IB Math AA HL — Unit A1: Sequences & Series — Solutions

Q1EASYPaper 1A1.2 Arithmetic Sequences[5 marks]

Arithmetic sequence with $u_5 = 23$ and $u_{15} = 53$ — find (a) $d$, (b) $u_1$, (c) $S_{10}$.已知等差数列满足 $u_5 = 23$、$u_{15} = 53$ —— 求 (a) $d$，(b) $u_1$，(c) $S_{10}$。

Answers:答案： (a) $d = 3$ · (b) $u_1 = 11$ · (c) $S_{10} = 245$

(a) Common difference M1·A1

Use the explicit formula $u_n = u_1 + (n-1)d$ to set up two equations: $$ u_5 = u_1 + 4d = 23, \qquad u_{15} = u_1 + 14d = 53. $$ Subtract to eliminate $u_1$: $$ (u_{15}) - (u_5) = 10d = 53 - 23 = 30 \quad \Longrightarrow \quad d = 3. $$

(b) First term M1·A1

Substitute $d = 3$ back into $u_5 = u_1 + 4d = 23$: $$ u_1 = 23 - 4(3) = 11. $$

(c) Sum of first ten terms A1

$$ S_{10} = \tfrac{10}{2}\bigl[\,2u_1 + 9d\,\bigr] = 5\bigl[\,22 + 27\,\bigr] = 5(49) = 245. $$

Shortcut. For any arithmetic sequence, $u_m - u_n = (m-n)d$ — so given any two terms you can read off $d$ by a single subtraction. The "two equations, eliminate $u_1$" approach is the IB-standard write-up, but for mental arithmetic the gap formula $d = (u_{15}-u_5)/(15-5) = 30/10 = 3$ is faster.

(a) 公差 M1·A1

使用通项公式（explicit formula）$u_n = u_1 + (n-1)d$，列出两个方程： $$ u_5 = u_1 + 4d = 23, \qquad u_{15} = u_1 + 14d = 53. $$ 两式相减以消去 $u_1$： $$ (u_{15}) - (u_5) = 10d = 53 - 23 = 30 \quad \Longrightarrow \quad d = 3. $$

(b) 首项 M1·A1

将 $d = 3$ 代回 $u_5 = u_1 + 4d = 23$： $$ u_1 = 23 - 4(3) = 11. $$

(c) 前十项之和 A1

$$ S_{10} = \tfrac{10}{2}\bigl[\,2u_1 + 9d\,\bigr] = 5\bigl[\,22 + 27\,\bigr] = 5(49) = 245. $$

速算技巧。对任意等差数列，$u_m - u_n = (m-n)d$ —— 所以只要给出任意两个项，就能用一次减法直接读出 $d$。"两个方程消去 $u_1$"是 IB 的标准书写格式，但若做心算，用项间差（gap formula） $d = (u_{15}-u_5)/(15-5) = 30/10 = 3$ 更快。

Q2EASYPaper 1A1.3 Geometric Sequences[4 marks]

Geometric sequence with $u_2 = 12$ and $u_5 = -96$ — find (a) $r$, (b) $u_1$ and $u_8$.已知等比数列满足 $u_2 = 12$、$u_5 = -96$ —— 求 (a) $r$，(b) $u_1$ 与 $u_8$。

Answers:答案： (a) $r = -2$ · (b) $u_1 = -6,\; u_8 = 768$

(a) Common ratio M1·A1

$u_n = u_1 r^{n-1}$, so the ratio of two terms collapses $u_1$: $$ \frac{u_5}{u_2} \;=\; \frac{u_1 r^{4}}{u_1 r^{1}} \;=\; r^{3} \;=\; \frac{-96}{12} \;=\; -8 \quad \Longrightarrow \quad r = \sqrt[3]{-8} = -2. $$ (Cube root is single-valued over the reals, so the sign is determined.)

(b) First and eighth term M1·A1

$u_1 = u_2 / r = 12 / (-2) = -6$. Then $$ u_8 \;=\; u_1\, r^{7} \;=\; (-6)\,(-2)^{7} \;=\; (-6)(-128) \;=\; 768. $$

Sign-check trap. When $r$ is negative, the terms alternate sign. Verify by working out $u_3$ and $u_4$: $u_3 = u_2 r = 12 \cdot (-2) = -24$, $u_4 = 48$, $u_5 = -96$ ✓. A common slip is to take $r = +2$ (forgetting the negative ratio), giving $u_5 = +96$ instead of $-96$. Always test your $r$ against a known term.

(a) 公比 M1·A1

$u_n = u_1 r^{n-1}$，两项相比即可消去 $u_1$： $$ \frac{u_5}{u_2} \;=\; \frac{u_1 r^{4}}{u_1 r^{1}} \;=\; r^{3} \;=\; \frac{-96}{12} \;=\; -8 \quad \Longrightarrow \quad r = \sqrt[3]{-8} = -2. $$ （实数立方根（cube root）为单值，正负号由此确定。）

(b) 首项与第八项 M1·A1

$u_1 = u_2 / r = 12 / (-2) = -6$。再得 $$ u_8 \;=\; u_1\, r^{7} \;=\; (-6)\,(-2)^{7} \;=\; (-6)(-128) \;=\; 768. $$

正负号陷阱。当 $r$ 为负数时，相邻项的正负号交替。可以算 $u_3$、$u_4$ 检验：$u_3 = u_2 r = 12 \cdot (-2) = -24$，$u_4 = 48$，$u_5 = -96$ ✓。常见失误是把 $r$ 取成 $+2$（忽略了 $r$ 的负号），那样得到 $u_5 = +96$ 而非 $-96$。每次求出 $r$ 后，都用已知项校验一遍。

Q3MEDIUMPaper 1A1.4 Sigma Notation[5 marks]

Evaluate $\displaystyle\sum_{k=1}^{20}(4k - 3)$ by recognising the summand as an arithmetic sequence.将求和项识别为等差数列后计算 $\displaystyle\sum_{k=1}^{20}(4k - 3)$。

Answer:答案： $\displaystyle\sum_{k=1}^{20}(4k - 3) = 780$

Identify the structure M1·A1

The summand $a_k = 4k - 3$ is linear in $k$, so $\{a_k\}$ is an arithmetic sequence. Extract first term, common difference, and last term: $$ a_1 = 4(1) - 3 = 1, \qquad d = a_{k+1} - a_k = 4, \qquad a_{20} = 4(20) - 3 = 77. $$

Apply the arithmetic sum formula M1·A1·A1

$$ \sum_{k=1}^{20}(4k-3) \;=\; \tfrac{n}{2}(a_1 + a_n) \;=\; \tfrac{20}{2}(1 + 77) \;=\; 10 \cdot 78 \;=\; 780. $$

Two routes, same answer. The linearity-of-sigma route gives $\displaystyle\sum(4k-3) = 4\sum k - \sum 3 = 4 \cdot \tfrac{20 \cdot 21}{2} - 60 = 840 - 60 = 780$. Both are method-mark legitimate; in Paper 1A the arithmetic-formula route is shorter when the summand is linear, while the linearity route is preferable when you must sum a polynomial like $\sum (k^2 + k)$.

识别结构 M1·A1

求和项 $a_k = 4k - 3$ 关于 $k$ 是线性的，所以 $\{a_k\}$ 是等差数列。提取首项、公差、末项： $$ a_1 = 4(1) - 3 = 1, \qquad d = a_{k+1} - a_k = 4, \qquad a_{20} = 4(20) - 3 = 77. $$

套用等差求和公式 M1·A1·A1

$$ \sum_{k=1}^{20}(4k-3) \;=\; \tfrac{n}{2}(a_1 + a_n) \;=\; \tfrac{20}{2}(1 + 77) \;=\; 10 \cdot 78 \;=\; 780. $$

两条路径，同一答案。利用 $\Sigma$ 的线性可得 $\displaystyle\sum(4k-3) = 4\sum k - \sum 3 = 4 \cdot \tfrac{20 \cdot 21}{2} - 60 = 840 - 60 = 780$。两种做法都给方法分；Paper 1A 中求和项为线性时，等差公式更短；但若要求和的是 $\sum (k^2 + k)$ 这种多项式形式，线性拆分法更稳妥。

Q4MEDIUMPaper 1A1.8 Infinite GP[6 marks]

Express $0.\overline{42}$ as a fraction in simplest form using the infinite GP sum, with a convergence check.用无穷等比级数求和公式（含收敛性检验）将 $0.\overline{42}$ 写为最简分数。

Answer:答案： $0.\overline{42} = \dfrac{14}{33}$

Decimal-as-GP set-up M1·A1

Write the repeating decimal as the sum of place-value contributions of each "$42$" block: $$ 0.\overline{42} \;=\; \frac{42}{100} + \frac{42}{10\,000} + \frac{42}{1\,000\,000} + \cdots $$ This is an infinite geometric series with $$ u_1 = \tfrac{42}{100}, \qquad r = \tfrac{1}{100}. $$

Convergence check R1

$|r| = \tfrac{1}{100} < 1$ ✓ — the series converges, so the closed-form sum is valid.

Apply $S_\infty = u_1/(1-r)$ M1·A1·A1

$$ S_\infty \;=\; \frac{u_1}{1 - r} \;=\; \frac{42/100}{1 - 1/100} \;=\; \frac{42/100}{99/100} \;=\; \frac{42}{99} \;=\; \frac{14}{33}. $$ (Divide numerator and denominator by $\gcd(42, 99) = 3$.)

Pattern. Any purely repeating decimal $0.\overline{a_1 a_2 \cdots a_k}$ equals $\dfrac{(a_1 a_2 \cdots a_k)}{10^k - 1}$ — the digits over a string of $k$ nines. So $0.\overline{42} = \tfrac{42}{99}$ directly. The GP derivation justifies this trick: $r = 10^{-k}$ gives $1 - r = (10^k - 1)/10^k$, and the $10^k$ cancels with the $u_1$ denominator. Worth memorising — comes up in Paper 1A and in the "show that this irrational equals…" style exploration questions.

把循环小数写成 GP M1·A1

把循环小数按"$42$"块的位值贡献展开为一个和： $$ 0.\overline{42} \;=\; \frac{42}{100} + \frac{42}{10\,000} + \frac{42}{1\,000\,000} + \cdots $$ 这是一个无穷等比级数（infinite GP），其中 $$ u_1 = \tfrac{42}{100}, \qquad r = \tfrac{1}{100}. $$

收敛性检验 R1

$|r| = \tfrac{1}{100} < 1$ ✓ —— 级数收敛，所以可使用闭合求和公式。

套用 $S_\infty = u_1/(1-r)$ M1·A1·A1

$$ S_\infty \;=\; \frac{u_1}{1 - r} \;=\; \frac{42/100}{1 - 1/100} \;=\; \frac{42/100}{99/100} \;=\; \frac{42}{99} \;=\; \frac{14}{33}. $$ （分子分母同除 $\gcd(42, 99) = 3$。）

规律。任何纯循环小数 $0.\overline{a_1 a_2 \cdots a_k}$ 都等于 $\dfrac{(a_1 a_2 \cdots a_k)}{10^k - 1}$ —— 即用 $k$ 个 9 作分母。所以 $0.\overline{42} = \tfrac{42}{99}$ 可直接套用。上面 GP 的推导正是这条捷径的依据：$r = 10^{-k}$ 给出 $1 - r = (10^k - 1)/10^k$，分母中的 $10^k$ 与 $u_1$ 的 $10^k$ 相消。值得背诵 —— 在 Paper 1A 与"证明某无理数等于⋯⋯"型探究题中常用。

Q5MEDIUMPaper 1A1.8 Infinite GP[6 marks]

Infinite GP with $u_1 = 18$ and $S_\infty = 24$. (a) Find $r$ as an exact fraction. (b) Find $S_5$ as an exact fraction.无穷等比级数 $u_1 = 18$、$S_\infty = 24$。(a) 求 $r$ 的精确分数。(b) 求 $S_5$ 的精确分数。

Answers:答案： (a) $r = \dfrac{1}{4}$ · (b) $S_5 = \dfrac{3069}{128}$

(a) Solve for $r$ from $S_\infty$ M1·A1·R1

$$ S_\infty \;=\; \frac{u_1}{1 - r} \quad \Longrightarrow \quad 24 \;=\; \frac{18}{1 - r} \quad \Longrightarrow \quad 1 - r \;=\; \tfrac{18}{24} \;=\; \tfrac{3}{4} \quad \Longrightarrow \quad r \;=\; \tfrac{1}{4}. $$ Convergence check: $|r| = \tfrac{1}{4} < 1$ ✓ (so $S_\infty$ formula was legitimate).

(b) Apply the finite GP sum formula M1·A1·A1

$$ S_5 \;=\; \frac{u_1\,(1 - r^{5})}{1 - r} \;=\; \frac{18\,\bigl(1 - (1/4)^{5}\bigr)}{1 - 1/4} \;=\; \frac{18\,\bigl(1 - \tfrac{1}{1024}\bigr)}{3/4}. $$ Simplify: $\dfrac{18}{3/4} = 18 \cdot \dfrac{4}{3} = 24$, and $1 - \tfrac{1}{1024} = \tfrac{1023}{1024}$. So $$ S_5 \;=\; 24 \cdot \tfrac{1023}{1024} \;=\; \tfrac{24 \cdot 1023}{1024} \;=\; \tfrac{24\,552}{1024} \;=\; \tfrac{3069}{128}. $$ (Divide top and bottom by $\gcd = 8$.)

Why $S_5$ < $S_\infty$. $S_5 = \tfrac{3069}{128} \approx 23.977$, while $S_\infty = 24$. The remaining "tail" $S_\infty - S_5 = u_1 r^{5}/(1-r) = 18 \cdot (1/4)^5 / (3/4) = 18 \cdot \tfrac{1}{1024} \cdot \tfrac{4}{3} = \tfrac{24}{1024} = \tfrac{3}{128} \approx 0.023$. Partial sums of a fast-converging GP overshoot/undershoot the limit by exactly the discarded tail — useful for error analysis in Paper 3 problems.

(a) 由 $S_\infty$ 解出 $r$ M1·A1·R1

$$ S_\infty \;=\; \frac{u_1}{1 - r} \quad \Longrightarrow \quad 24 \;=\; \frac{18}{1 - r} \quad \Longrightarrow \quad 1 - r \;=\; \tfrac{18}{24} \;=\; \tfrac{3}{4} \quad \Longrightarrow \quad r \;=\; \tfrac{1}{4}. $$ 收敛性检验：$|r| = \tfrac{1}{4} < 1$ ✓（说明 $S_\infty$ 公式可用）。

(b) 套用有限项 GP 求和公式 M1·A1·A1

$$ S_5 \;=\; \frac{u_1\,(1 - r^{5})}{1 - r} \;=\; \frac{18\,\bigl(1 - (1/4)^{5}\bigr)}{1 - 1/4} \;=\; \frac{18\,\bigl(1 - \tfrac{1}{1024}\bigr)}{3/4}. $$ 化简：$\dfrac{18}{3/4} = 18 \cdot \dfrac{4}{3} = 24$，且 $1 - \tfrac{1}{1024} = \tfrac{1023}{1024}$。因此 $$ S_5 \;=\; 24 \cdot \tfrac{1023}{1024} \;=\; \tfrac{24 \cdot 1023}{1024} \;=\; \tfrac{24\,552}{1024} \;=\; \tfrac{3069}{128}. $$ （分子分母同除 $\gcd = 8$。）

$S_5$ 为何小于 $S_\infty$。$S_5 = \tfrac{3069}{128} \approx 23.977$，而 $S_\infty = 24$。剩余"尾部"$S_\infty - S_5 = u_1 r^{5}/(1-r) = 18 \cdot (1/4)^5 / (3/4) = 18 \cdot \tfrac{1}{1024} \cdot \tfrac{4}{3} = \tfrac{24}{1024} = \tfrac{3}{128} \approx 0.023$。快速收敛的 GP 的部分和与极限差正好等于被截断的尾部 —— Paper 3 误差分析题里很常用。

Q6MEDIUMPaper 1A1.2 / 1.3 Mixed[5 marks]

For the sequence $4, x, 9$ — (a) arithmetic value of $x$, (b) all geometric values, (c) is there an $x$ for which it is both?对数列 $4, x, 9$ —— (a) 等差时 $x$ 的值，(b) 等比时 $x$ 的所有值，(c) 是否存在使其同时为等差与等比的 $x$。

Answers:答案： (a) $x = \tfrac{13}{2}$ · (b) $x = \pm 6$ · (c) No — would require the same $x$ to satisfy both, but $\tfrac{13}{2} \ne \pm 6$.不存在 —— 需要同一个 $x$ 同时满足两条件，但 $\tfrac{13}{2} \ne \pm 6$。

(a) Arithmetic middle term A1

In an arithmetic sequence the middle term is the arithmetic mean of its neighbours: $x = \tfrac{4 + 9}{2} = \tfrac{13}{2}$.

(b) Geometric middle term M1·A1

In a geometric sequence the middle term satisfies $x^{2} = 4 \cdot 9 = 36$, so $x = \pm 6$. (Both signs give a valid GP: $4, 6, 9$ has $r = \tfrac{3}{2}$, and $4, -6, 9$ has $r = -\tfrac{3}{2}$.)

(c) Can it be both? M1·R1

The sequence is simultaneously arithmetic and geometric iff $x$ satisfies both conditions. From (a), arithmetic requires $x = \tfrac{13}{2}$. From (b), geometric requires $x = \pm 6$. Since $\tfrac{13}{2} \notin \{-6, +6\}$, no value of $x$ satisfies both.

Sharper argument. A three-term sequence $a, b, c$ is both arithmetic and geometric iff $2b = a + c$ AND $b^2 = ac$. Eliminating $b$: $(\tfrac{a+c}{2})^2 = ac \Rightarrow (a+c)^2 = 4ac \Rightarrow (a-c)^2 = 0 \Rightarrow a = c$. So the only three-term sequences that are both arithmetic and geometric are the constant ones. Since $a = 4 \ne 9 = c$, this sequence cannot be both. $\square$

The bigger picture. The "constant sequence" lemma above generalises: a sequence is both arithmetic and geometric iff it is constant. Equivalently, common difference $d = 0$ and common ratio $r = 1$ are forced together. This is a high-value Paper 1 distractor — students who mechanically solve part (c) by "checking compatibility" will get the answer, but understanding why the constraint forces constancy turns the question into a one-line write-up.

(a) 等差中项 A1

在等差数列中，中间项等于两端项的算术平均：$x = \tfrac{4 + 9}{2} = \tfrac{13}{2}$。

(b) 等比中项 M1·A1

在等比数列中，中间项满足 $x^{2} = 4 \cdot 9 = 36$，故 $x = \pm 6$。（两个值都对应有效的 GP：$4, 6, 9$ 对应 $r = \tfrac{3}{2}$；$4, -6, 9$ 对应 $r = -\tfrac{3}{2}$。）

(c) 能否同时？ M1·R1

数列同时为等差与等比当且仅当 $x$ 同时满足两个条件。由 (a)，等差要求 $x = \tfrac{13}{2}$；由 (b)，等比要求 $x = \pm 6$。因 $\tfrac{13}{2} \notin \{-6, +6\}$，不存在使两条件同时成立的 $x$。

更深一步。三项数列 $a, b, c$ 既等差又等比 $\Longleftrightarrow$ $2b = a + c$ 且 $b^2 = ac$。消去 $b$：$(\tfrac{a+c}{2})^2 = ac \Rightarrow (a+c)^2 = 4ac \Rightarrow (a-c)^2 = 0 \Rightarrow a = c$。所以同时为等差与等比的三项数列只能是常数列。因 $a = 4 \ne 9 = c$，本题数列不可能同时是两者。$\square$

更大的图景。上面"常数列引理"可以推广：一个数列既等差又等比 $\Longleftrightarrow$ 它是常数列。等价地，公差 $d = 0$ 与公比 $r = 1$ 必须同时成立。这是 Paper 1 中高价值的干扰项 —— 机械地"逐一核对兼容性"也能得到答案，但理解为什么该约束逼出常数列，能让答题压缩到一行。

Section B — Worked SolutionsB 节 —— 详细解析

Q7HARDPaper 1B1.2 / 1.4 Multi-skill[16 marks]

$S_n = n^2 + 3n$ — (a) first three terms via $u_n = S_n - S_{n-1}$; (b) show $u_n = 2n+2$ for $n \ge 2$; (c) verify at $n=1$; (d) hence arithmetic, state $d$; (e) closed form for $\sum_{k=1}^n u_{2k}$; (f) value at $n = 15$.$S_n = n^2 + 3n$ —— (a) 用 $u_n = S_n - S_{n-1}$ 求前三项；(b) 证明 $n \ge 2$ 时 $u_n = 2n+2$；(c) 在 $n=1$ 处验证；(d) 由此得等差，写出 $d$；(e) $\sum_{k=1}^n u_{2k}$ 的闭合形式；(f) 求 $n = 15$ 时的值。

Answers:答案： (a) $u_1 = 4,\; u_2 = 6,\; u_3 = 8$ · (d) $d = 2$ · (e) $\displaystyle\sum_{k=1}^{n} u_{2k} = 2n^{2} + 4n$ · (f) $510$

(a) First three terms M1·A1·A1

$u_1 = S_1 = 1 + 3 = 4$. $u_2 = S_2 - S_1 = (4 + 6) - 4 = 6$. $u_3 = S_3 - S_2 = (9 + 9) - 10 = 8$.

(b) Closed form for $u_n$ when $n \ge 2$ M1·A1·A1

For $n \ge 2$: $$ u_n \;=\; S_n - S_{n-1} \;=\; (n^{2} + 3n) - \bigl[\,(n-1)^{2} + 3(n-1)\,\bigr]. $$ Expand $(n-1)^{2} + 3(n-1) = n^{2} - 2n + 1 + 3n - 3 = n^{2} + n - 2$. Hence $$ u_n \;=\; (n^{2} + 3n) - (n^{2} + n - 2) \;=\; 2n + 2. \qquad \blacksquare $$

(c) Verify at $n = 1$ A1

The formula gives $u_1 = 2(1) + 2 = 4$, which matches $S_1 = 4$ from part (a). So the formula $u_n = 2n+2$ is valid for all $n \ge 1$.

(d) Arithmetic with common difference M1·A1

For all $n \ge 1$, $u_{n+1} - u_n = (2(n+1)+2) - (2n+2) = 2$. The difference is constant $\Rightarrow \{u_n\}$ is arithmetic with common difference $d = 2$.

(e) Closed form for $\sum u_{2k}$ M1·M1·A1·A1

Substitute $n = 2k$ in $u_n = 2n + 2$: $$ u_{2k} \;=\; 2(2k) + 2 \;=\; 4k + 2. $$ So $\{u_{2k}\}_{k = 1}^{n}$ is itself an arithmetic sequence with first term $u_2 = 6$, common difference $4$, $n$ terms. Apply $S = \tfrac{n}{2}(2a + (n-1)d)$: $$ \sum_{k=1}^{n} u_{2k} \;=\; \tfrac{n}{2}\bigl[\,2(6) + (n-1)(4)\,\bigr] \;=\; \tfrac{n}{2}(4n + 8) \;=\; 2n(n + 2) \;=\; 2n^{2} + 4n. $$

(f) Evaluate at $n = 15$ M1·A1·A1

$$ \sum_{k=1}^{15} u_{2k} \;=\; 2(15)(17) \;=\; 510. $$

Two-level structure. Whenever a problem gives you $S_n$ as a polynomial in $n$, the sequence $\{u_n\}$ is one degree lower — and if $S_n$ is quadratic, then $\{u_n\}$ is linear i.e. arithmetic. (Quick check: $S_n = an^2 + bn + c$ forces $u_1 = a+b+c$ and $u_n = 2an + (b-a)$ for $n \ge 2$, then $c = 0$ is required for the formula to match at $n=1$.) Here $c = 0$, $a = 1$, $b = 3$, giving $d = 2a = 2$ and $u_1 = a + b = 4$ instantly — the IB-style write-up above is the safe long version.

(a) 前三项 M1·A1·A1

$u_1 = S_1 = 1 + 3 = 4$。 $u_2 = S_2 - S_1 = (4 + 6) - 4 = 6$。 $u_3 = S_3 - S_2 = (9 + 9) - 10 = 8$。

(b) $n \ge 2$ 时 $u_n$ 的闭合形式 M1·A1·A1

当 $n \ge 2$ 时： $$ u_n \;=\; S_n - S_{n-1} \;=\; (n^{2} + 3n) - \bigl[\,(n-1)^{2} + 3(n-1)\,\bigr]. $$ 展开 $(n-1)^{2} + 3(n-1) = n^{2} - 2n + 1 + 3n - 3 = n^{2} + n - 2$。故 $$ u_n \;=\; (n^{2} + 3n) - (n^{2} + n - 2) \;=\; 2n + 2. \qquad \blacksquare $$

(c) 在 $n = 1$ 处验证 A1

公式给出 $u_1 = 2(1) + 2 = 4$，与 (a) 中 $S_1 = 4$ 一致。所以 $u_n = 2n+2$ 对一切 $n \ge 1$ 都成立。

(d) 等差与公差 M1·A1

对一切 $n \ge 1$，$u_{n+1} - u_n = (2(n+1)+2) - (2n+2) = 2$。差恒为常数 $\Rightarrow \{u_n\}$ 是等差数列，公差 $d = 2$。

(e) $\sum u_{2k}$ 的闭合形式 M1·M1·A1·A1

在 $u_n = 2n + 2$ 中取 $n = 2k$： $$ u_{2k} \;=\; 2(2k) + 2 \;=\; 4k + 2. $$ 所以 $\{u_{2k}\}_{k = 1}^{n}$ 本身就是等差数列：首项 $u_2 = 6$，公差 $4$，共 $n$ 项。套用 $S = \tfrac{n}{2}(2a + (n-1)d)$： $$ \sum_{k=1}^{n} u_{2k} \;=\; \tfrac{n}{2}\bigl[\,2(6) + (n-1)(4)\,\bigr] \;=\; \tfrac{n}{2}(4n + 8) \;=\; 2n(n + 2) \;=\; 2n^{2} + 4n. $$

(f) 取 $n = 15$ M1·A1·A1

$$ \sum_{k=1}^{15} u_{2k} \;=\; 2(15)(17) \;=\; 510. $$

两层结构。只要题目给的 $S_n$ 是关于 $n$ 的多项式，$\{u_n\}$ 的次数必降一阶 —— 若 $S_n$ 为二次，$\{u_n\}$ 必为一次（即等差）。（速验：$S_n = an^2 + bn + c$ 强迫 $u_1 = a+b+c$，而 $n \ge 2$ 时 $u_n = 2an + (b-a)$；若要公式在 $n=1$ 也成立，必须 $c = 0$。）本题 $c = 0$、$a = 1$、$b = 3$，立得 $d = 2a = 2$ 与 $u_1 = a + b = 4$ —— 但 IB 评分仍要求上面的标准长写法。

Paper 2 — Worked Solutions第二卷 —— 详细解析

Q8MEDIUMPaper 21.4 Compound Interest[8 marks]

€$5000$ at $4.2\%$ p.a. compounded quarterly from 1 Jan 2025. (a) Value on 1 Jan 2031. (b) First 1 Jan date the balance exceeds €$7500$. (c) Comparison with annual compounding on the same date.2025 年 1 月 1 日起，€$5000$ 按年利率 $4.2\%$ 季度复利。(a) 2031 年 1 月 1 日的金额。(b) 首次余额超过 €$7500$ 的 1 月 1 日所在年份。(c) 同日与年复利方案的对比。

Answers:答案： (a) $\approx$ €$6424.54$ · (b) 1 January 20352035 年 1 月 1 日 · (c) Lily exceeds Jenny by $\approx$ €$24.58$.Lily 比 Jenny 多约 €$24.58$。

Compound-interest formula R1

$$ V \;=\; P\left(1 + \frac{r}{n}\right)^{nt}, \qquad P = 5000,\; r = 0.042,\; n = 4 \text{ (quarterly)},\; t = \text{years elapsed}. $$

(a) Value on 1 Jan 2031 ($t = 6$) M1·A1

$$ V \;=\; 5000\left(1 + \tfrac{0.042}{4}\right)^{4 \cdot 6} \;=\; 5000\,(1.0105)^{24} \;\stackrel{\text{GDC}}{\approx}\; 6424.54. $$ Rounded to the nearest cent: €$6424.54$.

(b) Find smallest $t$ with $V > 7500$ M1·M1·A1

We need $5000(1.0105)^{4t} > 7500$, i.e. $(1.0105)^{4t} > 1.5$. Take logs: $$ 4t\,\ln(1.0105) > \ln(1.5) \quad \Longrightarrow \quad 4t > \frac{\ln(1.5)}{\ln(1.0105)} \;\stackrel{\text{GDC}}{\approx}\; 38.808. $$ So $t > 9.702$. The smallest integer year is $t = 10$, i.e. 1 Jan 2035.

(c) Jenny (annual) versus Lily (quarterly), $t = 6$ M1·A1·A1

Jenny: $V_J = 5000\,(1.042)^{6} \stackrel{\text{GDC}}{\approx}$ €$6399.96$. Difference: $V_L - V_J \approx 6424.54 - 6399.96 = $ €$24.58$.

Why Lily wins. Quarterly compounding pays interest on accrued interest within each year, so the effective annual rate $(1.0105)^4 - 1 \approx 4.267\%$ slightly exceeds the nominal $4.2\%$ annual rate. The same nominal rate plus more frequent compounding always produces strictly greater growth.

The "effective rate" trick. When comparing schemes, collapse to the effective annual rate $i_{\text{eff}} = (1 + r/n)^n - 1$. For Lily, $i_{\text{eff}} \approx 4.267\%$; for Jenny, $i_{\text{eff}} = 4.2\%$. Then $V = P(1 + i_{\text{eff}})^t$ in both cases — and the comparison is reduced to comparing the effective rates. As $n \to \infty$ (continuous compounding), $i_{\text{eff}} \to e^{r} - 1 \approx 4.290\%$ for $r = 4.2\%$ — so the gap between Lily and continuous compounding is much smaller than the gap between Jenny and Lily.

复利公式 R1

$$ V \;=\; P\left(1 + \frac{r}{n}\right)^{nt}, \qquad P = 5000,\; r = 0.042,\; n = 4,\; t \in \mathbb{R}_{\ge 0}. $$ 其中 $n = 4$ 为按季复利，$t$ 为经过年数。

(a) 2031 年 1 月 1 日（$t = 6$）的金额 M1·A1

$$ V \;=\; 5000\left(1 + \tfrac{0.042}{4}\right)^{4 \cdot 6} \;=\; 5000\,(1.0105)^{24} \;\stackrel{\text{GDC}}{\approx}\; 6424.54. $$ 四舍五入到分：€$6424.54$。

(b) 求满足 $V > 7500$ 的最小 $t$ M1·M1·A1

要求 $5000(1.0105)^{4t} > 7500$，即 $(1.0105)^{4t} > 1.5$。两边取对数： $$ 4t\,\ln(1.0105) > \ln(1.5) \quad \Longrightarrow \quad 4t > \frac{\ln(1.5)}{\ln(1.0105)} \;\stackrel{\text{GDC}}{\approx}\; 38.808. $$ 故 $t > 9.702$。最小整数年为 $t = 10$，即 2035 年 1 月 1 日。

(c) 同期 Jenny（年复利） vs. Lily（季度复利），$t = 6$ M1·A1·A1

Jenny：$V_J = 5000\,(1.042)^{6} \stackrel{\text{GDC}}{\approx}$ €$6399.96$。差：$V_L - V_J \approx 6424.54 - 6399.96 = $ €$24.58$。

Lily 为何更大。季度复利在年内便把已计的利息再加入本金生息，故有效年利率（effective annual rate） $(1.0105)^4 - 1 \approx 4.267\%$ 略高于名义年利率 $4.2\%$。在名义利率相同的情况下，复利频率越高，增长越快。

"有效利率"小技巧。比较不同方案时，先化为有效年利率 $i_{\text{eff}} = (1 + r/n)^n - 1$。Lily 的 $i_{\text{eff}} \approx 4.267\%$、Jenny 的 $i_{\text{eff}} = 4.2\%$。然后两边都写成 $V = P(1 + i_{\text{eff}})^t$，比较就归结为比较有效利率。当 $n \to \infty$（连续复利，continuous compounding），$i_{\text{eff}} \to e^{r} - 1 \approx 4.290\%$（$r = 4.2\%$）—— 所以 Lily 与连续复利之差，远小于 Jenny 与 Lily 之差。

Q9MEDIUMPaper 21.4 / 1.8 Real-World GP[7 marks]

Ball dropped from $10$ m, bounces to $75\%$ of previous height. (a) Height of 3rd bounce. (b) Total vertical distance before resting. (c) Why is the total finite?小球从 $10$ m 下落，每次反弹至上一次高度的 $75\%$。(a) 第 3 次反弹高度。(b) 静止前的总垂直距离。(c) 为何总距离有限？

Answers:答案： (a) $4.21875 \approx 4.22$ m · (b) $70$ m · (c) $|r| < 1$ ⇒ infinite GP converges.$|r| < 1$ ⇒ 无穷 GP 收敛。

(a) Height of bounce $n$ M1·A1

Bounce heights form a GP: $h_n = 10 \cdot (0.75)^{n}$ (the initial drop is $h_0 = 10$, and after bounce $n$ the ball rises to a fraction $0.75^{n}$ of the original drop height). The third bounce: $$ h_3 \;=\; 10 \cdot (0.75)^{3} \;=\; 10 \cdot 0.421875 \;=\; 4.21875 \;\approx\; 4.22 \text{ m (3 s.f.).} $$

(b) Total vertical distance M1·M1·A1·A1

The ball moves $10$ m down on the initial drop, then for every bounce $k \ge 1$ it moves $h_k$ up and the same $h_k$ down (until it comes to rest). So $$ D \;=\; 10 \;+\; 2\sum_{k=1}^{\infty} h_k \;=\; 10 \;+\; 2\sum_{k=1}^{\infty} 10\,(0.75)^{k}. $$ The infinite GP $\sum_{k=1}^{\infty} 10\,(0.75)^{k}$ has first term $a = 10(0.75) = 7.5$ and common ratio $r = 0.75$, with $|r| < 1$. So $$ \sum_{k=1}^{\infty} h_k \;=\; \frac{7.5}{1 - 0.75} \;=\; \frac{7.5}{0.25} \;=\; 30. $$ Therefore $D = 10 + 2(30) = 70$ m.

(c) Why finite A1

The bounce heights form a geometric sequence with common ratio $|r| = 0.75 < 1$, so the infinite series converges and the total distance — a finite constant plus twice a convergent series — is finite.

The "up + down" doubling. The factor of $2$ in $D = 10 + 2 \sum h_k$ catches students out — they forget that each bounce contributes both an upward leg and an equal downward leg. The initial drop is special: it's only downward, so it sits outside the doubled sum. Drawing the first three bounces on a time-vs-height sketch makes the accounting visual and removes the slip.

(a) 第 $n$ 次反弹高度 M1·A1

反弹高度构成等比数列：$h_n = 10 \cdot (0.75)^{n}$（初次下落 $h_0 = 10$；第 $n$ 次反弹后到达原下落高度的 $0.75^{n}$）。第 3 次反弹： $$ h_3 \;=\; 10 \cdot (0.75)^{3} \;=\; 10 \cdot 0.421875 \;=\; 4.21875 \;\approx\; 4.22 \; \mathrm{m}. $$ （保留 3 位有效数字。）

(b) 总垂直距离 M1·M1·A1·A1

首次下落 $10$ m，之后每次反弹 $k \ge 1$ 都包含向上 $h_k$ 与同等长度的向下 $h_k$（直至静止）。所以 $$ D \;=\; 10 \;+\; 2\sum_{k=1}^{\infty} h_k \;=\; 10 \;+\; 2\sum_{k=1}^{\infty} 10\,(0.75)^{k}. $$ 无穷 GP $\sum_{k=1}^{\infty} 10\,(0.75)^{k}$ 的首项 $a = 10(0.75) = 7.5$，公比 $r = 0.75$，且 $|r| < 1$。故 $$ \sum_{k=1}^{\infty} h_k \;=\; \frac{7.5}{1 - 0.75} \;=\; \frac{7.5}{0.25} \;=\; 30. $$ 因此 $D = 10 + 2(30) = 70$ m。

(c) 为何有限 A1

反弹高度构成的等比数列公比 $|r| = 0.75 < 1$，故无穷级数收敛，总距离（一个有限常数加上两倍的收敛级数）为有限值。

"上 + 下"翻倍计法。$D = 10 + 2 \sum h_k$ 中的因子 $2$ 是学生常踩的坑 —— 容易忘记每次反弹同时贡献向上与向下两段同长。首次下落是特例：只有向下，所以独立写在翻倍求和之外。把前三次反弹画成时间‑高度（time vs height）草图，可视化即可去除该失误。

Q10HARDPaper 21.2 / 1.3 / 1.4 Hybrid[9 marks]

30-year salary trajectory: year-1 €$42\,000$, year-30 €$71\,000$. (a) AP increment $d$. (b) AP total earnings. (c)(i) GP common ratio $r$ (4 dp). (c)(ii) Compare GP vs AP total without explicit summation.30 年薪酬轨迹：第 1 年 €$42\,000$、第 30 年 €$71\,000$。(a) AP 年增量 $d$。(b) AP 总收入。(c)(i) GP 公比 $r$（保留 4 位小数）。(c)(ii) 不必算总和，比较 GP 与 AP 的总收入。

Answers:答案： (a) $d = $ €$1000$/yr · (b) €$1\,695\,000$ · (c)(i) $r \approx 1.0183$ · (c)(ii) GP scheme total < AP total (convexity).GP 总收入 < AP 总收入（凸性）。

(a) Annual increment M1·A1

$u_{30} = u_1 + 29 d \Rightarrow 71\,000 = 42\,000 + 29d \Rightarrow d = \tfrac{29\,000}{29} = $ €$1000$ per year.

(b) 30-year AP total M1·A1·A1

$$ S_{30} \;=\; \tfrac{30}{2}(u_1 + u_{30}) \;=\; 15 \cdot (42\,000 + 71\,000) \;=\; 15 \cdot 113\,000 \;=\; \text{€}1\,695\,000. $$

(c)(i) GP ratio M1·A1

$v_n = v_1\, r^{n-1}$ with $v_1 = 42\,000$ and $v_{30} = 71\,000$: $$ r^{29} \;=\; \frac{v_{30}}{v_1} \;=\; \frac{71\,000}{42\,000} \;=\; \frac{71}{42} \;\approx\; 1.6905. $$ So $r = (71/42)^{1/29} \stackrel{\text{GDC}}{\approx} 1.01828$, i.e. $r \approx 1.0183$ (4 dp).

(c)(ii) GP vs AP total — convexity argument M1·A1

Both schemes share endpoints: $v_1 = u_1$ and $v_{30} = u_{30}$. The AP $u_n = 42\,000 + 1000(n-1)$ is the linear chord joining the points $(1, 42\,000)$ and $(30, 71\,000)$. The GP $v_n = 42\,000 \cdot r^{n-1}$ with $r > 1$ is a strictly convex exponential function of $n$. A strictly convex curve through two fixed endpoints lies strictly below its chord on the open interval. Hence $v_n < u_n$ for all $1 < n < 30$, and therefore $$ \sum_{n=1}^{30} v_n \;<\; \sum_{n=1}^{30} u_n \;=\; \text{€}1\,695\,000. $$ So the geometric scheme produces lower total earnings than the arithmetic scheme.

The convexity intuition. Fixing two endpoints constrains where the function can go. A convex (positively-curved) function "sags down" in the middle relative to the linear chord; a concave one "bulges up." For salary growth, the GP scheme back-loads pay (small early raises, larger later raises) while the AP scheme spreads raises evenly. Same final salary, but the GP worker earns less over the career because each early salary is below the AP equivalent. This trade-off is the same principle behind "compound interest is back-loaded" — and it explains why a 50-year retirement saver who starts late ends up far behind, even with higher contribution rates.

(a) 年增量 M1·A1

$u_{30} = u_1 + 29 d \Rightarrow 71\,000 = 42\,000 + 29d \Rightarrow d = \tfrac{29\,000}{29} = $ €$1000$/年。

(b) 30 年 AP 总收入 M1·A1·A1

$$ S_{30} \;=\; \tfrac{30}{2}(u_1 + u_{30}) \;=\; 15 \cdot (42\,000 + 71\,000) \;=\; 15 \cdot 113\,000 \;=\; \text{€}1\,695\,000. $$

(c)(i) GP 公比 M1·A1

$v_n = v_1\, r^{n-1}$，$v_1 = 42\,000$、$v_{30} = 71\,000$： $$ r^{29} \;=\; \frac{v_{30}}{v_1} \;=\; \frac{71\,000}{42\,000} \;=\; \frac{71}{42} \;\approx\; 1.6905. $$ 故 $r = (71/42)^{1/29} \stackrel{\text{GDC}}{\approx} 1.01828$，即 $r \approx 1.0183$（4 位小数）。

(c)(ii) GP vs AP 总收入 —— 凸性论证 M1·A1

两套方案端点相同：$v_1 = u_1$、$v_{30} = u_{30}$。等差 $u_n = 42\,000 + 1000(n-1)$ 即为连接点 $(1, 42\,000)$ 与 $(30, 71\,000)$ 的线段（linear chord）。等比 $v_n = 42\,000 \cdot r^{n-1}$ 当 $r > 1$ 时是严格凸（strictly convex）的指数函数。固定两端点的严格凸曲线在开区间上严格低于其弦。故 $v_n < u_n$ 对一切 $1 < n < 30$ 成立，于是 $$ \sum_{n=1}^{30} v_n \;<\; \sum_{n=1}^{30} u_n \;=\; \text{€}1\,695\,000. $$ 所以等比方案的总收入低于等差方案。

凸性的直觉。固定两端等价于约束函数能走到哪里。凸函数（正曲率）在中间会"下凹"低于弦；凹函数则"上凸"。对薪酬来说，GP 方案是"后置"加薪（前期涨得少、后期涨得多），AP 方案则均匀分布。终薪相同，但 GP 员工整个职业生涯赚得更少 —— 因为前期每个年薪都低于等差对应值。这一权衡和"复利后置"是同一个原理 —— 也解释了为何起步晚的退休储蓄者即使提高缴费率仍会被早起步者甩开。

Paper 3 — Worked Solutions第三卷 —— 详细解析

Q11HARDPaper 31.8 Infinite GP & Power Series[16 marks]

Build the link between $\frac{1}{1-x}$ and $\frac{1}{(1-x)^2}$ via term-by-term differentiation, evaluate $\sum k\,(1/2)^{k-1}$, then conjecture a closed form for the triangular-number power series.用逐项求导建立 $\frac{1}{1-x}$ 与 $\frac{1}{(1-x)^2}$ 的联系，计算 $\sum k\,(1/2)^{k-1}$，再猜测三角形数幂级数的闭合形式。

Answers:答案： (a) $S_\infty = \dfrac{u_1}{1-r}$, $|r| < 1$ · (b) $f(x) = \dfrac{1}{1-x}$ · (c) $g(x) = \dfrac{1}{(1-x)^{2}}$ · (d) $4$ · (e) $\approx 3.9766$ · (f) $h(x) = \dfrac{1}{(1-x)^{3}}$

(a) Infinite GP sum + convergence condition A1·R1

$$ S_\infty \;=\; \frac{u_1}{1 - r}, \qquad \text{valid iff } |r| < 1. $$

(b) Closed form for $f(x) = 1 + x + x^2 + x^3 + \cdots$ A1

Apply (a) with $u_1 = 1$ and $r = x$ (so $|x| < 1$): $$ f(x) \;=\; \frac{1}{1 - x}, \qquad |x| < 1. $$

(c) Derive $g(x) = 1 + 2x + 3x^2 + 4x^3 + \cdots$ M1·M1·A1·A1

Differentiate both sides of $f(x) = 1/(1-x)$ with respect to $x$.

RHS, by chain rule. $\dfrac{d}{dx}(1-x)^{-1} = -1 \cdot (1-x)^{-2} \cdot (-1) = (1-x)^{-2} = \dfrac{1}{(1-x)^{2}}$.

LHS, term-by-term. $\dfrac{d}{dx}\bigl[\,1 + x + x^{2} + x^{3} + \cdots\bigr] = 0 + 1 + 2x + 3x^{2} + 4x^{3} + \cdots = g(x)$.

Equating and noting that termwise differentiation of a power series is valid inside its radius of convergence ($|x| < 1$): $$ g(x) \;=\; \frac{1}{(1 - x)^{2}}, \qquad |x| < 1. $$

(d) Evaluate $\sum_{k=1}^\infty k\,(1/2)^{k-1}$ M1·M1·A1

Recognise $g(x) = \sum_{k=1}^{\infty} k\,x^{k-1}$. Set $x = \tfrac{1}{2}$ (well inside the radius of convergence): $$ \sum_{k=1}^{\infty} k\,(1/2)^{k-1} \;=\; g(1/2) \;=\; \frac{1}{(1 - 1/2)^{2}} \;=\; \frac{1}{(1/2)^{2}} \;=\; 4. $$

(e) GDC partial sum to $k = 10$ A1·R1

$$ \sum_{k=1}^{10} k\,(1/2)^{k-1} \;=\; 1 + 1 + \tfrac{3}{4} + \tfrac{1}{2} + \tfrac{5}{16} + \tfrac{3}{16} + \tfrac{7}{64} + \tfrac{1}{16} + \tfrac{9}{256} + \tfrac{5}{256} \;\stackrel{\text{GDC}}{\approx}\; 3.9766. $$ Compared to the predicted limit $4$, the partial sum is about $0.023$ short — the series is converging fast (ratio $1/2$ deep inside the radius), so even $10$ terms get us to roughly $99.4\%$ of the limit.

(f) Conjecture closed form for $h(x) = 1 + 3x + 6x^2 + 10x^3 + 15x^4 + \cdots$ M1·M1·A1·A1

Differentiate $g(x) = 1/(1-x)^{2}$ term-by-term: $$ g'(x) \;=\; \frac{d}{dx}\Bigl[\,(1 - x)^{-2}\,\Bigr] \;=\; -2 \cdot (1-x)^{-3} \cdot (-1) \;=\; \frac{2}{(1-x)^{3}}. $$ Termwise: $g'(x) = 0 + 2 + 6x + 12x^{2} + 20x^{3} + 30x^{4} + \cdots$. Group out the factor of $2$: $$ g'(x) \;=\; 2\,\bigl(\,1 + 3x + 6x^{2} + 10x^{3} + 15x^{4} + \cdots\bigr) \;=\; 2\,h(x). $$ Solve: $$ h(x) \;=\; \tfrac{1}{2}\,g'(x) \;=\; \frac{1}{(1-x)^{3}}, \qquad |x| < 1. $$

The differentiation tower. The three identities you just derived sit in a tower: $$ \frac{1}{1-x} = \sum x^{k}, \qquad \frac{1}{(1-x)^{2}} = \sum k\,x^{k-1}, \qquad \frac{1}{(1-x)^{3}} = \sum \tfrac{k(k+1)}{2}\,x^{k-1}. $$ One more differentiation gives $\frac{1}{(1-x)^{4}} = \sum \binom{k+2}{3} x^{k-1}$ — the tetrahedral-number generating function. The general rule (HL exploration material): $\dfrac{1}{(1-x)^{n+1}} = \sum_{k=0}^{\infty} \binom{k+n}{n} x^{k}$. These negative binomial series show up in physics (vibrational mode counts), probability (negative-binomial distributions), and combinatorics (compositions with $n$ parts). The Paper 3 strategy: derive the tower-of-three explicitly, then use it to evaluate weighted sums by clever choice of $x$.

(a) 无穷 GP 求和 + 收敛条件 A1·R1

$$ S_\infty \;=\; \frac{u_1}{1 - r}, \qquad |r| < 1. $$ 当且仅当 $|r| < 1$ 时该公式成立。

(b) $f(x) = 1 + x + x^2 + x^3 + \cdots$ 的闭合形式 A1

在 (a) 中取 $u_1 = 1$、$r = x$（要求 $|x| < 1$）： $$ f(x) \;=\; \frac{1}{1 - x}, \qquad |x| < 1. $$

(c) 推导 $g(x) = 1 + 2x + 3x^2 + 4x^3 + \cdots$ M1·M1·A1·A1

对 $f(x) = 1/(1-x)$ 两边关于 $x$ 求导。

右边，链式法则。$\dfrac{d}{dx}(1-x)^{-1} = -1 \cdot (1-x)^{-2} \cdot (-1) = (1-x)^{-2} = \dfrac{1}{(1-x)^{2}}$。

左边，逐项求导。$\dfrac{d}{dx}\bigl[\,1 + x + x^{2} + x^{3} + \cdots\bigr] = 0 + 1 + 2x + 3x^{2} + 4x^{3} + \cdots = g(x)$。

两边相等；幂级数的逐项求导在其收敛半径（radius of convergence，$|x| < 1$）内合法： $$ g(x) \;=\; \frac{1}{(1 - x)^{2}}, \qquad |x| < 1. $$

(d) 计算 $\sum_{k=1}^\infty k\,(1/2)^{k-1}$ M1·M1·A1

识别 $g(x) = \sum_{k=1}^{\infty} k\,x^{k-1}$。代入 $x = \tfrac{1}{2}$（位于收敛半径内）： $$ \sum_{k=1}^{\infty} k\,(1/2)^{k-1} \;=\; g(1/2) \;=\; \frac{1}{(1 - 1/2)^{2}} \;=\; \frac{1}{(1/2)^{2}} \;=\; 4. $$

(e) GDC 部分和（取到 $k = 10$） A1·R1

$$ \sum_{k=1}^{10} k\,(1/2)^{k-1} \;=\; 1 + 1 + \tfrac{3}{4} + \tfrac{1}{2} + \tfrac{5}{16} + \tfrac{3}{16} + \tfrac{7}{64} + \tfrac{1}{16} + \tfrac{9}{256} + \tfrac{5}{256} \;\stackrel{\text{GDC}}{\approx}\; 3.9766. $$ 与所预测极限 $4$ 比较，部分和约差 $0.023$ —— 级数收敛极快（公比 $1/2$ 远在收敛半径内），$10$ 项即可达到极限的约 $99.4\%$。

(f) 猜测 $h(x) = 1 + 3x + 6x^2 + 10x^3 + 15x^4 + \cdots$ 的闭合形式 M1·M1·A1·A1

对 $g(x) = 1/(1-x)^{2}$ 逐项求导： $$ g'(x) \;=\; \frac{d}{dx}\Bigl[\,(1 - x)^{-2}\,\Bigr] \;=\; -2 \cdot (1-x)^{-3} \cdot (-1) \;=\; \frac{2}{(1-x)^{3}}. $$ 逐项：$g'(x) = 0 + 2 + 6x + 12x^{2} + 20x^{3} + 30x^{4} + \cdots$。提出因子 $2$： $$ g'(x) \;=\; 2\,\bigl(\,1 + 3x + 6x^{2} + 10x^{3} + 15x^{4} + \cdots\bigr) \;=\; 2\,h(x). $$ 解出： $$ h(x) \;=\; \tfrac{1}{2}\,g'(x) \;=\; \frac{1}{(1-x)^{3}}, \qquad |x| < 1. $$

求导塔。刚刚推出的三条恒等式构成一座"塔"： $$ \frac{1}{1-x} = \sum x^{k}, \qquad \frac{1}{(1-x)^{2}} = \sum k\,x^{k-1}, \qquad \frac{1}{(1-x)^{3}} = \sum \tfrac{k(k+1)}{2}\,x^{k-1}. $$ 再求导一次可得 $\frac{1}{(1-x)^{4}} = \sum \binom{k+2}{3} x^{k-1}$ —— 即四面体数（tetrahedral numbers）的母函数。一般公式（HL 探究范围）：$\dfrac{1}{(1-x)^{n+1}} = \sum_{k=0}^{\infty} \binom{k+n}{n} x^{k}$。这类负二项级数（negative binomial series）出现于物理（振动模式计数）、概率（负二项分布）、组合数学（$n$ 段的组合分拆）之中。Paper 3 思路：先显式推导这座"塔"，再通过巧选 $x$ 来求各类加权和。