Sequences
and Series数列与级数

A function $f : A \to B$ assigns to each input in $A$ a single output in $B$. A sequence $(a_n)_{n \ge 1}$ does exactly this with $A = \mathbb{N}$ and $B = \mathbb{R}$; the subscript $a_n$ is shorthand for $a(n)$. Everything you know about function notation carries over: domain, range, equality, composition. The recursive-vs-explicit dichotomy is the same as the difference between an iteration $x_{n+1} = g(x_n)$ and a closed-form $x_n = h(n)$ , iterations are easy to define from a rule, closed forms are easy to evaluate. CCSSM HSF-BF.A.2 asks for fluency both directions on the two families where the closed form is mechanical.函数 $f : A \to B$ 把 $A$ 中的每个输入唯一对应到 $B$ 中的输出。数列 $(a_n)_{n \ge 1}$ 正是取 $A = \mathbb{N}$、$B = \mathbb{R}$ 的特例；下标 $a_n$ 是 $a(n)$ 的简写。函数那一套（定义域、值域、相等、复合）全部继承下来。递推与通项之分，就是迭代 $x_{n+1} = g(x_n)$ 与闭式 $x_n = h(n)$ 之分 , 迭代易由规则写出，闭式易于求值。CCSSM HSF-BF.A.2 要求在两类闭式可机械写出的家族中熟练双向转换。

Plot $(n, a_n)$ for $n = 1, 2, 3, \dots$. Because $a_{n+1} - a_n = d$ is constant, the dots lie on a line of slope $d$; the continuous extension is $f(x) = d x + (a_1 - d)$. Ontario MCR3U C1.1 names this explicitly ("arithmetic sequences correspond to discrete linear functions"); BC FMP&PC10 names the same picture as "connecting arithmetic sequences to linear relations." Whichever curriculum you sit inside, the two ideas are the same idea wearing different vocabulary.对 $n = 1, 2, 3, \dots$ 描点 $(n, a_n)$。因为 $a_{n+1} - a_n = d$ 恒定，所有点位于斜率为 $d$ 的直线上；其连续延拓为 $f(x) = d x + (a_1 - d)$。安大略 MCR3U C1.1 明确指出："等差数列对应离散线性函数"；BC FMP&PC10 描述为"将等差数列与线性关系联系起来"。无论你身处哪一大纲，这两种说法是同一概念的不同表达。

Write the sum forwards and backwards, then add columnwise:把求和正序与逆序写出，再按列相加：

For arithmetic sequences each "balanced" pair sums to the same value: $a_k + a_{n-k+1} = 2 a_1 + (n - 1) d = a_1 + a_n$. So adding the two lines gives $n$ copies of $a_1 + a_n$:对等差数列而言，每一组"对称"配对之和相同：$a_k + a_{n-k+1} = 2 a_1 + (n - 1) d = a_1 + a_n$。两行相加便得 $n$ 个 $a_1 + a_n$：

The apocryphal story: young Gauss summed $1 + 2 + \dots + 100$ by spotting that $1 + 100 = 2 + 99 = \dots = 101$, with $50$ pairs, giving $5050$.流传的轶事：少年高斯察觉 $1 + 100 = 2 + 99 = \dots = 101$，凑出 $50$ 对，得 $5050$，从而瞬间算出 $1 + 2 + \dots + 100$。

BC PC12's Connecting and reflecting Curricular Competency reads: "examine the structure of and connections between mathematical ideas (e.g., exponential functions to geometric sequences)." Start from a continuous exponential $f(x) = A \cdot b^{x}$ and sample at integer inputs:BC PC12 的"联系与反思"课程胜任力写道："考察数学思想的结构与联系（例如指数函数与等比数列）。"自连续指数函数 $f(x) = A \cdot b^{x}$ 出发，仅在整数输入处取样：

That is a geometric sequence with first term $A$ and common ratio $b$; re-indexing to start at $n = 1$ gives $a_n = A \cdot b^{n - 1}$. Conversely, every geometric sequence $a_n = a_1 r^{n - 1}$ is a sample of the exponential $f(x) = (a_1 / r) \cdot r^{x}$. The per-step multiplier $r$ is the exponential base.这是首项 $A$、公比 $b$ 的等比数列；重置索引从 $n = 1$ 起便得 $a_n = A \cdot b^{n - 1}$。反过来，任一等比数列 $a_n = a_1 r^{n - 1}$ 都是指数函数 $f(x) = (a_1 / r) \cdot r^{x}$ 的取样。每步的乘数 $r$ 就是指数底数。

This is why compound interest (Ontario MCR3U C3.1) sits naturally in the discrete-functions strand: principal $P$ grows by factor $(1 + r)$ per compounding period, a geometric sequence in the period number and an exponential in continuous time. The same picture re-appears in Unit 5 and in IB Math HL B1.这正是为何复利（安大略 MCR3U C3.1）自然落入"离散函数"链：本金 $P$ 每个计息周期乘以 $(1 + r)$，按周期编号看是等比数列，按连续时间看是指数函数。此图象在第 5 单元及 IB Math HL B1 再次出现。

The "shift, subtract, factor" trick. Write"错位相减"技巧。写出

Subtract: every term in $r S_n$ matches a term in $S_n$ shifted by one, so only the first term of $S_n$ and the last of $r S_n$ survive:相减：$r S_n$ 中每一项都与 $S_n$ 错位一项后对应，仅剩 $S_n$ 的首项与 $r S_n$ 的末项：

This is the derivation HSA-SSE.B.4 asks for. The argument works for any $r \ne 1$. The infinite series in §6 is this formula's limit as $n \to \infty$ , well-defined only when $r^{n} \to 0$, i.e. $|r| < 1$.这正是 HSA-SSE.B.4 要求的推导。对任意 $r \ne 1$ 都成立。§6 中的无穷级数即此式 $n \to \infty$ 的极限 , 仅当 $r^{n} \to 0$（即 $|r| < 1$）时该极限有定义。

The finite-sum formula $S_n = a_1 (1 - r^n)/(1 - r)$ works for any $r \ne 1$. The infinite-sum $S_\infty = a_1/(1 - r)$ is a limit: the number $S_n$ approaches as $n \to \infty$. Whether it exists depends on $r^n$.有限求和公式 $S_n = a_1 (1 - r^n)/(1 - r)$ 对任意 $r \ne 1$ 都成立。无穷和 $S_\infty = a_1/(1 - r)$ 则是极限：$n \to \infty$ 时 $S_n$ 趋近的值。是否存在取决于 $r^n$。

• $|r| < 1$: $r^{n} \to 0$, so $S_n \to a_1 / (1 - r)$. Converges.$|r| < 1$：$r^{n} \to 0$，故 $S_n \to a_1 / (1 - r)$。收敛。
• $r = 1$: partial sums $a_1, 2 a_1, 3 a_1, \dots$ unbounded. Diverges.$r = 1$：部分和 $a_1, 2 a_1, 3 a_1, \dots$ 无界。发散。
• $r = -1$: partial sums oscillate $a_1, 0, a_1, 0, \dots$ Diverges.$r = -1$：部分和振荡 $a_1, 0, a_1, 0, \dots$。发散。
• $|r| > 1$: $|r|^{n}$ grows without bound; $S_n$ grows in magnitude. Diverges.$|r| > 1$：$|r|^{n}$ 无界增长；$S_n$ 模长发散。发散。

This is the simplest non-trivial convergence test. It re-appears as the first named test in AP Calc BC Unit 10 and as the prototype for the ratio test. The Maclaurin series of $1/(1 - x)$ in IB Math HL E6 is literally an infinite geometric series with $a_1 = 1$, $r = x$ , valid for $|x| < 1$ by exactly this argument.这是最简单的非平凡收敛判别。它在 AP Calc BC Unit 10 作为第一个命名判别法再次登场，也是比值判别法的原型。IB Math HL E6 中 $1/(1 - x)$ 的麦克劳林级数就是 $a_1 = 1$、$r = x$ 的无穷等比 , 其 $|x| < 1$ 收敛域恰由此论证。

The summation index is a dummy variable , renamable and shiftable. Compare求和变量是哑指标 , 可重命名、可平移。对照

Substitute $j = k - 1$: $k = 1 \to j = 0$; $k = n \to j = n - 1$; $r^{k - 1} = r^{j}$. Different bounds, same value. This trick is essential when combining two sums whose indices don't quite line up , AP Calc BC's term-by-term differentiation of power series, IB Math HL E6's Maclaurin re-indexing, linear algebra's summation conventions. Get it clean here.令 $j = k - 1$：$k = 1 \to j = 0$；$k = n \to j = n - 1$；$r^{k - 1} = r^{j}$。上下界不同，数值相同。当两个求和的索引并不对齐时，此技巧不可或缺 , AP Calc BC 幂级数的逐项求导、IB Math HL E6 麦克劳林展开的重新编号、线性代数的求和约定皆然。此处先练熟。