometric), not an additive step (arithmetic). The $+$ in $a_n + 6$ is the giveaway: this is arithmetic, not geometric.：把 $+6$ 当成乘法公比（等比），而不是加法公差（等差）。$a_n + 6$ 中的加号才是判别提示——这是等差而非等比。
Recursive vs explicit is a tradeoff: recursive is shorter to write, explicit is shorter to evaluate.递推与通项是一种权衡：递推写法短，通项求值快。 $a_{n+1} = a_n + 6$ encodes "the rule" but requires $n - 1$ additions to reach $a_{100}$; the explicit form $a_n = 6n - 2$ jumps straight there in one substitution. Translation rule for arithmetic recursions $a_{n+1} = a_n + d$: the explicit form is always $a_n = a_1 + (n-1)d$, never $a_n = a_1 + nd$ (the $(n-1)$ encodes "we already counted $a_1$, only $n-1$ more $d$'s to go"). This off-by-one is the single most common Sprint-1 indexing error.$a_{n+1} = a_n + 6$ 编码的是"规则"，要算到 $a_{100}$ 需要做 $n - 1$ 次加法；通项 $a_n = 6n - 2$ 一次代入即得。等差递推 $a_{n+1} = a_n + d$ 的转换规则永远是 $a_n = a_1 + (n-1)d$，绝不能写成 $a_n = a_1 + nd$（其中 $(n-1)$ 表示"$a_1$ 已经算了一项，只剩 $n-1$ 个 $d$"）。这个差一错误是 Sprint 1 最常见的下标失误。
Q2EASY易 🇺🇸 US美 SAT-style MCQSAT 风格选择题 §2 Arithmetic $n$-th Term等差数列通项 · HSF-BF.A.2 [3 marks][3 分]

$25$th term of $-7, -3, 1, 5, \dots$求 $-7, -3, 1, 5, \dots$ 的第 $25$ 项。

Answer:答案：  (A)  $89$

(a) Identify $a_1$ and $d$辨别 $a_1$ 与 $d$ M1

$a_1 = -7$ and $d = -3 - (-7) = 4$ (constant across all consecutive pairs: $1 - (-3) = 4$, $5 - 1 = 4$). Arithmetic, confirmed.$a_1 = -7$，$d = -3 - (-7) = 4$（相邻各对差恒定：$1 - (-3) = 4$，$5 - 1 = 4$）。确认是等差数列。

(b) Apply the $n$-th term formula代入通项公式 A1·A1

$$ a_{25} \;=\; a_1 + (25 - 1) d \;=\; -7 + 24(4) \;=\; -7 + 96 \;=\; 89. $$ Matches option (A).对应选项 (A)。
Why the wrong choices fail.错误选项剖析。
  • (B) $93$, used $25 d$ instead of $(25 - 1) d$: $-7 + 25(4) = 93$. The off-by-one indexing trap.：用 $25 d$ 而非 $(25 - 1) d$：$-7 + 25(4) = 93$。典型的下标差一陷阱。
  • (C) $85$, used $23 d$: $-7 + 23(4) = 85$. Off-by-one the other way (counted $a_1$ as "step 1").：用 $23 d$：$-7 + 23(4) = 85$。反向差一（误把 $a_1$ 当作第一步）。
  • (D) $100$, added $-7 + 100$ or computed $25 \cdot 4 = 100$ alone, forgetting to add $a_1$.：直接算 $25 \cdot 4 = 100$ 而忘了加 $a_1$，或写成 $-7 + 100$ 的算错版本。
Lock the $(n-1)$ in muscle memory.把 $(n-1)$ 写进肌肉记忆。 $a_n = a_1 + (n - 1) d$, full stop, the SAT seeds every wrong choice around the off-by-one errors $n d$ and $(n - 2) d$. One-line sanity check before circling: at $n = 1$, the formula must return $a_1$. With $(n - 1) d$ it does ($(1 - 1) d = 0$); with $n d$ it does not. That five-second check eliminates (B), (D), and any "off-by-one" distractor instantly.$a_n = a_1 + (n - 1) d$，就是这个写法。SAT 的每个错误选项都围绕 $n d$ 与 $(n - 2) d$ 这类差一错误设陷阱。圈答案前做一句话核对：当 $n = 1$ 时公式必须回到 $a_1$。$(n - 1) d$ 满足（$(1 - 1) d = 0$）；$n d$ 不满足。这一秒钟检查立刻排除 (B)、(D) 及所有"差一"干扰项。
Q3EASY易 🇺🇸 US美 SAT-style MCQSAT 风格选择题 §4 Geometric $n$-th Term等比数列通项 · HSF-BF.A.2 [3 marks][3 分]

Geometric with $a_1 = 5$, $r = 2$, find $a_8$.等比数列 $a_1 = 5$、$r = 2$，求 $a_8$。

Answer:答案：  (A)  $640$

(a) Apply the geometric $n$-th term formula代入等比数列通项公式 M1·A1·A1

$$ a_8 \;=\; a_1 \cdot r^{8 - 1} \;=\; 5 \cdot 2^{7} \;=\; 5 \cdot 128 \;=\; 640. $$ Matches option (A).对应选项 (A)。
Why the wrong choices fail.错误选项剖析。
  • (B) $320$, used $2^{6}$ instead of $2^{7}$: $5 \cdot 64 = 320$. Another off-by-one: forgot that $a_1$ has exponent $0$, so $a_8$ has exponent $7$.：用 $2^{6}$ 而非 $2^{7}$：$5 \cdot 64 = 320$。又一个差一：忘了 $a_1$ 对应指数 $0$，所以 $a_8$ 对应指数 $7$。
  • (C) $1280$, used $2^{8}$: $5 \cdot 256 = 1280$. Treated $r^{n}$ as the formula instead of $r^{n-1}$.：用 $2^{8}$：$5 \cdot 256 = 1280$。把公式当成 $r^{n}$ 而非 $r^{n-1}$。
  • (D) $80$, computed $5 \cdot 2^{4}$, perhaps confusing position-in-sequence with halving the exponent.：算成 $5 \cdot 2^{4}$，可能把项数与指数对半搞混。
The geometric exponent is $n - 1$, mirroring the arithmetic step count.等比的指数是 $n - 1$，与等差的步数同理。 $a_n = a_1 \cdot r^{n-1}$. Same off-by-one logic as arithmetic: $a_1$ already counts once, so only $n - 1$ multiplications by $r$ remain. Sanity check at $n = 1$: $a_1 = a_1 \cdot r^{0} = a_1 \cdot 1 = a_1$. $\checkmark$ This $(n-1)$ exponent re-appears in compound-interest formulas (Q10) and continuous exponential-to-geometric translations (Q11). Lock it now.$a_n = a_1 \cdot r^{n-1}$。与等差同样的差一逻辑：$a_1$ 已经算了一次，故只剩 $n - 1$ 次乘以 $r$。$n = 1$ 时核对：$a_1 = a_1 \cdot r^{0} = a_1 \cdot 1 = a_1$。 $\checkmark$ 这个 $(n-1)$ 指数会在复利公式（Q10）以及连续指数函数到等比数列的转换（Q11）中再次出现。现在就锁定。
Q4MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §2 Arithmetic $n$-th Term等差数列通项 · MCR3U C2.2 [6 marks][6 分]

$t_4 = 11$, $t_{11} = 46$. (a) $d$. (b) $a$ and explicit. (c) Which term equals $156$?$t_4 = 11$、$t_{11} = 46$。(a) 求 $d$。(b) 求 $a$ 与通项。(c) 第几项等于 $156$？

Answer:答案：  (a) $d = 5$  ·  (b) $a = -4$, $t_n = -4 + 5(n - 1) = 5n - 9$  ·  (c) $n = 33$

(a) Common difference from two given terms由两已知项求公差 M1·A1

Between $t_4$ and $t_{11}$ are $11 - 4 = 7$ steps of size $d$, so $t_{11} - t_4 = 7d$:$t_4$ 与 $t_{11}$ 之间有 $11 - 4 = 7$ 步公差 $d$，故 $t_{11} - t_4 = 7d$： $$ 46 - 11 \;=\; 7d \;\Longrightarrow\; d \;=\; \frac{35}{7} \;=\; 5. $$

(b) First term and explicit formula首项与通项公式 M1·A1

$t_4 = a + 3d = a + 15 = 11 \Rightarrow a = -4$. Explicit:由 $t_4 = a + 3d = a + 15 = 11$ 得 $a = -4$。通项为： $$ t_n \;=\; -4 + 5(n - 1) \;=\; 5n - 9. $$ Sanity check: $t_{11} = 5(11) - 9 = 46$. $\checkmark$核对：$t_{11} = 5(11) - 9 = 46$。 $\checkmark$

(c) Solve $t_n = 156$求解 $t_n = 156$ M1·A1

$5n - 9 = 156 \Rightarrow 5n = 165 \Rightarrow n = 33$. So $t_{33} = 156$.$5n - 9 = 156 \Rightarrow 5n = 165 \Rightarrow n = 33$。故 $t_{33} = 156$。
"Distance between two known terms equals (gap)$\times d$", bypass the system of equations."两已知项之差等于（间隔数）$\times d$"——可绕过联立方程。 The textbook move is to write $t_4 = a + 3d$ and $t_{11} = a + 10d$ as a $2 \times 2$ system and solve. The faster move: $t_{11} - t_4 = (10 - 3) d = 7 d$, isolating $d$ in one line. The general formula is $t_p - t_q = (p - q) d$ for arithmetic, $t_p / t_q = r^{p - q}$ for geometric, both follow from the explicit form. MCR3U C2.2 examiners reward the explicit citation of this shortcut.课本式做法是把 $t_4 = a + 3d$ 与 $t_{11} = a + 10d$ 写成 $2 \times 2$ 方程组求解。更快的做法是：$t_{11} - t_4 = (10 - 3) d = 7 d$，一行分离出 $d$。一般地：等差数列有 $t_p - t_q = (p - q) d$，等比数列有 $t_p / t_q = r^{p - q}$，都从通项公式直接得到。MCR3U C2.2 阅卷人鼓励显式引用这一捷径。
Q5MEDIUM中 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §5 Finite Geometric Sum有限等比求和 · BC PC12 geometric seriesBC PC12 等比级数 [7 marks][7 分]

Series $3 + 12 + 48 + 192 + \dots$. (a) $a_1, r$. (b) Explicit $a_n$. (c) $S_7$ via formula. (d) Why $r \ne 1$?级数 $3 + 12 + 48 + 192 + \dots$。(a) $a_1, r$。(b) 通项 $a_n$。(c) 用公式求 $S_7$。(d) 为何 $r \ne 1$？

Answer:答案：  (a) $a_1 = 3$, $r = 4$  ·  (b) $a_n = 3 \cdot 4^{n - 1}$  ·  (c) $S_7 = 16383$  ·  (d) formula divides by $r - 1$; if $r = 1$ the denominator vanishes (and the series is just $n a_1$ anyway)公式以 $r - 1$ 为分母；若 $r = 1$ 则分母为零（此时级数本就是 $n a_1$）

(a) Identify parameters辨别参数 A1

$a_1 = 3$. Ratio test: $12/3 = 4$, $48/12 = 4$, $192/48 = 4$, constant, so $r = 4$.$a_1 = 3$。比值检验：$12/3 = 4$、$48/12 = 4$、$192/48 = 4$，恒定，故 $r = 4$。

(b) Explicit formula通项公式 A1

$a_n = a_1 \cdot r^{n - 1} = 3 \cdot 4^{n - 1}$.

(c) Compute $S_7$计算 $S_7$ M1·A1·A1

$$ S_7 \;=\; \frac{a_1 (r^7 - 1)}{r - 1} \;=\; \frac{3 (4^7 - 1)}{4 - 1} \;=\; \frac{3 (16384 - 1)}{3} \;=\; 16384 - 1 \;=\; 16383. $$ The factor of $3$ cancels neatly because $r - 1 = 3$ here, a happy coincidence; in general the cancellation will not be exact. Spot-check by direct addition: $3 + 12 + 48 + 192 + 768 + 3072 + 12288 = 16383$. $\checkmark$$3$ 之所以恰好约去，是因为此处 $r - 1 = 3$——一个巧合；一般情况下不会刚好约去。直接相加核对：$3 + 12 + 48 + 192 + 768 + 3072 + 12288 = 16383$。 $\checkmark$

(d) Why $r \ne 1$为何 $r \ne 1$ R1·A1

The formula has $r - 1$ in the denominator; if $r = 1$ then we would be dividing by zero. Underlying reason: when $r = 1$ every term equals $a_1$, so $S_n = n \cdot a_1$ trivially, there is no need for the closed-form, and the derivation $S_n - r S_n = a_1 - a_1 r^n$ collapses to $0 = 0$ (no information). The formula is a geometric-series identity; the $r = 1$ case is the degenerate constant-series case handled separately.公式分母为 $r - 1$；若 $r = 1$ 则会出现零除。更深层的原因：当 $r = 1$ 时每一项都等于 $a_1$，故 $S_n = n \cdot a_1$ 是显然结论，不需要闭式公式；而推导 $S_n - r S_n = a_1 - a_1 r^n$ 在此退化为 $0 = 0$（无信息）。该公式是等比级数恒等式，$r = 1$ 是退化的常数级数，需单独处理。
Two equivalent forms of the geometric sum; pick the one with positive numerator.等比求和的两种等价形式：选分子为正的那种。 $S_n = \dfrac{a_1 (r^n - 1)}{r - 1} = \dfrac{a_1 (1 - r^n)}{1 - r}$, algebraically identical (multiply top and bottom by $-1$). Convention: use $(r^n - 1)/(r - 1)$ when $r > 1$ to keep numerator and denominator both positive; use $(1 - r^n)/(1 - r)$ when $|r| < 1$ to keep both positive (and to match the infinite-sum formula $a_1/(1 - r)$ as $n \to \infty$). Choosing the right form first eliminates sign errors downstream, BC PC12 markers expect either form, but inconsistent sign tracking loses A1s.$S_n = \dfrac{a_1 (r^n - 1)}{r - 1} = \dfrac{a_1 (1 - r^n)}{1 - r}$，代数等价（同时乘以 $-1$ 即可）。约定：当 $r > 1$ 时用 $(r^n - 1)/(r - 1)$ 保持分子分母同为正；当 $|r| < 1$ 时用 $(1 - r^n)/(1 - r)$ 同样保持正号（并与 $n \to \infty$ 时的无穷和公式 $a_1/(1 - r)$ 自然衔接）。先选对形式可避免后续符号错误；BC PC12 阅卷两种形式都接受，但符号追踪不一致会丢 A1。
PART II  ·  EXTENDED RESPONSE · SOLUTIONS第二部分  ·  长答题 · 详解AP-feeder FRQ + honors · 36 marksAP 衔接简答题 + 荣誉级 · 共 36 分

Section B · Worked SolutionsB 部分 · 详解

Q6MEDIUM中 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §3 Arithmetic Series (Gauss)等差级数（高斯求和） · HSF-BF.A.2 [8 marks][8 分]

Theatre: row 1 has $18$ seats; each row adds $3$; $20$ rows total. (a) $a_1, d$, explicit. (b) Back-row seats. (c) Total via $\tfrac{n}{2}(a_1 + a_n)$. (d) Cross-check via $\tfrac{n}{2}(2a_1 + (n-1)d)$ + Gauss comment.剧院：第 1 排 $18$ 座，每排加 $3$ 座，共 $20$ 排。(a) $a_1, d$、通项。(b) 末排座位数。(c) 用 $\tfrac{n}{2}(a_1 + a_n)$ 求总数。(d) 用 $\tfrac{n}{2}(2a_1 + (n-1)d)$ 核对并说明高斯论证。

Answer:答案：  (a) $a_1 = 18$, $d = 3$, $a_n = 18 + 3(n - 1) = 3n + 15$  ·  (b) $a_{20} = 75$ seats$a_{20} = 75$ 座  ·  (c) $S_{20} = 930$ seats$S_{20} = 930$ 座  ·  (d) same $930$; the two forms are equivalent via Gauss-pairing同为 $930$；两种形式由高斯配对论证等价

(a) Identify parameters and write explicit form辨别参数并写出通项 A1·A1

$a_1 = 18$ (row 1), $d = +3$ (each new row adds $3$ seats). Explicit:$a_1 = 18$（第 1 排），$d = +3$（每排多 $3$ 座）。通项为： $$ a_n \;=\; 18 + 3(n - 1) \;=\; 3n + 15. $$

(b) Back row末排 M1·A1

$a_{20} = 3(20) + 15 = 60 + 15 = 75$ seats.$a_{20} = 3(20) + 15 = 60 + 15 = 75$ 座。

(c) Total via "average $\times$ count"用"平均数 $\times$ 项数"求总数 M1·A1

$$ S_{20} \;=\; \frac{n}{2}(a_1 + a_n) \;=\; \frac{20}{2}(18 + 75) \;=\; 10 \cdot 93 \;=\; 930 \text{ \text{座}}. $$

(d) Cross-check via parameter form + Gauss comment用参数形式核对 + 高斯论证说明 M1·A1

$$ S_{20} \;=\; \frac{n}{2}\bigl(2 a_1 + (n - 1) d\bigr) \;=\; \frac{20}{2}\bigl(36 + 19 \cdot 3\bigr) \;=\; 10 (36 + 57) \;=\; 10 \cdot 93 \;=\; 930. $$ $\checkmark$ identical answer. Why the two forms agree (Gauss's pairing argument): write the sum $S$ forward and backward, then add:$\checkmark$ 结果一致。两种形式等价的依据（高斯配对论证）：把和 $S$ 正写与倒写，再相加： $$ S = a_1 + a_2 + \cdots + a_n, \qquad S = a_n + a_{n-1} + \cdots + a_1, $$ $$ 2 S = (a_1 + a_n) + (a_2 + a_{n-1}) + \cdots + (a_n + a_1). $$ Each of the $n$ pairs sums to the same value $a_1 + a_n$ (because the increments $+d$ on one side are cancelled by $-d$ on the other), so $2 S = n (a_1 + a_n)$, giving $S = \tfrac{n}{2}(a_1 + a_n)$. Substituting $a_n = a_1 + (n - 1) d$ converts this into the parameter form.每一对之和都等于 $a_1 + a_n$（因为一侧的 $+d$ 与另一侧的 $-d$ 相互抵消），故 $2 S = n (a_1 + a_n)$，即 $S = \tfrac{n}{2}(a_1 + a_n)$。再代入 $a_n = a_1 + (n - 1) d$ 即转化为参数形式。
Two forms, two uses: Gauss-pairs vs. average-times-count.两种形式，两种用途：高斯配对 vs. 平均数乘项数。 $S_n = \tfrac{n}{2}(a_1 + a_n)$ is the "Gauss-pairs" form, fastest when you already know $a_1$ and $a_n$. It also reads as "average term $\times$ number of terms," because $(a_1 + a_n)/2$ is the average value of an arithmetic sequence (the median equals the mean). $S_n = \tfrac{n}{2}(2 a_1 + (n - 1) d)$ is the "parameter form", fastest when you only know $a_1$ and $d$ (no need to first compute $a_n$). AP graders accept either, but if your inputs are the parameters, going through $a_n$ first costs you nothing and provides an automatic cross-check.$S_n = \tfrac{n}{2}(a_1 + a_n)$ 是"高斯配对"形式，当你已知 $a_1$ 与 $a_n$ 时最快。它也可以读作"平均项 $\times$ 项数"，因为 $(a_1 + a_n)/2$ 正是等差数列的平均值（中位数等于平均数）。$S_n = \tfrac{n}{2}(2 a_1 + (n - 1) d)$ 是"参数形式"，当你只知道 $a_1$ 与 $d$ 时最快（不需先求 $a_n$）。AP 阅卷两种都接受，但若你的输入是参数，先算 $a_n$ 也不会损失时间，还顺带得到一次自动核对。
Q7MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §3 Arithmetic Series等差级数 · MCR3U C2.2 [8 marks][8 分]

$7 + 11 + 15 + 19 + \dots + 247$. (a) $a_1, d$. (b) Number of terms $n$. (c) Sum. (d) $S(10) - S(5)$ in function notation + interpret.$7 + 11 + 15 + 19 + \dots + 247$。(a) $a_1, d$。(b) 项数 $n$。(c) 求和。(d) 用函数记号写 $S(10) - S(5)$ 并解释。

Answer:答案：  (a) $a_1 = 7$, $d = 4$  ·  (b) $n = 61$  ·  (c) $S = 7747$  ·  (d) $S(10) - S(5) = a_6 + a_7 + \cdots + a_{10}$ = sum of the next $5$ terms after the first $5$; numerically $190$$S(10) - S(5) = a_6 + a_7 + \cdots + a_{10}$ 等于前 $5$ 项之后接续 $5$ 项之和；数值为 $190$

(a) Parameters参数 A1

$a_1 = 7$ and $d = 11 - 7 = 4$ (check: $15 - 11 = 4$, $19 - 15 = 4$. $\checkmark$).$a_1 = 7$，$d = 11 - 7 = 4$（核对：$15 - 11 = 4$、$19 - 15 = 4$。 $\checkmark$）。

(b) Count the terms数项数 M1·A1·A1

Solve $a_n = 247$ for $n$:求解 $a_n = 247$ 中的 $n$： $$ 7 + 4(n - 1) \;=\; 247 \;\Longrightarrow\; 4(n - 1) \;=\; 240 \;\Longrightarrow\; n - 1 \;=\; 60 \;\Longrightarrow\; n \;=\; 61. $$ So there are $61$ terms. Sigma-bound warning: note we solved for $n$ such that $a_n$ equals the last term, not the term "just past it", off-by-one in either direction is the textbook trap.故共有 $61$ 项。求和上下限提醒：这里求的是使 $a_n$ 等于末项的 $n$，而非"再下一项"。两个方向的差一都是常见陷阱。

(c) Sum求和 M1·A1

$$ S_{61} \;=\; \frac{61}{2}(7 + 247) \;=\; \frac{61}{2} \cdot 254 \;=\; 61 \cdot 127 \;=\; 7747. $$

(d) $S(10) - S(5)$ in function notation用函数记号写 $S(10) - S(5)$ M1·A1

$S(10) = \tfrac{10}{2}(2 \cdot 7 + 9 \cdot 4) = 5 \cdot 50 = 250$ and $S(5) = \tfrac{5}{2}(2 \cdot 7 + 4 \cdot 4) = \tfrac{5}{2} \cdot 30 = 75$. Hence$S(10) = \tfrac{10}{2}(2 \cdot 7 + 9 \cdot 4) = 5 \cdot 50 = 250$，$S(5) = \tfrac{5}{2}(2 \cdot 7 + 4 \cdot 4) = \tfrac{5}{2} \cdot 30 = 75$。故 $$ S(10) - S(5) \;=\; 250 - 75 \;=\; 190. $$ Interpretation:含义： $S(10) - S(5) = a_6 + a_7 + a_8 + a_9 + a_{10}$, the sum of the five terms from position $6$ through position $10$, a "partial-sum window," analogous to a definite-integral difference in calculus.$S(10) - S(5) = a_6 + a_7 + a_8 + a_9 + a_{10}$，即第 $6$ 项至第 $10$ 项这 $5$ 项之和——一个"部分和窗口"，类似微积分中的定积分之差。
Counting terms in $a + (a + d) + \cdots + L$: always $(L - a)/d + 1$.数 $a + (a + d) + \cdots + L$ 的项数：永远是 $(L - a)/d + 1$。 The single most common error in finite arithmetic series is forgetting the $+1$, the difference $(L - a)/d$ counts the gaps, not the terms. Mnemonic: "fenceposts." Five fence panels need six fenceposts; sixty gaps of size $4$ need sixty-one terms. Also note the parallel with the partial-sum-difference idea in (d): $S(b) - S(a)$ telescopes to the terms with positions in $\{a+1, \ldots, b\}$, the discrete analogue of the fundamental theorem of calculus, a key concept students will see again in sigma notation (Q9) and series convergence.有限等差级数最常见的错误就是漏掉 $+1$——$(L - a)/d$ 数的是间隔数而非项数。口诀："栅栏柱"：五块栅栏板需要六根栅栏柱；六十个 $4$ 的间隔需要六十一项。再注意它与 (d) 中部分和之差的对应：$S(b) - S(a)$ 抵消后剩下下标在 $\{a+1, \ldots, b\}$ 的若干项，正是微积分基本定理的离散对应——学生会在西格玛求和（Q9）与级数收敛中反复遇到这一思想。
Q8HARD难Honors荣誉级 🇨🇦 BC卑 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §6 Infinite Geometric无穷等比级数 · BC PC12 / HSA-SSE.B.4 [10 marks][10 分]

$a_1 = 18$, unknown $r$. (a) Convergence test. (b) $r = 2/3$: verify, compute $S_\infty$. (c) $S_\infty = 24$: find $r$. (d) $0.\overline{45}$ as a fraction.$a_1 = 18$、$r$ 未知。(a) 收敛判别。(b) $r = 2/3$：验证后求 $S_\infty$。(c) $S_\infty = 24$：求 $r$。(d) 把 $0.\overline{45}$ 写为分数。

Answer:答案：  (a) converges iff $|r| < 1$当且仅当 $|r| < 1$ 时收敛  ·  (b) $S_\infty = 54$  ·  (c) $r = 1/4$  ·  (d) $0.\overline{45} = 5/11$

(a) Convergence test收敛判别 R1

An infinite geometric series $\sum_{k=0}^{\infty} a_1 \cdot r^{k}$ converges if and only if $|r| < 1$; in that case $S_\infty = a_1/(1 - r)$. If $|r| \ge 1$ the partial sums grow without bound (or oscillate without limit when $r = -1$), and the series diverges.无穷等比级数 $\sum_{k=0}^{\infty} a_1 \cdot r^{k}$ 当且仅当 $|r| < 1$ 时收敛；此时 $S_\infty = a_1/(1 - r)$。若 $|r| \ge 1$，部分和无界（或当 $r = -1$ 时无极限振荡），级数发散。

(b) $r = 2/3$取 $r = 2/3$ M1·A1

$|r| = 2/3 < 1$. $\checkmark$ Convergent.$|r| = 2/3 < 1$。 $\checkmark$ 收敛。 $$ S_\infty \;=\; \frac{a_1}{1 - r} \;=\; \frac{18}{1 - 2/3} \;=\; \frac{18}{1/3} \;=\; 54. $$

(c) Solve $S_\infty = 24$ for $r$由 $S_\infty = 24$ 求 $r$ M1·A1·A1

$$ \frac{18}{1 - r} \;=\; 24 \;\Longrightarrow\; 1 - r \;=\; \frac{18}{24} \;=\; \frac{3}{4} \;\Longrightarrow\; r \;=\; 1 - \frac{3}{4} \;=\; \frac{1}{4}. $$ Convergence check: $|r| = 1/4 < 1$. $\checkmark$收敛核对：$|r| = 1/4 < 1$。 $\checkmark$

(d) Repeating decimal as a fraction循环小数化为分数 M1·A1·A1·A1

Expand $0.\overline{45} = 0.454545\ldots$ as把 $0.\overline{45} = 0.454545\ldots$ 展开为 $$ 0.\overline{45} \;=\; 0.45 + 0.0045 + 0.000045 + \cdots \;=\; \sum_{k = 0}^{\infty} 0.45 \cdot (0.01)^{k}. $$ This is geometric with $a_1 = 0.45$ and $r = 0.01$. Verify $|r| = 0.01 < 1$. $\checkmark$ Apply the formula:这是等比级数，$a_1 = 0.45$，$r = 0.01$。验证 $|r| = 0.01 < 1$。 $\checkmark$ 代入公式： $$ S_\infty \;=\; \frac{0.45}{1 - 0.01} \;=\; \frac{0.45}{0.99} \;=\; \frac{45}{99} \;=\; \frac{5}{11}. $$ (Reduce by $\gcd(45, 99) = 9$.) Final: $0.\overline{45} = 5/11$, in lowest terms because $\gcd(5, 11) = 1$.（除以 $\gcd(45, 99) = 9$ 约分。）最终：$0.\overline{45} = 5/11$，因 $\gcd(5, 11) = 1$，已是最简分数。
$|r| < 1$ is the trap, not a footnote.$|r| < 1$ 是陷阱所在，不是脚注。 Every infinite-geometric question on AP-feeder or BC PC12 papers checks whether the student wrote "$|r| < 1$ so the series converges" before using $a_1/(1 - r)$. Skipping the check costs an A1 (R1 on rigour). The classic distractor: an exam-style problem where $|r| = 1$ or $|r| > 1$, and the formula still yields a finite number, but that number is the formula's algebraic value, not the series' sum (which doesn't exist). The deeper insight for (d): every repeating decimal is an infinite geometric series with $r = 10^{-k}$ where $k$ is the repeat-block length, so every repeating decimal is rational. This is the high-school proof of the "repeating-decimal $\Leftrightarrow$ rational" equivalence (HSA-SSE.B.4).AP 衔接与 BC PC12 的每一道无穷等比题都会检查考生是否在使用 $a_1/(1 - r)$ 之前先写下了"$|r| < 1$ 故级数收敛"。跳过这步会损失一个 A1（严密性 R1）。典型干扰：题中 $|r| = 1$ 或 $|r| > 1$，公式仍能给出一个有限数，但那只是公式的代数值，并非级数之和（后者不存在）。(d) 的更深层洞察：每个循环小数都是 $r = 10^{-k}$ 的无穷等比级数（$k$ 为循环节长度），故每个循环小数都是有理数。这就是"循环小数 $\Leftrightarrow$ 有理数"等价性的高中证明（HSA-SSE.B.4）。
Q9HARD难Honors荣誉级 🇺🇸 US美 🇨🇦 BC卑 AP-feeder FRQAP 衔接简答题 §7 Sigma Notation西格玛求和符号 · HSA-SSE.B.4 / BC PC12 [10 marks][10 分]

(a) $\sum_{k=1}^{30}(2k+5)$. (b) $\sum_{k=1}^{6} 4 \cdot 3^{k-1}$. (c) $\sum_{k=0}^{\infty} 5 (1/4)^{k}$. (d) Split $\sum_{k=1}^{10}(7 \cdot 2^{k-1} - 3)$.(a) $\sum_{k=1}^{30}(2k+5)$。(b) $\sum_{k=1}^{6} 4 \cdot 3^{k-1}$。(c) $\sum_{k=0}^{\infty} 5 (1/4)^{k}$。(d) 拆分 $\sum_{k=1}^{10}(7 \cdot 2^{k-1} - 3)$。

Answer:答案：  (a) $1080$  ·  (b) $1456$  ·  (c) $20/3$  ·  (d) $7131$

(a) Arithmetic $\sum_{k=1}^{30}(2k + 5)$等差 $\sum_{k=1}^{30}(2k + 5)$ M1·A1·A1

Expanded: $7, 9, 11, \ldots, 65$. Arithmetic with $a_1 = 7$, $d = 2$, $n = 30$, $a_{30} = 2(30) + 5 = 65$. Sum:展开：$7, 9, 11, \ldots, 65$。等差，$a_1 = 7$、$d = 2$、$n = 30$、$a_{30} = 2(30) + 5 = 65$。求和： $$ \sum_{k=1}^{30}(2k+5) \;=\; \frac{30}{2}(7 + 65) \;=\; 15 \cdot 72 \;=\; 1080. $$

(b) Finite geometric $\sum_{k=1}^{6} 4 \cdot 3^{k-1}$有限等比 $\sum_{k=1}^{6} 4 \cdot 3^{k-1}$ M1·A1·A1

$a_1 = 4 \cdot 3^{0} = 4$, $r = 3$, $n = 6$.$a_1 = 4 \cdot 3^{0} = 4$、$r = 3$、$n = 6$。 $$ S_6 \;=\; \frac{a_1 (r^n - 1)}{r - 1} \;=\; \frac{4 (3^{6} - 1)}{3 - 1} \;=\; \frac{4 (729 - 1)}{2} \;=\; 2 \cdot 728 \;=\; 1456. $$

(c) Infinite geometric $\sum_{k=0}^{\infty} 5 \cdot (1/4)^{k}$无穷等比 $\sum_{k=0}^{\infty} 5 \cdot (1/4)^{k}$ M1·A1

$a_1 = 5 \cdot (1/4)^{0} = 5$, $r = 1/4$. Check $|r| = 1/4 < 1$. $\checkmark$$a_1 = 5 \cdot (1/4)^{0} = 5$、$r = 1/4$。核对 $|r| = 1/4 < 1$。 $\checkmark$ $$ S_\infty \;=\; \frac{a_1}{1 - r} \;=\; \frac{5}{1 - 1/4} \;=\; \frac{5}{3/4} \;=\; \frac{20}{3}. $$ Sigma-bound watch:求和上下限提醒： the sum starts at $k = 0$, so the first term is $5$ (not $5 \cdot 1/4$); confirm by plugging in $k = 0$.求和从 $k = 0$ 开始，因此首项为 $5$（而不是 $5 \cdot 1/4$）；可代入 $k = 0$ 验证。

(d) Split $\sum_{k=1}^{10}(7 \cdot 2^{k-1} - 3)$拆分 $\sum_{k=1}^{10}(7 \cdot 2^{k-1} - 3)$ M1·A1

Linearity of sigma: $\sum (A_k - B_k) = \sum A_k - \sum B_k$, and $\sum c \cdot A_k = c \sum A_k$.西格玛的线性性：$\sum (A_k - B_k) = \sum A_k - \sum B_k$，以及 $\sum c \cdot A_k = c \sum A_k$。 $$ \sum_{k=1}^{10}(7 \cdot 2^{k-1} - 3) \;=\; 7 \sum_{k=1}^{10} 2^{k-1} \;-\; \sum_{k=1}^{10} 3. $$ First piece (geometric, $a_1 = 1$, $r = 2$, $n = 10$): $\sum 2^{k-1} = (2^{10} - 1)/(2 - 1) = 1023$, so $7 \cdot 1023 = 7161$.
Second piece (constant): $\sum_{k=1}^{10} 3 = 10 \cdot 3 = 30$.
Total: $7161 - 30 = 7131$.第一部分（等比，$a_1 = 1$、$r = 2$、$n = 10$）：$\sum 2^{k-1} = (2^{10} - 1)/(2 - 1) = 1023$，故 $7 \cdot 1023 = 7161$。
第二部分（常数）：$\sum_{k=1}^{10} 3 = 10 \cdot 3 = 30$。
总和：$7161 - 30 = 7131$。
Always start sigma at the smallest index and ask "what is the first term and how many terms am I summing?"永远从最小下标开始处理西格玛，并自问"首项是什么？求和多少项？" $\sum_{k=0}^{\infty}$ and $\sum_{k=1}^{\infty}$ differ by one term, the $k = 0$ contribution. $\sum_{k=1}^{n}$ has $n$ terms; $\sum_{k=0}^{n}$ has $n + 1$ terms. This off-by-one is the sigma-notation trap that haunts every honors student. Diagnostic: plug $k = $ (lower bound) into the summand to get the first term; plug $k = $ (upper bound) to get the last; count terms as (upper $-$ lower $+ 1$). The linearity move in (d), split into a constant times one sum plus another sum, is the high-school precursor to the integral linearity $\int (f + g) = \int f + \int g$ that calculus students will use thousands of times.$\sum_{k=0}^{\infty}$ 与 $\sum_{k=1}^{\infty}$ 相差一项，即 $k = 0$ 的贡献。$\sum_{k=1}^{n}$ 共有 $n$ 项；$\sum_{k=0}^{n}$ 共有 $n + 1$ 项。这是困扰每位荣誉生的西格玛差一陷阱。诊断法：把 $k = $（下限）代入被求和项得首项；代入 $k = $（上限）得末项；项数 $= $（上限 $-$ 下限 $+ 1$）。(d) 中"常数倍 + 拆分"的线性操作正是高中阶段的预演，对应未来微积分中将反复使用的 $\int (f + g) = \int f + \int g$。
PART III  ·  MODELING / APPLIED · SOLUTIONS第三部分  ·  建模 / 应用题 · 详解Universal · 32 marks通用 · 共 32 分

Section C · Modeling and ApplicationsC 部分 · 建模与应用

Q10MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §4 Compound Interest复利 · MCR3U C3.1 [10 marks][10 分]

$A_0 = 2000$, $4\%$ annual compounding. (a) Why geometric? State $a_1, r$. (b) $A_n = 2000(1.04)^n$; compute $A_{10}$. (c) When does $A_n$ first exceed $3000$? (d) Geometric $\leftrightarrow$ exponential connection.$A_0 = 2000$，年复利 $4\%$。(a) 为什么是等比？写出 $a_1, r$。(b) $A_n = 2000(1.04)^n$；求 $A_{10}$。(c) $A_n$ 何时首次超过 $3000$？(d) 等比 $\leftrightarrow$ 指数的联系。

Answer:答案：  (a) geometric with $a_1 = A_1 = 2080$ and $r = 1.04$等比数列，$a_1 = A_1 = 2080$，$r = 1.04$  ·  (b) $A_{10} \approx \$2960.49$  ·  (c) $n = 11$ (first full year balance exceeds \$3000)$n = 11$（余额首次超过 \$3000 的整年）  ·  (d) the geometric sequence $A_n$ is the integer-sampled version of the continuous exponential $f(t) = 2000 \cdot 1.04^{t}$等比数列 $A_n$ 是连续指数函数 $f(t) = 2000 \cdot 1.04^{t}$ 在整数时刻的离散取样

(a) Why geometric?为何是等比？ R1·A1

Each year's balance is the previous balance plus $4\%$ of it: $A_n = A_{n - 1} + 0.04 A_{n - 1} = 1.04 \cdot A_{n - 1}$. So the ratio $A_n / A_{n - 1} = 1.04$ is constant, which is precisely the definition of a geometric sequence. First-term-after-one-year and common ratio:每年的余额等于上一年余额加上其 $4\%$：$A_n = A_{n - 1} + 0.04 A_{n - 1} = 1.04 \cdot A_{n - 1}$。故比值 $A_n / A_{n - 1} = 1.04$ 恒定，正是等比数列的定义。一年后的首项与公比： $$ a_1 \;=\; A_1 \;=\; 2000 \cdot 1.04 \;=\; 2080, \qquad r \;=\; 1.04. $$

(b) Balance after $10$ years$10$ 年后的余额 M1·A1·A1

Indexing from $A_0$: $A_n = 2000 \cdot 1.04^{n}$. (Equivalently, $A_n = a_1 \cdot r^{n - 1} = 2080 \cdot 1.04^{n - 1}$ for $n \ge 1$.)以 $A_0$ 起算：$A_n = 2000 \cdot 1.04^{n}$。（等价地，对 $n \ge 1$ 有 $A_n = a_1 \cdot r^{n - 1} = 2080 \cdot 1.04^{n - 1}$。） $$ A_{10} \;=\; 2000 \cdot 1.04^{10}. $$ Compute $1.04^{10}$: $1.04^{2} = 1.0816$; $1.04^{4} = 1.0816^{2} \approx 1.16986$; $1.04^{5} \approx 1.21665$; $1.04^{10} = (1.04^{5})^{2} \approx 1.21665^{2} \approx 1.48024$. So计算 $1.04^{10}$：$1.04^{2} = 1.0816$；$1.04^{4} = 1.0816^{2} \approx 1.16986$；$1.04^{5} \approx 1.21665$；$1.04^{10} = (1.04^{5})^{2} \approx 1.21665^{2} \approx 1.48024$。故 $$ A_{10} \;\approx\; 2000 \cdot 1.48024 \;\approx\; \$2960.49. $$

(c) First year balance exceeds \$3000余额首次超过 \$3000 的整年 M1·A1·A1

Solve $2000 \cdot 1.04^{n} > 3000$, i.e. $1.04^{n} > 1.5$:求解 $2000 \cdot 1.04^{n} > 3000$，即 $1.04^{n} > 1.5$： $$ n > \frac{\ln 1.5}{\ln 1.04} \;\approx\; \frac{0.4055}{0.03922} \;\approx\; 10.34. $$ The smallest integer $n$ satisfying this is $\mathbf{n = 11}$. Sanity check: $A_{10} \approx \$2960.49 < \$3000$ and $A_{11} = 1.04 \cdot 2960.49 \approx \$3078.91 > \$3000$. $\checkmark$满足条件的最小整数 $n$ 为 $\mathbf{n = 11}$。核对：$A_{10} \approx \$2960.49 < \$3000$，$A_{11} = 1.04 \cdot 2960.49 \approx \$3078.91 > \$3000$。 $\checkmark$

(d) Connection to exponential function与指数函数的联系 R1·A1

The continuous exponential $f(t) = 2000 \cdot 1.04^{t}$ is defined for all real $t \ge 0$ and describes balance at any moment; the geometric sequence $A_n = 2000 \cdot 1.04^{n}$ is its restriction to integer times. Said another way, $A_n = f(n)$, the sequence is a discrete sample of the continuous function. Ontario MCR3U C3.1 explicitly names this bridge; BC PC12 names the same idea.连续指数函数 $f(t) = 2000 \cdot 1.04^{t}$ 对所有 $t \ge 0$ 都有定义，刻画任意时刻的余额；等比数列 $A_n = 2000 \cdot 1.04^{n}$ 是其在整数时刻的限制。换言之，$A_n = f(n)$——数列即连续函数的离散取样。Ontario MCR3U C3.1 明确点名这一桥梁；BC PC12 表述相同的思想。
Compound interest is the cleanest place to see geometric $=$ exponential at integer times.复利是看清"等比 $=$ 整数时刻的指数"的最佳场合。 The growth factor $(1 + r) = 1.04$ is the common ratio of the geometric sequence and the base of the exponential function, same number, two interpretations. Students often write $A_n = 2000(0.04)^{n}$ (using the interest rate $0.04$ as the ratio, forgetting that "the new balance includes the principal you already had"). The fix is the algebraic identity $A_n = A_{n - 1} + 0.04 \, A_{n - 1} = 1.04 \, A_{n - 1}$: the ratio is $1 + r$, not $r$. This is the single most common compound-interest error.增长因子 $(1 + r) = 1.04$ 既是等比数列的公比，也是指数函数的底——同一个数，两种解读。学生常误写成 $A_n = 2000(0.04)^{n}$（把利率 $0.04$ 当公比，忘了"新余额还包含原本的本金"）。修正方法是代数恒等式 $A_n = A_{n - 1} + 0.04 \, A_{n - 1} = 1.04 \, A_{n - 1}$：公比是 $1 + r$，而非 $r$。这是复利问题最常见的错误。
Q11MEDIUM中 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §4 Geometric Decay等比衰减 · BC PC12 exponential ↔ geometricBC PC12 指数 ↔ 等比 [10 marks][10 分]

Isotope: $a_1 = 40$ mg, $r = 0.85$. (a) Why geometric? (b) $a_8$. (c) Smallest $n$ with $a_n < 5$. (d) Continuous-decay connection. (e) Constraint on $n$.同位素：$a_1 = 40$ mg、$r = 0.85$。(a) 为何是等比？(b) $a_8$。(c) 使 $a_n < 5$ 的最小 $n$。(d) 连续衰减联系。(e) $n$ 的限制条件。

Answer:答案：  (a) ratio $r = 0.85$ is constant; geometric公比 $r = 0.85$ 恒定，故为等比  ·  (b) $a_8 \approx 12.82$ mg  ·  (c) $n = 14$  ·  (d) $a_n = f(n)$ where $f(t) = 40 \cdot 0.85^{t-1}$ is continuous decay$a_n = f(n)$，其中 $f(t) = 40 \cdot 0.85^{t-1}$ 为连续衰减函数  ·  (e) $n \in \mathbb{Z}^{+}$ (whole days), $n \ge 1$$n \in \mathbb{Z}^{+}$（整天），$n \ge 1$

(a) Why geometric?为何是等比？ R1·A1

"Each day, the amount remaining is $85\%$ of the previous day's amount" means $a_n = 0.85 \, a_{n - 1}$, i.e. the ratio $a_n / a_{n - 1} = 0.85$ is constant. Hence $(a_n)$ is geometric with common ratio $r = 0.85$ (and $|r| < 1$, so the sequence decays toward zero)."每天剩余量是前一天的 $85\%$"意味着 $a_n = 0.85 \, a_{n - 1}$，即比值 $a_n / a_{n - 1} = 0.85$ 恒定。故 $(a_n)$ 是公比 $r = 0.85$ 的等比数列（且 $|r| < 1$，数列趋向于零）。

(b) Compute $a_8$计算 $a_8$ M1·A1

$a_n = a_1 \cdot r^{n - 1} = 40 \cdot 0.85^{n - 1}$. At $n = 8$:$a_n = a_1 \cdot r^{n - 1} = 40 \cdot 0.85^{n - 1}$。在 $n = 8$： $$ a_8 \;=\; 40 \cdot 0.85^{7}. $$ Compute $0.85^{7}$: $0.85^{2} = 0.7225$; $0.85^{4} = 0.7225^{2} \approx 0.5220$; $0.85^{7} = 0.85^{4} \cdot 0.85^{2} \cdot 0.85 \approx 0.5220 \cdot 0.7225 \cdot 0.85 \approx 0.3206$. So算 $0.85^{7}$：$0.85^{2} = 0.7225$；$0.85^{4} = 0.7225^{2} \approx 0.5220$；$0.85^{7} = 0.85^{4} \cdot 0.85^{2} \cdot 0.85 \approx 0.5220 \cdot 0.7225 \cdot 0.85 \approx 0.3206$。故 $$ a_8 \;\approx\; 40 \cdot 0.3206 \;\approx\; 12.82 \text{ mg}. $$

(c) When does $a_n < 5$?$a_n < 5$ 何时成立？ M1·A1·A1

Solve $40 \cdot 0.85^{n - 1} < 5$, i.e. $0.85^{n - 1} < 0.125$.求解 $40 \cdot 0.85^{n - 1} < 5$，即 $0.85^{n - 1} < 0.125$。 $$ (n - 1) \ln 0.85 \;<\; \ln 0.125. $$ Because $\ln 0.85 < 0$, dividing flips the inequality:因 $\ln 0.85 < 0$，除以负数时不等号反向： $$ n - 1 \;>\; \frac{\ln 0.125}{\ln 0.85} \;\approx\; \frac{-2.0794}{-0.16252} \;\approx\; 12.79. $$ Smallest integer: $n - 1 = 13$, so $\mathbf{n = 14}$. Sanity check: $a_{13} \approx 40 \cdot 0.85^{12} \approx 40 \cdot 0.1422 \approx 5.69$ mg ($\ge 5$); $a_{14} \approx 40 \cdot 0.85^{13} \approx 40 \cdot 0.1209 \approx 4.83$ mg ($< 5$). $\checkmark$最小整数 $n - 1 = 13$，故 $\mathbf{n = 14}$。核对：$a_{13} \approx 40 \cdot 0.85^{12} \approx 40 \cdot 0.1422 \approx 5.69$ mg（$\ge 5$）；$a_{14} \approx 40 \cdot 0.85^{13} \approx 40 \cdot 0.1209 \approx 4.83$ mg（$< 5$）。 $\checkmark$

(d) Continuous-decay connection连续衰减联系 R1·A1

The continuous decay function $f(t) = 40 \cdot 0.85^{t - 1}$ (or equivalently $f(t) = 40 \, e^{-\lambda (t - 1)}$ with $\lambda = -\ln 0.85 \approx 0.1625$ per day) describes the amount remaining at any real time $t$; the geometric sequence $a_n$ is the value of $f$ at integer days. BC PC12 names this connection explicitly: "exponential functions to geometric sequences".连续衰减函数 $f(t) = 40 \cdot 0.85^{t - 1}$（等价地 $f(t) = 40 \, e^{-\lambda (t - 1)}$，$\lambda = -\ln 0.85 \approx 0.1625$ 每天）描述任意实数时刻 $t$ 的剩余量；等比数列 $a_n$ 即 $f$ 在整数天的取值。BC PC12 显式点名这一联系："指数函数与等比数列"。

(e) Constraint on $n$$n$ 的限制 A1

Since $n$ counts days, $n$ must be a positive integer: $n \in \mathbb{Z}^{+}$, $n \ge 1$.由于 $n$ 表示天数，$n$ 必须为正整数：$n \in \mathbb{Z}^{+}$，$n \ge 1$。
Sign of the inequality flips when you divide by a negative log.除以负对数时不等号方向必须反向。 The most common error in solving $r^{n} < \varepsilon$ with $0 < r < 1$: taking logs but forgetting that $\ln r < 0$. The correct chain: $r^{n} < \varepsilon \Leftrightarrow n \ln r < \ln \varepsilon \Leftrightarrow n > \ln \varepsilon / \ln r$, inequality reverses on division. If you write $n < \ln 0.125 / \ln 0.85$, you'll get $n < 12.79$ and conclude $n = 12$, wrong by two terms. Train the habit: before dividing, write down the sign of the divisor; if negative, flip. BC provincial markers explicitly probe this.求解 $0 < r < 1$ 时 $r^{n} < \varepsilon$ 的最常见错误：取对数后忘了 $\ln r < 0$。正确链条：$r^{n} < \varepsilon \Leftrightarrow n \ln r < \ln \varepsilon \Leftrightarrow n > \ln \varepsilon / \ln r$，除法时不等号反向。如果写成 $n < \ln 0.125 / \ln 0.85$，就会得到 $n < 12.79$ 并误答 $n = 12$，相差两项。养成习惯：在除法之前先标记除数的符号，若为负则反向。BC 省考阅卷专门检查这一点。
Q12HARD难 🇨🇦 BC卑 🇺🇸 US美 BC Provincial-style卑诗省考风格 §5 Finite Geometric Sum有限等比求和 · HSA-SSE.B.4 [12 marks][12 分]

Salary: $a_1 = 30000$, $r = 1.05$. (a) Why geometric? (b) $a_{10}$. (c) $S_{10}$. (d) Smallest $n$ with $S_n > 500000$. (e) Compare to flat \$42000/yr for $10$ years.薪酬：$a_1 = 30000$、$r = 1.05$。(a) 为何是等比？(b) $a_{10}$。(c) $S_{10}$。(d) 使 $S_n > 500000$ 的最小 $n$。(e) 与固定年薪 \$42000、共 $10$ 年方案比较。

Answer:答案：  (a) ratio $r = 1.05$ constant; geometric公比 $r = 1.05$ 恒定，为等比  ·  (b) $a_{10} \approx \$46{,}539.85$  ·  (c) $S_{10} \approx \$377{,}336.78$  ·  (d) $n = 13$  ·  (e) flat \$42000 wins by $\approx \$42{,}663$ over $10$ years固定 \$42000 在 $10$ 年内多出约 \$42{,}663

(a) Why geometric?为何是等比？ R1·A1

Each year's salary is $1.05$ times the previous year's: $a_n = 1.05 \, a_{n - 1}$, so $a_n / a_{n - 1} = 1.05$ is constant. Geometric with $a_1 = 30000$ and $r = 1.05$.每年薪酬等于前一年的 $1.05$ 倍：$a_n = 1.05 \, a_{n - 1}$，故 $a_n / a_{n - 1} = 1.05$ 恒定。等比数列，$a_1 = 30000$、$r = 1.05$。

(b) Salary in year $10$第 $10$ 年的薪酬 M1·A1

$a_{10} = 30000 \cdot 1.05^{9}$. Compute $1.05^{9}$: $1.05^{2} = 1.1025$; $1.05^{4} \approx 1.21551$; $1.05^{8} \approx 1.21551^{2} \approx 1.47746$; $1.05^{9} = 1.05^{8} \cdot 1.05 \approx 1.55133$. So$a_{10} = 30000 \cdot 1.05^{9}$。算 $1.05^{9}$：$1.05^{2} = 1.1025$；$1.05^{4} \approx 1.21551$；$1.05^{8} \approx 1.21551^{2} \approx 1.47746$；$1.05^{9} = 1.05^{8} \cdot 1.05 \approx 1.55133$。故 $$ a_{10} \;\approx\; 30000 \cdot 1.55133 \;\approx\; \$46{,}539.85. $$

(c) Cumulative earnings $S_{10}$累计收入 $S_{10}$ M1·A1·A1

Use $S_n = a_1 (r^n - 1)/(r - 1)$ with $a_1 = 30000$, $r = 1.05$, $n = 10$:用 $S_n = a_1 (r^n - 1)/(r - 1)$，$a_1 = 30000$、$r = 1.05$、$n = 10$： $$ S_{10} \;=\; \frac{30000 (1.05^{10} - 1)}{1.05 - 1} \;=\; \frac{30000 (1.62889 - 1)}{0.05} \;=\; \frac{30000 \cdot 0.62889}{0.05}. $$ Compute $1.05^{10} = 1.05^{9} \cdot 1.05 \approx 1.55133 \cdot 1.05 \approx 1.62889$. Then算 $1.05^{10} = 1.05^{9} \cdot 1.05 \approx 1.55133 \cdot 1.05 \approx 1.62889$。继续： $$ S_{10} \;\approx\; \frac{18{,}866.78}{0.05} \;\approx\; \$377{,}336.78. $$

(d) Smallest $n$ with $S_n > 500000$使 $S_n > 500000$ 的最小 $n$ M1·A1·A1

Set up the inequality:列不等式： $$ \frac{30000 (1.05^{n} - 1)}{0.05} \;>\; 500000 \;\Longrightarrow\; 1.05^{n} - 1 \;>\; \frac{500000 \cdot 0.05}{30000} \;=\; \frac{25000}{30000} \;=\; \frac{5}{6}. $$ So $1.05^{n} > 1 + 5/6 = 11/6 \approx 1.8333$. Take logs:故 $1.05^{n} > 1 + 5/6 = 11/6 \approx 1.8333$。取对数： $$ n > \frac{\ln(11/6)}{\ln 1.05} \;\approx\; \frac{0.6061}{0.04879} \;\approx\; 12.42. $$ Smallest integer: $\mathbf{n = 13}$. Sanity check: $S_{12} \approx 30000(1.05^{12} - 1)/0.05$ with $1.05^{12} \approx 1.79586$, giving $S_{12} \approx 477{,}517$ ($< 500000$); $S_{13}$ with $1.05^{13} \approx 1.88565$, giving $S_{13} \approx 531{,}390$ ($> 500000$). $\checkmark$最小整数 $\mathbf{n = 13}$。核对：$1.05^{12} \approx 1.79586$ 时 $S_{12} \approx 477{,}517$（$< 500000$）；$1.05^{13} \approx 1.88565$ 时 $S_{13} \approx 531{,}390$（$> 500000$）。 $\checkmark$

(e) Compare to flat \$42000/yr与固定 \$42000 / 年比较 A1·A1

Flat-salary total over $10$ years: $10 \cdot 42000 = \$420{,}000$. Geometric-salary total: $\approx \$377{,}336.78$.固定方案 $10$ 年总收入：$10 \cdot 42000 = \$420{,}000$。等比方案总收入：$\approx \$377{,}336.78$。 $$ \$420{,}000 \;-\; \$377{,}336.78 \;\approx\; \$42{,}663. $$ Conclusion:结论： the flat-\$42000 offer pays approximately \$42{,}663 more over the first $10$ years. (Beyond year $10$ the geometric offer eventually overtakes, at $n \approx 19$ the geometric salary first exceeds \$72{,}000 and the cumulative crossover occurs around year $18$, depending on what range is chosen, but in the first decade alone the flat offer wins.)在前 $10$ 年内，固定 \$42000 方案多出约 \$42{,}663。（$10$ 年之后等比方案最终会反超：$n \approx 19$ 时单年薪酬首次超过 \$72{,}000，累计金额的交叉点约在第 $18$ 年附近——视区间选择而定。但仅就头十年而言，固定方案胜出。）
Geometric-vs-flat is the classic "raise-vs-bonus" tradeoff, and the timeframe matters more than the rate."等比 vs. 固定"是经典的"加薪 vs. 奖金"权衡，且时间窗口比百分比更重要。 A $5\%$ raise on \$30000 sounds great, but over $10$ years it's still behind a flat \$42000, because the geometric series builds slowly when $r$ is close to $1$. The crossover year depends on $r$ and the flat-vs-initial gap: roughly $n^{*} \approx \ln(F/a_1) / \ln(r) + 1$ where $F$ is the flat salary. Here $\ln(42/30)/\ln(1.05) \approx 0.336/0.0488 \approx 6.9$, so by year $7$ the geometric annual salary exceeds the flat, but the cumulative total takes longer to catch up. Lesson: compounding wins eventually, but "eventually" can be longer than the contract.在 \$30000 基础上 $5\%$ 的加薪听起来很美，但 $10$ 年内仍输给固定 \$42000，因为当 $r$ 接近 $1$ 时等比级数累积缓慢。交叉年份取决于 $r$ 与"固定值 - 起点"的差：约为 $n^{*} \approx \ln(F/a_1) / \ln(r) + 1$，其中 $F$ 为固定年薪。此处 $\ln(42/30)/\ln(1.05) \approx 0.336/0.0488 \approx 6.9$，所以第 $7$ 年起等比年薪超过固定年薪，但累计总额追上需更久。教训：复利终将胜出，但"终将"可能比合同期还久。