📚 Binomial Expansion | IGCSE WJEC 数学：二项式展开 考点精讲
Binomial expansion is a core topic in the IGCSE WJEC Mathematics syllabus, requiring you to expand expressions of the form (a + b)n where n is a positive integer. Mastering Pascal’s Triangle, binomial coefficients, and the general term formula will enable you to tackle coefficient, constant term, and specific term questions confidently in your exam.
二项式展开是 IGCSE WJEC 数学大纲中的核心专题，要求你展开形如 (a + b)n（n 为正整数）的表达式。掌握帕斯卡三角形、二项式系数和通项公式，将使你能够自信地应对考试中有关系数、常数项和特定项的问题。
1. What Is Binomial Expansion? | 什么是二项式展开？
An algebraic expression with two terms, such as (x + y) or (2a – 3b), is called a binomial. Binomial expansion refers to multiplying a binomial by itself repeatedly to express it as a sum of terms. For example, (x + y)2 = x2 + 2xy + y2 and (x + y)3 = x3 + 3x2y + 3xy2 + y3.
含有两项的代数式，比如 (x + y) 或 (2a – 3b)，称为二项式。二项式展开是指将二项式反复自乘，将其表示为一个多项式的和。例如 (x + y)2 = x2 + 2xy + y2，以及 (x + y)3 = x3 + 3x2y + 3xy2 + y3。
2. The Role of Pascal’s Triangle | 帕斯卡三角形的作用
Pascal’s Triangle provides the coefficients for expansions of (a + b)n quickly. The top row corresponds to n = 0, and each new row is built by adding the two numbers above. The coefficients for n = 5 are 1, 5, 10, 10, 5, 1.
帕斯卡三角形可以快速给出 (a + b)n 展开式的系数。最顶行对应 n = 0，每一新行由其上方两数相加得到。n = 5 时的系数为 1、5、10、10、5、1。
Here is a section of Pascal’s Triangle up to n = 6:
以下是直到 n = 6 的帕斯卡三角形：
| 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 | ||||||
| n = 6 | 1 6 15 20 15 6 1 | ||||||
3. Expanding (a + b)n Using Pascal’s Triangle | 利用帕斯卡三角形展开 (a + b)n
To expand (x + 2)4, take the coefficients from row n = 4: 1, 4, 6, 4, 1. Then write terms with powers of x descending from 4 to 0 and powers of 2 ascending from 0 to 4. The expansion is 1·x4·20 + 4·x3·21 + 6·x2·22 + 4·x1·23 + 1·x0·24, which simplifies to x4 + 8x3 + 24x2 + 32x + 16.
要展开 (x + 2)4，从第 4 行取出系数：1、4、6、4、1。然后写出各项：x 的幂从 4 递减到 0，2 的幂从 0 递增到 4。展开式为 1·x4·20 + 4·x3·21 + 6·x2·22 + 4·x1·23 + 1·x0·24，化简得 x4 + 8x3 + 24x2 + 32x + 16。
Remember to handle negative signs carefully: for (3x – y)3, treat ‘b’ as -y. The coefficients from row 3 are 1, 3, 3, 1. The terms become 1·(3x)3(-y)0 + 3·(3x)2(-y)1 + 3·(3x)1(-y)2 + 1·(3x)0(-y)3 = 27x3 – 27x2y + 9xy2 – y3.
切记小心处理负号：对于 (3x – y)3，将 ‘b’ 视为 -y。第 3 行系数为 1、3、3、1。各项为 1·(3x)3(-y)0 + 3·(3x)2(-y)1 + 3·(3x)1(-y)2 + 1·(3x)0(-y)3 = 27x3 – 27x2y + 9xy2 – y3。
4. Binomial Coefficients and the nCr Notation | 二项式系数与 nCr 表示法
When n becomes large, drawing Pascal’s Triangle is time-consuming. Instead, we use binomial coefficients, written as nCr or sometimes as C(n, r), which equals n! / (r! (n – r)!). Here, r is the term number starting from r = 0. For example, 5C2 = 5! / (2! 3!) = 10, matching the third entry in row 5 of the triangle.
当 n 很大时，画出帕斯卡三角形非常耗时。这时我们改用二项式系数，记作 nCr 或有时写作 C(n, r)，它等于 n! / (r! (n – r)!)。这里的 r 是从 0 开始的项序号。例如 5C2 = 5! / (2! 3!) = 10，恰好与三角形第 5 行第三个数字吻合。
Key properties to remember: nC0 = 1, nCn = 1, and nCr = nCn-r. Your calculator has an nCr function, usually accessed via a ‘MATH’ or ‘PROB’ menu, which is essential for exam efficiency.
需要记住的关键性质：nC0 = 1，nCn = 1，以及 nCr = nCn-r。你的计算器具有 nCr 功能，通常通过 ‘MATH’ 或 ‘PROB’ 菜单调出，这对提高考试效率至关重要。
5. The General Term in the Binomial Expansion | 二项展开式的通项
The expansion of (a + b)n consists of (n + 1) terms. The general term, often denoted as Tr+1, gives the (r+1)th term and is expressed as:
(a + b)n 的展开式由 (n + 1) 项组成。通项通常记作 Tr+1，它给出第 (r+1) 项，表达式为：
Tr+1 = nCr an-r br
Here, r runs from 0 to n. The first term corresponds to r = 0, the second to r = 1, and so on. The sum of the exponents of a and b in each term always equals n.
这里的 r 取值范围从 0 到 n。第一项对应 r = 0，第二项对应 r = 1，以此类推。每一项中 a 和 b 的指数之和始终等于 n。
For negative b, simply substitute b with -b; the general term becomes nCr an-r (-b)r, and the sign alternates depending on whether r is even or odd.
对于负的 b，只需用 -b 代入；通项变为 nCr an-r (-b)r，正负号将根据 r 的奇偶而交替变化。
6. Finding a Specific Term | 求特定项
WJEC questions often ask for a term with a given power of x. For example, in the expansion of (2x – 3)5, find the term in x3. Write the general term: Tr+1 = 5Cr (2x)5-r (-3)r. The power of x is 5 – r. Set 5 – r = 3, so r = 2. Then T3 = 5C2 (2x)3 (-3)2 = 10 · 8x3 · 9 = 720x3.
WJEC 的试题常常要求找出含特定 x 幂次的项。例如，在 (2x – 3)5 的展开式中，求含 x3 的项。先写出通项：Tr+1 = 5Cr (2x)5-r (-3)r。x 的指数是 5 – r。令 5 – r = 3，解得 r = 2。于是 T3 = 5C2 (2x)3 (-3)2 = 10 · 8x3 · 9 = 720x3。
Always check which term number is required. If the question asks for ‘the 4th term’, use r = 3, because the first term corresponds to r = 0.
务必确认所需的是第几项。如果试题问“第 4 项”，应使用 r = 3，因为第一项对应于 r = 0。
7. Expanding (1 + x)n Efficiently | 高效展开 (1 + x)n
A special and exam-friendly case is (1 + x)n. Here a = 1 and b = x, so the expansion simplifies to 1 + nC1 x + nC2 x2 + … + xn. This form is particularly useful when combined with algebraic manipulation, such as expanding (2 + 3x)4 by factoring out a power.
(1 + x)n 是一种对考试很有帮助的特殊情况。此时 a = 1，b = x，展开式简化为 1 + nC1 x + nC2 x2 + … + xn。这种形式在与代数技巧结合时特别有用，例如通过提取公因数来展开 (2 + 3x)4。
Using the expansion of (1 + y)n and then substituting y = kx can speed up calculations. For instance, (1 + 2x)6 = 1 + 6·2x + 15·(2x)2 + 20·(2x)3 + … which gives coefficients directly without intricate powers of a and b separately.
使用 (1 + y)n 的展开式然后代入 y = kx 可以加快计算。例如 (1 + 2x)6 = 1 + 6·2x + 15·(2x)2 + 20·(2x)3 + …，这样就可以直接得到系数，无需分别处理 a、b 的复杂次幂。
8. Expanding (a + bx)n Using Substitution | 用代入法展开 (a + bx)n
When the first term is not 1, rewrite the expression to make use of the (1 + X)n pattern. For (2 + 5x)3, factor out 23 to get 8(1 + (5x/2))3. Then expand (1 + (5x/2))3 and multiply by 8 afterwards. This avoids combining powers of ‘a’ and ‘b’ incorrectly.
当首项不是 1 时，可重写表达式以套用 (1 + X)n 的模式。对于 (2 + 5x)3，提取公因数 23 得到 8(1 + (5x/2))3。然后展开 (1 + (5x/2))3 再乘以 8。这可以避免错误地组合 a 和 b 的幂次。
Let’s see a worked example: (3 – 2x)4 = 34 (1 – (2x/3))4 = 81[1 + 4(-2x/3) + 6(-2x/3)2 + 4(-2x/3)3 + (-2x/3)4]. Simplify term by term to obtain 81 – 216x + 216x2 – 96x3 + 16x4.
我们看一个完整例题：(3 – 2x)4 = 34 (1 – (2x/3))4 = 81[1 + 4(-2x/3) + 6(-2x/3)2 + 4(-2x/3)3 + (-2x/3)4]。逐项化简得到 81 – 216x + 216x2 – 96x3 + 16x4。
9. Finding Coefficients Without Full Expansion | 不完整展开而求系数
Many exam questions ask for the coefficient of a specific power of x. There is no need to expand the whole expression. Use the general term directly, set the power equal to the target, solve for r, and then compute nCr multiplied by the appropriate constants.
很多考题要求找出 x 的某个特定次幂的系数。无需展开整个式子。直接使用通项，令指数等于目标次幂，解出 r，然后计算 nCr 乘以相应的常数值。
For example, find the coefficient of x2 in (5 – 3x)7. Here Tr+1 = 7Cr 57-r (-3x)r. Power of x is r. So r = 2. Coefficient = 7C2 · 55 · (-3)2 = 21 · 3125 · 9 = 590625.
例如，求 (5 – 3x)7 中 x2 的系数。通项 Tr+1 = 7Cr 57-r (-3x)r。x 的指数为 r，故令 r = 2。系数 = 7C2 · 55 · (-3)2 = 21 · 3125 · 9 = 590625。
10. Constant Term and Independent of x | 常数项与 x 无关的项
A common WJEC problem is to find the term independent of x, i.e., the constant term. This occurs when the power of x equals zero. For an expansion like (x2 + 1/x)6, the general term is 6Cr (x2)6-r (1/x)r = 6Cr x12-2r-r = 6Cr x12-3r. Set 12 – 3r = 0 → r = 4. The constant term is 6C4 = 15.
WJEC 常见的一类问题是求与 x 无关的项，即常数项。此时 x 的指数等于零。对于像 (x2 + 1/x)6 这样的展开式，通项为 6Cr (x2)6-r (1/x)r = 6Cr x12-2r-r = 6Cr x12-3r。令 12 – 3r = 0 → r = 4。常数项为 6C4 = 15。
In more complex cases, such as (axp + b/xq)n, the exponent of x in the general term is p(n – r) – qr. Set this equal to 0 and solve for r. Always ensure r is an integer between 0 and n; otherwise, there is no constant term.
在更复杂的情形下，比如 (axp + b/xq)n，通项中 x 的指数为 p(n – r) – qr。令其等于 0 并解出 r。务必确保 r 是 0 到 n 之间的整数；否则就没有常数项。
11. Common Pitfalls and How to Avoid Them | 常见易错点及避免方法
Misidentifying the value of r is the number one mistake. Remember, the term with xk has r determined by the power equation; if the binomial is (axm + bxt)n, the power of x is m(n – r) + tr. Solve carefully and check that r is a whole number.
错误判断 r 的值是头号易错点。请记住，含 xk 的项须由指数方程确定 r；如果二项式为 (axm + bxt)n，x 的指数为 m(n – r) + tr。仔细求解并检查 r 是否为整数。
Another common error is forgetting to raise the coefficient of x to the required power. In (2 – 5x)4, when r = 3, the term involves ( -5x )3, giving (-125)x3, not just -5. Also, negative signs must be carried through: (-3)2 = 9, while (-3)3 = -27.
另一个常见错误是忘记将 x 的系数也进行乘方。在 (2 – 5x)4 中，当 r = 3 时，项中包含 ( -5x )3，得到 (-125)x3，而不仅仅是 -5。此外，负号必须全程跟随：(-3)2 = 9，而 (-3)3 = -27。
Finally, when using Pascal’s Triangle by hand, it is easy to miscopy a row. Double-check the symmetry and always verify that the sum of the coefficients equals 2n. For (1 + 1)n the sum is 2n, which acts as a quick sanity check for any expansion of (a + b)n when a = b = 1.
最后，手动使用帕斯卡三角形时很容易抄错数字。请再次确认对称性，并始终核验系数总和是否等于 2n。对 (1 + 1)n 求和应为 2n，这可以作为任何 (a + b)n 展开式在 a = b = 1 时快速自查的检验。
12. Exam Tips and WJEC-Style Advice | 考试技巧与 WJEC 风格建议
WJEC IGCSE papers often include binomial expansion in the non-calculator sections where Pascal’s Triangle is quick for n ≤ 6, or in calculator sections where nCr is essential for larger n. Always read the question carefully: does it ask for a specific term, a coefficient, or the full expansion?
WJEC IGCSE 试卷中，二项式展开常出现在不可用计算器的部分（此时 n ≤ 6 用帕斯卡三角形更快），或者可用计算器的部分（此时 nCr 对较大的 n 必不可少）。务必仔细读题：问题是要特定项、系数，还是完整的展开式？
If asked for the coefficient of x3, state just the number (e.g., 720) and not the ‘x3‘ part. If the question says ‘hence find the coefficient of x3 in (2x – 3)(…)’ after an expansion, multiply out only the required term, not the entire product.
如果要求 x3 的系数，只需给出数字（例如 720），不要带上 ‘x3‘。如果试题在展开后接着问“由此求 (2x – 3)(…) 中 x3 的系数”，只需乘出所需的项，不必展开整个乘积。
Practice past WJEC questions focusing on constant term and coefficient extraction. Show your general term formula and the r-value clearly; method marks are awarded for correct reasoning even if the final arithmetic has a slip.
多练习 WJEC 历年试题，重点放在常数项和系数提取。清晰地展示你的通项公式和 r 值；即使最终计算有小错，正确的推理也能拿到方法分。
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