数列与规律

考什么 · What it tests

需要先会 · Prerequisites

• 等差数列通项 $a_n = a_1 + (n-1)d$ / The arithmetic term formula $a_n = a_1 + (n-1)d$
• 四则运算与代数式代入 / Arithmetic and substituting into expressions
• 余数（周期数列要用） / Remainders (needed for periodic sequences)

常见套路 · Common moves

1. 等差求和用「首尾配对」：和 =\frac{(\text{首项}+\text{末项})\times\text{项数}}{2}，本质是把两端往中间配成相等的对——最常用也最快。

   Sum an arithmetic series by pairing ends: sum $\displaystyle =\frac{(\text{first}+\text{last})\times\text{count}}{2}$, which really just pairs the two ends inward into equal sums — the most common and fastest move.
2. 周期数列先找周期，再用余数定位：数列每 $k$ 项循环一次，第 $n$ 项就看 $n$ 除以 $k$ 的余数落在周期哪个位置。

   For periodic sequences, find the period then use a remainder to locate: if it repeats every $k$ terms, the $n$-th term is fixed by $n$ divided by $k$.
3. 项数别数错：从第 $m$ 到第 $n$ 项一共 n-m+1 项，「加 1」容易漏。

   Don't miscount the number of terms: from term $m$ to term $n$ there are n-m+1 terms — the "+1" is easy to drop.

易错点 · Common pitfalls

• 项数漏「加 1」：1 到 100 是 100 个数不是 99 个，求和时代错项数满盘皆输。

  Dropping the "+1": 1 to 100 is 100 numbers, not 99, and a wrong count wrecks the whole sum.
• 把非等差的数列硬套等差公式：先验证「相邻两项差是否恒定」，差不恒定不能用等差求和。

  Forcing the arithmetic formula onto a non-arithmetic sequence: first check that consecutive differences are constant — if not, the sum formula doesn't apply.