📚 A General Formula: Sequences and Binomial Expansion | 通用公式：数列与二项式展开
In A-Level Mathematics, a "general formula" is a single expression that allows you to calculate any term in a sequence or any specific term in a binomial expansion without listing all previous terms. Mastering general formulas saves time in exams and deepens your understanding of patterns.
在 A-Level 数学中，“通用公式”是一个单一表达式，让你无需逐一列出前面的项，就能计算数列中任意一项或二项式展开式中某一特定项。掌握通用公式不仅能节省考试时间，还能加深你对规律的理解。
1. What Is a General Formula? | 什么是通用公式？
tten as uₙ = f(n); for binomial expansion it gives each term in the expansion of (a + b)ⁿ.
通用公式直接用 n（或 r）表示第 n 项（或第 r 项）。对于数列，通常写作 uₙ = f(n)；对于二项式展开，它给出 (a + b)ⁿ 展开式中的每一项。
lation such as uₙ₊₁ = uₙ + d, because you can substitute n directly and obtain the answer.
使用通用公式比使用递推关系（如 uₙ₊₁ = uₙ + d）快得多，因为你可以直接代入 n 得到答案。
2. Arithmetic Sequences: The nth Term Formula | 等差数列：第 n 项公式
An arithmetic sequence has a constant difference d between consecutive terms. The general formula for the nth term is:
等差数列相邻两项之差为常数 d。第 n 项的通用公式为：
uₙ = a + (n − 1)d
the sequence 3, 7, 11, 15, ... we have a = 3 and d = 4, so uₙ = 3 + (n − 1) × 4 = 4n − 1.
其中 a 是首项，d 是公差，n 是项的位置。例如，在数列 3, 7, 11, 15, ... 中，a = 3，d = 4，因此 uₙ = 3 + (n − 1) × 4 = 4n − 1。
This formula comes from starting at a and adding d a total of (n − 1) times to reach the nth term.
这个公式来源于从首项 a 开始，加上 (n − 1) 次公差 d 来得到第 n 项。
3. Arithmetic Series: Sum Formula | 等差数列：求和公式
To add the first n terms of an arithmetic sequence, use the sum formula. There are two useful versions:
要计算等差数列前 n 项的和，可以使用求和公式。有两个常用形式：
Sₙ = n/2 [2a + (n − 1)d]
Sₙ = n/2 (a + l)
Here l is the last term. The second version is very quick if you already know the first and last terms.
其中 l 是最后一项。如果你已经知道首项和末项，第二个公式会非常快捷。
of the first 20 terms of 2, 5, 8, 11, ... is S₂₀ = 20/2 [2×2 + (20−1)×3] = 10 × 61 = 610.
例如，数列 2, 5, 8, 11, ... 前 20 项的和为 S₂₀ = 20/2 [2×2 + (20−1)×3] = 10 × 61 = 610。
4. Geometric Sequences: The nth Term Formula | 等比数列：第 n 项公式
A geometric sequence has a constant ratio r between consecutive terms. The general formula for the nth term is:
等比数列相邻两项之比为常数 r。第 n 项的通用公式为：
uₙ = arⁿ⁻¹
For example, in the sequence 5, 10, 20, 40, ... we have a = 5 and r = 2, so uₙ = 5 × 2ⁿ⁻¹.
其中 a 是首项，r 是公比。例如，在数列 5, 10, 20, 40, ... 中，a = 5，r = 2，因此 uₙ = 5 × 2ⁿ⁻¹。
Be careful with the power: the first term has r⁰, so the exponent is n − 1, not n.
注意指数：第一项对应 r⁰，因此指数是 n − 1，而不是 n。
5. Geometric Series: Finite and Infinite Sums | 等比数列：有限与无限求和
The sum of the first n terms of a geometric sequence is:
等比数列前 n 项的和为：
Sₙ = a(1 − rⁿ) / (1 − r) for r ≠ 1
= an. The formula is derived by multiplying Sₙ by r and subtracting to cancel most terms.
如果 r = 1，则 Sₙ = an。这个公式的推导方法是把 Sₙ 乘以 r 后相减，使大部分项相互抵消。
When |r| < 1, the infinite geometric series converges to:
当 |r| < 1 时，无穷等比数列收敛于：
S∞ = a / (1 − r)
the infinite series 8 + 4 + 2 + 1 + ... has a = 8 and r = 1/2, so S∞ = 8 / (1 − 1/2) = 16.
例如，无穷级数 8 + 4 + 2 + 1 + ... 中 a = 8，r = 1/2，因此 S∞ = 8 / (1 − 1/2) = 16。
6. Finding a Formula from Given Terms | 从已知项求公式
In exams you may be given two terms and asked to find a general formula. For arithmetic sequences, use the fact that the difference between terms is a multiple of d.
考试中可能会给出两项，要求你求出通用公式。对于等差数列，可以利用两项之差是 d 的倍数这一事实。
d = 3. Since u₃ = a + 2d, we have 10 = a + 6, so a = 4. Hence uₙ = 4 + 3(n − 1) = 3n + 1.
例如，如果 u₃ = 10 且 u₇ = 22，那么 4d = 12，所以 d = 3。由 u₃ = a + 2d 得 10 = a + 6，因此 a = 4。于是 uₙ = 4 + 3(n − 1) = 3n + 1。
For geometric sequences, divide one term by another to eliminate a and solve for r first.
对于等比数列，可以用两项相除消去 a，先求出 r。
7. The Binomial Theorem: A General Formula | 二项式定理：通用公式
The binomial theorem gives a general formula for expanding (a + b)ⁿ where n is a positive integer:
二项式定理给出了展开 (a + b)ⁿ（n 为正整数）的通用公式：
(a + b)ⁿ = nC0 aⁿ + nC1 aⁿ⁻¹b + nC2 aⁿ⁻²b² + ... + nCr aⁿ⁻ʳ bʳ + ... + nCn bⁿ
Here nCr = n! / [r!(n − r)!] is the binomial coefficient. It tells you how many ways to choose r items from n items.
其中 nCr = n! / [r!(n − r)!] 是二项式系数，表示从 n 个物品中选出 r 个的组合数。
xample, (x + 2)⁴ = x⁴ + 4C1 x³·2 + 4C2 x²·2² + 4C3 x·2³ + 2⁴ = x⁴ + 8x³ + 24x² + 32x + 16.
例如，(x + 2)⁴ = x⁴ + 4C1 x³·2 + 4C2 x²·2² + 4C3 x·2³ + 2⁴ = x⁴ + 8x³ + 24x² + 32x + 16。
8. The General Term in a Binomial Expansion | 二项式展开中的通项
The (r + 1)th term in the expansion of (a + b)ⁿ is given by:
(a + b)ⁿ 展开式中的第 (r + 1) 项为：
Tᵣ₊₁ = nCr aⁿ⁻ʳ bʳ
Here r starts at 0 for the first term and goes up to n for the last term. This is extremely useful when you only need one specific term, not the whole expansion.
这里 r 从第一项的 0 开始，一直到第 n 项。当你只需要某一个特定项而无需完整展开时，这个公式非常有用。
he expansion of (2x − 3)⁵, set a = 2x, b = −3 and solve for r so that the power of x is 3.
例如，要在 (2x − 3)⁵ 的展开式中求 x³ 项，设 a = 2x，b = −3，并求出使 x 的指数为 3 的 r 值。
9. Using the General Formula to Find Specific Terms | 用通项公式求特定项
Let us find the coefficient of x³ in (2x − 3)⁵. The general term is Tᵣ₊₁ = 5Cr (2x)⁵⁻ʳ (−3)ʳ. The power of x is 5 − r, so we set 5 − r = 3, giving r = 2.
我们来求 (2x − 3)⁵ 中 x³ 的系数。通项为 Tᵣ₊₁ = 5Cr (2x)⁵⁻ʳ (−3)ʳ。x 的指数是 5 − r，因此令 5 − r = 3，得 r = 2。
Then T₃ = 5C2 (2x)³ (−3)² = 10 × 8x³ × 9 = 720x³, so the coefficient is 720.
于是 T₃ = 5C2 (2x)³ (−3)² = 10 × 8x³ × 9 = 720x³，因此系数为 720。
f x is not 1, remember to apply the power to the whole coefficient, as we did with 2³ = 8.
当 x 的系数不是 1 时，记得要把整个系数进行乘方，就像我们计算 2³ = 8 一样。
10. Common Mistakes and Exam Tips | 常见错误与考试技巧
- Mistake: Using uₙ = arⁿ instead of arⁿ⁻¹. Tip: Check by substituting n = 1; you must get a.
- 错误：使用 uₙ = arⁿ 而不是 arⁿ⁻¹。技巧：代入 n = 1 检查，结果必须是 a。
- Mistake: Forgetting that the binomial general term is Tᵣ₊₁, not Tᵣ. Tip: Write Tᵣ₊₁ every time and label r clearly.
- 错误：忘记二项式通项是 Tᵣ₊₁ 而不是 Tᵣ。技巧：每次都写出 Tᵣ₊₁，并清楚标注 r。
- Mistake: Using the sum formula for arithmetic series with n/2 but forgetting to multiply the whole bracket. Tip: Use brackets carefully.
- 错误：等差数列求和公式中用了 n/2 但忘记乘以整个括号。技巧：仔细使用括号。
- Mistake: Assuming an infinite geometric series always converges. Tip: Always check |r| < 1 first.
- 错误：认为无穷等比数列总是收敛。技巧：首先检查 |r| < 1。
11. Worked Example | 例题解析
A geometric sequence has third term 12 and sixth term 96. Find the general formula for the nth term and the sum of the first 8 terms.
一个等比数列的第三项为 12，第六项为 96。求第 n 项的通用公式以及前 8 项的和。
u₆ = ar⁵ = 96, dividing gives r³ = 8, so r = 2. Then ar² = 12 gives a × 4 = 12, so a = 3.
由 u₃ = ar² = 12 和 u₆ = ar⁵ = 96，相除得 r³ = 8，所以 r = 2。再由 ar² = 12 得 a × 4 = 12，因此 a = 3。
2ⁿ⁻¹. The sum of the first 8 terms is S₈ = 3(1 − 2⁸) / (1 − 2) = 3(1 − 256) / (−1) = 765.
因此 uₙ = 3 × 2ⁿ⁻¹。前 8 项的和为 S₈ = 3(1 − 2⁸) / (1 − 2) = 3(1 − 256) / (−1) = 765。
by checking a few terms: n = 1 gives 3, n = 2 gives 6, n = 3 gives 12, matching the data.
最后用几项验证公式：n = 1 得 3，n = 2 得 6，n = 3 得 12，与已知条件相符。
12. Summary | 总结
and uₙ = arⁿ⁻¹; for series, remember Sₙ = n/2 [2a + (n − 1)d] and Sₙ = a(1 − rⁿ)/(1 − r).
通用公式是 A-Level 数学中强大的工具。对于数列，记住 uₙ = a + (n − 1)d 和 uₙ = arⁿ⁻¹；对于级数，记住 Sₙ = n/2 [2a + (n − 1)d] 和 Sₙ = a(1 − rⁿ)/(1 − r)。
For binomial expansion, the general term Tᵣ₊₁ = nCr aⁿ⁻ʳ bʳ rm without full expansion. Practise these formulas until you can apply them automatically.
对于二项式展开，通项 Tᵣ₊₁ = nCr aⁿ⁻ʳ bʳ 可以让你无需完整展开就能求出任意一项。反复练习这些公式，直到能够熟练运用。
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