The Binomial Theorem | 二项式定理

📚 The Binomial Theorem | 二项式定理

The binomial theorem is one of the most elegant and powerful tools in algebra. It allows us to expand expressions of the form (a + b)ⁿ without having to multiply the binomial repeatedly. Whether n is a positive integer, a fraction, or a negative number, the theorem reveals a beautiful pattern of coefficients tied to combinations. For IB Mathematics, mastering the binomial theorem is essential not only for algebraic manipulation but also for understanding sequences, probability, and later calculus topics such as series expansions.

二项式定理是代数学中最优美且强大的工具之一。它使我们能够展开形如 (a + b)ⁿ 的表达式，而无需反复进行二项式相乘。无论 n 是正整数、分数还是负数，该定理都展示出与组合数紧密相关的优美系数规律。对于 IB 数学而言，掌握二项式定理不仅是代数运算的关键，也有助于理解数列、概率以及后续微积分中的级数展开等内容。

1. What is the Binomial Theorem? | 什么是二项式定理？

Suppose we wish to expand (x + y)². By multiplication we obtain x² + 2xy + y². For (x + y)³, we get x³ + 3x²y + 3xy² + y³. The binomial theorem generalises this pattern for any power n. It states that for a positive integer n, the expansion is a sum of terms of the form ⁿC_r a^n−r b^r, where ⁿC_r (read 'n choose r') is a binomial coefficient. This coefficient counts the number of ways to choose r items from n, giving the expansion a close link with combinatorics.

假设我们要展开 (x + y)²。通过乘法可得 x² + 2xy + y²。对于 (x + y)³，我们得到 x³ + 3x²y + 3xy² + y³。二项式定理将这一规律推广到任意幂次 n。它指出，对于正整数 n，展开式是各项形如 ⁿC_r a^n−r b^r 的和，其中 ⁿC_r（读作 “n 选 r”）是二项式系数。该系数表示从 n 个元素中选出 r 个的方法数，使展开式与组合数学紧密相连。

For IB Mathematics AA students, the theorem appears both at Standard Level (positive integer powers) and at Higher Level (extended to rational and negative powers). It provides a systematic structure that makes finding specific terms far quicker than manual multiplication. The coefficients follow a symmetric pattern easily remembered via Pascal's Triangle.

对于 IB 数学 AA 的学生来说，该定理既出现在标准水平（正整数幂）也出现在高级水平（推广至有理数和负数幂）。它提供了系统化的结构，使得寻找特定项远比手动相乘快捷。系数呈现出对称模式，借助杨辉三角可以轻松记忆。

2. Binomial Coefficients and Pascal's Triangle | 二项式系数与杨辉三角

The binomial coefficient ⁿC_r can be computed using the formula ⁿC_r = n! / (r! (n−r)! s correspond to n = 0, 1, 2, ... and the entries in row n give the coefficients of (a + b)ⁿ. For instance, row 4 reads 1, 4, 6, 4, 1 and matches (a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴.

二项式系数 ⁿC_r 可以用公式 ⁿC_r = n! / (r! (n−r)!) 计算。对于较小的 n，使用杨辉三角更为简便。在三角形中，每个数都是其上方两数之和。各行对应 n = 0, 1, 2, ...，第 n 行的各项给出 (a + b)ⁿ 的系数。例如，第4行为 1, 4, 6, 4, 1，与 (a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴ 相匹配。

eepens our number sense. The entry in row n, position r (starting from r = 0) is precisely ⁿC_r. This relationship becomes invaluable when handling binomial probabilities in the IB Statistics and Probability topics, where coefficients represent the number of favourable outcomes.

理解杨辉三角与组合数之间的关系能够增强数感。第 n 行、第 r 个位置（从 r = 0 开始）的数值正是 ⁿC_r。当处理 IB 统计与概率单元中的二项分布概率时，这一关系极为宝贵，因为系数代表有利结果的数目。

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3. The General Formula for (a + b)ⁿ with Positive Integer n | (a + b)ⁿ 的一般公式 (n 为正整数)

For any positive integer n, the binomial theorem states:

(a + b)ⁿ = ∑_r=0^{n n}C_r a^n−r b^r

对于任意正整数 n，二项式定理表示为：

(a + b)ⁿ = ∑_r=0^{n n}C_r a^n−r b^r

uns over r = 0, 1, 2, ..., n. The first term corresponds to r = 0 (coefficient 1, giving aⁿ), and the last term corresponds to r = n (coefficient 1, giving bⁿ). The expression ⁿC_r is often written as C(n, r) or in vector notation. It can be calculated by the factorial formula above or by using the 'nCr' button on your GDC.

这里求和指标 r 从 0 到 n 变化。首项对应 r = 0（系数为 1，得到 aⁿ），末项对应 r = n（系数为 1，得到 bⁿ）。表达式 ⁿC_r 经常写成 C(n, r) 或两竖括号形式。它可以通过上述阶乘公式计算，也可以使用图形计算器上的 'nCr' 按键求得。

To apply the formula correctly, treat a and b as the two terms, keeping signs intact. For example, to expand (2 − x)⁵, set a = 2, b = −x, and n = 5. The expansion then becomes ∑ ⁵C_r 2^5−r (−x)^r, where the alternating signs emerge naturally from (−x)^r. This technique avoids sign errors that occur when rewriting the binomial manually.

要正确应用公式，应将 a 和 b 视为含有符号的两个项。例如，展开 (2 − x)⁵ 时，令 a = 2，b = −x，n = 5。展开式即为 ∑ ⁵C_r 2^5−r (−x)^r，其中符号交替变化由 (−x)^r 自然产生。这一技巧避免了手动重写二项式时容易出现的符号错误。

4. The General Term (the r-th term) | 通项公式 (第 r 项)

When exam questions ask for 'the term in x^k' or 'the constant term', it is inefficient to write out the whole expansion. Instead, we use the general term, often denoted T_r+1:

T_r+1 = ⁿC_r a^n−r b^r

当考题要求找出“含 x^k 的项”或“常数项”时，写出整个展开式效率低下。这时我们使用通项公式，常记为 T_r+1：

T_r+1 = ⁿC_r a^n−r b^r

This notation means that the first term (r = 0) is T₁, the second term (r = 1) is T₂, and so on. To find a specific term, substitute a, b, and n into the expression, group the powers of the variable, and set the exponent equal to the desired value. Solving for r then identifies the correct coefficient and term.

该记号意味着首项 (r = 0) 为 T₁，第二项 (r = 1) 为 T₂，依此类推。要找到特定项，将 a、b 和 n 代入表达式，合并变量的幂次，然后令指数等于所需的值。解出 r，即可确定正确的系数和对应项。

For instance, to find the coefficient of x³ in (1 + 2x)⁸, we write T_r+1 = ⁸C_r 1^8−r (2x)^r = ⁸C_r 2^r x^r. Setting r = 3 gives the term, and the coefficient is ⁸C₃ × 2³ = 56 × 8 = 448. Always check the range of r: r must be an integer between 0 and n.

例如，求 (1 + 2x)⁸ 中 x³ 的系数，写出通项 T_r+1 = ⁸C_r 1^8−r (2x)^r = ⁸C_r 2^r x^r。令 r = 3 即得该项，系数为 ⁸C₃ × 2³ = 56 × 8 = 448。务必检查 r 的取值范围：r 必须是介于 0 到 n 的整数。

5. Finding Specific Coefficients and Constant Terms | 求特定系数与常数项

Many IB questions ask for the coefficient of a particular power of x or for the term independent of x (the constant term). The strategy is to write the general term with all powers expressed in the variable, then enforce the exponent condition. For example, in (x² − 2/x)⁹, the general term is ⁹C_r (x²)^9−r (−2/x)^r = ⁹C_r (−2)^r x^18−3r. To find the constant term, set 18 − 3r = 0, giving r = 6. The constant term is then ⁹C₆ (−2)⁶ = 84 × 64 = 5376.

许多 IB 题目要求找出 x 的某次幂的系数或与 x 无关的项（常数项）。策略是写出通项，将所有幂次用变量表示，然后令指数满足条件。例如，在 (x² − 2/x)⁹ 中，通项为 ⁹C_r (x²)^9−r (−2/x)^r = ⁹C_r (−2)^r x^18−3r。求常数项时，令 18 − 3r = 0，解得 r = 6。常数项即为 ⁹C₆ (−2)⁶ = 84 × 64 = 5376。

If a question asks for two consecutive terms or for the greatest coefficient, you may need to set up equations or inequalities involving ⁿC_r. For the greatest coefficient, compare T_r+1 and T_r, often using the ratio ⁿC_r / ⁿC_r−1 (n−r+1)/r. The coefficient increases while this ratio exceeds 1, giving the turning point.

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