📚 IB Math: Binomial Expansion – Definition and General Term | IB数学：二项式展开的定义与通项公式
The binomial theorem is one of the most powerful tools in algebra, allowing us to expand expressions of the form (a + b)ⁿ without performing repeated multiplication. In IB Mathematics, a clear understanding of the definition and the general term is essential for solving problems across Analysis and Approaches (AA) and Applications and Interpretation (AI).
二项式定理是代数中最强大的工具之一，它使我们能够展开形如 (a + b)ⁿ 的表达式，而无需进行逐次乘法。在 IB 数学中，清晰理解其定义和通项公式，对于解决 Analysis and Approaches（AA）与 Applications and Interpretation（AI）中的问题都至关重要。
1. What Is a Binomial? | 什么是二项式
A binomial is a polynomial expression that contains exactly two terms, usually connected by a plus or minus sign. Examples include x + y, 2x − 3, and a + 2b. The word "binomial" comes from the Latin prefix "bi-" meaning two, and the Greek word "nomos" meaning part or term.
二项式是恰好包含两项的多项式表达式，通常用加号或减号连接。例如 x + y、2x − 3 和 a + 2b。单词“binomial”源自拉丁语前缀“bi-”表示“二”，以及希腊语“nomos”表示“部分”或“项”。
When we raise a binomial to a positive integer power n, such as (a + b)ⁿ, the result is a polynomial with n + 1 terms. Each term has a specific structure involving powers of a and b, multiplied by a binomial coefficient.
当我们将一个二项式提升到正整数次幂 n 时，例如 (a + b)ⁿ，结果是一个含有 n + 1 项的多项式。每一项都有特定的结构，包含 a 和 b 的幂次，并乘以一个二项式系数。
2. Expanding by Direct Multiplication | 通过直接乘法展开
²b + 3ab² + b³. These expansions show that as n increases, the number of terms grows and the coefficients follow a triangular pattern known as Pascal's triangle.
在引入定理之前，先通过直接乘法观察规律是很有帮助的。例如，(a + b)² = a² + 2ab + b²，(a + b)³ = a³ + 3a²b + 3ab² + b³。这些展开式表明，随着 n 增大，项数增加，系数遵循被称为杨辉三角（帕斯卡三角）的三角形规律。
ovides a systematic formula that gives every term without the need for tedious multiplication.
对于较大的 n 值，例如 n = 10 或 n = 20，直接乘法变得不切实际。二项式定理提供了一个系统化的公式，无需繁琐的乘法即可给出每一项。
3. The Binomial Coefficient | 二项式系数
The binomial coefficient is written as ⁿCᵣ, C(n, r), or ⎛⎝nr⎞⎠. It represents the number of ways to choose r items from n items without regard to order. This coefficient is defined by the formula:
二项式系数写作 ⁿCᵣ、C(n, r) 或 ⎛⎝nr⎞⎠。它表示从 n 个物件中无序选取 r 个物件的方法数。该系数的定义公式为：
ⁿCᵣ = n! / (r!(n − r)!)
Here, n! denotes the factorial of n, meaning the product n × (n − 1) × (n − 2) × … × 1. For example, 5! = 5 × 4 × 3 × 2 × 1 = 120. The binomial coefficient is always a non-negative integer.
这里 n! 表示 n 的阶乘，即 n × (n − 1) × (n − 2) × … × 1。例如，5! = 5 × 4 × 3 × 2 × 1 = 120。二项式系数始终为非负整数。
cients possess important symmetries. In particular, ⁿCᵣ = ⁿCₙ₋ᵣ, which explains why Pascal's triangle is symmetric. These coefficients also satisfy Pascal's identity: ⁿCᵣ + ⁿCᵣ₊₁ = ⁿ⁺¹Cᵣ₊₁.
这些系数具有重要的对称性。特别地，ⁿCᵣ = ⁿCₙ₋ᵣ，这解释了杨辉三角为何是对称的。这些系数还满足杨辉恒等式：ⁿCᵣ + ⁿCᵣ₊₁ = ⁿ⁺¹Cᵣ₊₁。
4. The Binomial Theorem Statement | 二项式定理的表述
The binomial theorem states that for any positive integer n and any real numbers a and b:
二项式定理指出，对于任何正整数 n 以及任意实数 a 和 b：
(a + b)ⁿ = ⁿC₀aⁿb⁰ + ⁿC₁aⁿ⁻¹b¹ + ⁿC₂aⁿ⁻²b² + … + ⁿCᵣaⁿ⁻ʳbʳ + … + ⁿCₙa⁰bⁿ
. This formula is valid when n is a non-negative integer, which is the focus of IB Mathematics.
展开式中的每一项对应不同的 r 值，r 从 0 到 n。总项数为 n + 1。当 n 为非负整数时，该公式成立，这也是 IB 数学的重点。
In summation notation, the binomial theorem can be written compactly as:
用求和记号表示时，二项式定理可以简洁地写作：
(a + b)ⁿ = Σᵣ₌₀ⁿ ⁿCᵣ aⁿ⁻ʳ bʳ
This compact form is especially useful when solving problems involving a specific term, as it allows us to extract the general term directly.
这种紧凑形式在解决涉及特定项的问题时尤其有用，因为它允许我们直接提取通项。
5. The General Term | 通项公式
m corresponds to r = 0, the second term to r = 1, and so on. The general term is given by:
二项式展开中的通项是包含 b 的 r 次幂的那一项。它记作 Tᵣ₊₁，因为第一项对应 r = 0，第二项对应 r = 1，依此类推。通项公式为：
Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ
For example, in the expansion of (x + 2)⁵, the general term is Tᵣ₊₁ = ⁵Cᵣ x⁵⁻ʳ 2ʳ. If we want the third term, we set r = 2, giving T₃ = ⁵C₂ x³ 2² = 10 × x³ × 4 = 40x³.
例如，在 (x + 2)⁵ 的展开式中，通项为 Tᵣ₊₁ = ⁵Cᵣ x⁵⁻ʳ 2ʳ。如果我们想要第三项，令 r = 2，得到 T₃ = ⁵C₂ x³ 2² = 10 × x³ × 4 = 40x³。
It is critical to remember the index shift: Tᵣ₊₁ corresponds to r, not r + 1. Many IB students lose marks by confusing the term number with the value of r. The term number is always one greater than r.
务必记住下标的偏移：Tᵣ₊₁ 对应的是 r，而不是 r + 1。许多 IB 学生因混淆项序号与 r 值而失分。项序号总是比 r 大 1。
6. Handling Subtraction Signs | 处理减号
ch as (a − b)ⁿ, we rewrite it as (a + (−b))ⁿ. The general term then becomes Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ (−b)ʳ = (−1)ʳ ⁿCᵣ aⁿ⁻ʳ bʳ.
当二项式包含减号时，例如 (a − b)ⁿ，我们将其改写为 (a + (−b))ⁿ。此时通项变为 Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ (−b)ʳ = (−1)ʳ ⁿCᵣ aⁿ⁻ʳ bʳ。
The factor (−1)ʳ causes the signs of the terms to alternate. For example, (x − 3)⁴ has terms with signs +, −, +, −, + as r goes from 0 to 4. This alternating pattern is a common feature in IB examination questions.
因子 (−1)ʳ 使得各项符号交替变化。例如，(x − 3)⁴ 的各项符号在 r 从 0 到 4 时依次为 +、−、+、−、+。这种交替模式是 IB 考试题中的常见特征。
A common error is to apply the negative sign only to b, not to account for the power. Remember that (−b)ʳ requires raising the entire expression −b to the power r, so the sign depends on whether r is even or odd.
一个常见错误是只将负号作用于 b，而没有考虑幂次。请记住，(−b)ʳ 需要对整个表达式 −b 进行 r 次幂运算，因此符号取决于 r 是偶数还是奇数。
7. Finding a Specific Term | 求特定项
6th term of (2x + 1)¹⁰, set r = 5. The term is T₆ = ¹⁰C₅ (2x)⁵ (1)⁵ = 252 × 32x⁵ = 8064x⁵.
要求特定项（例如第 k 项），在通项中令 r = k − 1。例如，求 (2x + 1)¹⁰ 的第 6 项，令 r = 5。该项为 T₆ = ¹⁰C₅ (2x)⁵ (1)⁵ = 252 × 32x⁵ = 8064x⁵。
, to find the term containing x⁷ in (x² + 3)⁸, we write the general term Tᵣ₊₁ = ⁸Cᵣ (x²)⁸⁻ʳ 3ʳ = ⁸Cᵣ x¹⁶⁻²ʳ 3ʳ Setting 16 − 2r = 7 gives r = 4.5, which is not an integer; therefore no such term exists.
当题目用变量的幂次而不是位置来描述项时，我们令 x 的指数等于目标值并解出 r。例如，在 (x² + 3)⁸ 中求含 x⁷ 的项，写出通项 Tᵣ₊₁ = ⁸Cᵣ (x²)⁸⁻ʳ 3ʳ = ⁸Cᵣ x¹⁶⁻²ʳ 3ʳ。令 16 − 2r = 7，得 r = 4.5，不是整数，因此这样的项不存在。
This method demonstrates an essential skill: translating the wording of a problem into an equation involving the exponent. IB examiners frequently test whether students can perform this translation correctly.
这一方法展示了一项关键技能：将问题的表述转化为涉及指数的方程。IB 考官经常考查学生能否正确完成这种转化。
8. The Constant Term | 常数项
The constant term in a binomial expansion is the term that does not contain x, meaning the power of x is zero. To find it, write the general term, identify the exponent of x, set that exponent equal to 0, and solve for r.
二项式展开中的常数项是不含 x 的项，即 x 的指数为零。求法为：写出通项，找出 x 的指数，令该指数等于 0，并解出 r。
For example, consider (x + 2/x²)⁹. The general term is Tᵣ₊₁ = ⁹Cᵣ x⁹⁻ʳ (2/x²)ʳ = ⁹Cᵣ 2ʳ x⁹⁻ʳ⁻²ʳ = ⁹Cᵣ 2ʳ x⁹⁻³ʳ. Setting 9 − 3r = 0 gives r = 3. The constant term is T₄ = ⁹C₃ 2³ = 84 × 8 = 672.
例如，考虑 (x + 2/x²)⁹。通项为 Tᵣ₊₁ = ⁹Cᵣ x⁹⁻ʳ (2/x²)ʳ = ⁹Cᵣ 2ʳ x⁹⁻ʳ⁻²ʳ = ⁹Cᵣ 2ʳ x⁹⁻³ʳ。令 9 − 3r = 0，得 r = 3。常数项为 T₄ = ⁹C₃ 2³ = 84 × 8 = 672。
Note that the index r must be an integer between 0 and n inclusive. If solving the exponent equation yields a non-integer or an out-of-range value, the constant term does not exist. Checking this condition is an excellent habit for IB examinations.
注意 r 必须是 0 到 n 之间的整数。如果解指数方程得到非整数或超出范围的 r 值，则常数项不存在。检查这一条件是 IB 考试中的良好习惯。
9. The Coefficient of a Term | 项的系数
es the variable part. In the expansion of (2x + 3)⁴, the general term is Tᵣ₊₁ = ⁴Cᵣ (2x)⁴⁻ʳ 3ʳ = ⁴Cᵣ 2⁴⁻ʳ 3ʳ x⁴⁻ʳ. The coefficient of x⁴⁻ʳ is therefore ⁿCᵣ 2⁴⁻ʳ 3ʳ, not simply ⁿCᵣ.
项的系数是乘以变量部分的数值因子。在 (2x + 3)⁴ 的展开式中，通项为 Tᵣ₊₁ = ⁴Cᵣ (2x)⁴⁻ʳ 3ʳ = ⁴Cᵣ 2⁴⁻ʳ 3ʳ x⁴⁻ʳ。因此 x⁴⁻ʳ 的系数是 ⁿCᵣ 2⁴⁻ʳ 3ʳ，而不仅仅是 ⁿCᵣ。
Students must remember to include contributions from any constant inside the binomial term. For example, in (3x² − 2)⁵, the coefficient of x⁴ is obtained by first finding the general term, then isolating the product of the numerical factors.
学生必须记得包括二项式项内部常数对系数的贡献。例如，在 (3x² − 2)⁵ 中，求 x⁴ 的系数时，要先写出通项，然后分离出数值因子的乘积。
Tᵣ₊₁ = ⁵Cᵣ (3x²)⁵⁻ʳ (−2)ʳ = ⁵Cᵣ 3⁵⁻ʳ (−2)ʳ x¹⁰⁻²ʳ
−720. This example shows why it is safer to write the entire general term before simplifying.
我们要求 10 − 2r = 4，所以 r = 3。系数为 ⁵C₃ × 3² × (−2)³ = 10 × 9 × (−8) = −720。这个例子表明，在化简之前写出完整通项更加安全。
10. Pascal's Triangle and Its Limits | 杨辉三角及其局限
Pascal' e top. Each entry is the sum of the two entries above it. For small values of n, this triangle is a quick and reliable way to find coefficients.
杨辉三角逐行列出二项式系数，从顶部的 n = 0 开始。每个数字等于其上方两个数字之和。对于较小的 n，这个三角形是快速且可靠的求系数方法。
| n = 0 | 1 |
| n = 1 | 1 1 |
| n = 2 | 1 2 1 |
| n = 3 | 1 3 3 1 |
| n = 4 | 1 4 6 4 1 |
| n = 5 | 1 5 10 10 5 1 |
However, Pascal' ial coefficient formula or a calculator with nCr functionality is far more practical in those cases. IB examinations allow the use of a GDC, so students should know how to compute ⁿCᵣ quickly on their device.
然而，对于较大的 n，例如 n = 20 或 n = 30，杨辉三角变得低效。此时二项式系数公式或带有 nCr 功能的计算器更为实用。IB 考试允许使用图形计算器（GDC），因此学生应知道如何在自己的设备上快速计算 ⁿCᵣ。
11. Binomial Theorem Where n Is a Positive Integer | 当 n 为正整数时的二项式定理
In the IB syllabus, the binomial theorem is formally introduced for positive integer exponents n. This restriction ensures that the expansion terminates after exactly n + 1 terms. For non-integer or negative exponents, the binomial theorem becomes an infinite series, which is covered in higher-level calculus contexts.
在 IB 课程大纲中，二项式定理正式引入时要求指数 n 为正整数。这一限制确保展开式在恰好 n + 1 项后终止。对于非整数或负指数，二项式定理变为无穷级数，这将在更高阶的微积分背景中讨论。
Nevertheless, the general term formula Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ s without expanding the entire polynomial, which saves time and reduces arithmetic errors.
尽管如此，通项公式 Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ 仍然是解决大多数 IB 考试问题的基石。它使学生无需展开整个多项式即可处理展开式，从而节省时间并减少算术错误。
12. Common IB Exam Mistakes and Tips | 常见 IB 考试错误与建议
A frequent mistake is writing Tᵣ₊₁ = ⁿCᵣ aʳ bⁿ⁻ʳ exponent of b increases as r increases. The sum of the two exponents must always equal n.
一个常见错误是写成 Tᵣ₊₁ = ⁿCᵣ aʳ bⁿ⁻ʳ，而正确形式应相反。为避免此错误，请记住：随着 r 增大，a 的指数递减，b 的指数递增，且两个指数之和始终等于 n。
Another common error is confusing the term number with r. The phrase "the r-th term" sometimes appears in questions; in such cases, it conventionally means Tᵣ, so you must set the index to r − 1. Reading the question carefully is half the battle.
另一个常见错误是混淆项序号与 r。“第 r 项”这一表述在题目中有时出现；在这种情况下，按惯例它指的是 Tᵣ，因此你必须令下标为 r − 1。仔细审题是成功的一半。
Finally, always simplify coefficients fully and check whether the sign is positive or negative when the binomial contains a minus sign. A well-structured solution that begins by writing the general term will help avoid lost marks. Practising with past IB questions is the most effective way to master this topic.
最后，始终完整化简系数，并检查当二项式含减号时符号为正还是为负。一种好的解题结构是从写出通项开始，这样有助于避免失分。通过练习历年 IB 真题来掌握这一主题是最有效的方法。
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