📚 IGCSE AQA Maths: Binomial Expansion Revision | IGCSE AQA 数学：二项式展开 考点精讲
Binomial expansion is a powerful algebraic tool that allows us to expand expressions of the form (a + b)n without performing repeated multiplication. For IGCSE AQA Mathematics, you are expected to understand and apply the binomial theorem for positive integer exponents, use Pascal’s triangle, work with binomial coefficients (nCr), and find specific terms in an expansion. This guide covers every key point you need to master the topic, with clear explanations, worked examples, and exam-focused tips.
二项式展开是一种强大的代数工具，能让我们展开形如 (a + b)n 的表达式，而无需反复相乘。在 IGCSE AQA 数学考试中，你需要理解并应用正整数指数的二项式定理，会使用帕斯卡三角，会处理二项式系数（nCr），并能找出展开式中的特定项。本指南涵盖了你需要掌握的所有关键考点，提供清晰的解释、详细例题和应试技巧。
1. What is Binomial Expansion? | 什么是二项式展开？
A binomial is an algebraic expression containing two terms, such as (x + 2) or (3a – 5b). Expanding a binomial means raising it to a power and writing the result as a sum of terms. For example, (x + y)2 = x2 + 2xy + y2. As the power increases, manual multiplication becomes tedious, so we use the binomial theorem to find the expansion directly.
二项式是含有两个项的代数表达式，例如 (x + 2) 或 (3a – 5b)。二项式展开就是将其乘方，并将结果写成若干个项的和。例如 (x + y)2 = x2 + 2xy + y2。随着指数增大，手算乘法变得繁琐，因此我们使用二项式定理直接得出展开式。
2. Pascal’s Triangle – The Foundation | 帕斯卡三角——基础
Pascal’s triangle is a triangular array of numbers where each number is the sum of the two numbers directly above it. The rows correspond to the powers of a binomial expansion. Row n gives the coefficients for (a + b)n. The first row is row 0: 1. Row 1: 1 1; Row 2: 1 2 1; Row 3: 1 3 3 1; Row 4: 1 4 6 4 1, and so on.
帕斯卡三角是一个三角形数列，其中每个数是它上方两数之和。每一行对应二项式展开的指数。第 n 行给出 (a + b)n 的系数。第 0 行：1；第 1 行：1 1；第 2 行：1 2 1；第 3 行：1 3 3 1；第 4 行：1 4 6 4 1，以此类推。
lways remember the triangle is symmetrical, which can save time when writing coefficients.
在 IGCSE 中，你可能会被要求补全帕斯卡三角的某一行，或利用它展开一个指数较小的二项式（通常 n ≤ 6）。请记住三角形是对称的，这可以帮你节省书写系数的时间。
3. Introducing Binomial Coefficients (nCr) | 引入二项式系数 (nCr)
Pascal’s triangle can also be found using the combination formula nCr, defined as nCr = n! / (r! (n – r)!). This is read as ‘n choose r’. nCr gives the coefficient of the (r + 1)th term in the expansion of (a + b)n, where r starts from 0. For example, the coefficient of the third term (r = 2) in (a + b)5 is ⁵C₂ = 10.
帕斯卡三角中的数也可以用组合公式 nCr 来求得，定义为 nCr = n! / (r! (n – r)!) ，读作 “n 选 r”。nCr 给出 (a + b)n 展开式中第 (r + 1) 项的系数，其中 r 从 0 开始。例如 (a + b)5 展开式的第三项 (r = 2) 系数为 ⁵C₂ = 10。
In AQA exams, you often use the nCr button on your calculator to find these coefficients quickly. Understanding the connection between Pascal’s triangle and nCr is essential, as it allows you to handle expansions for any positive integer n without drawing a huge triangle.
在 AQA 考试中，你经常需要使用计算器上的 nCr 按钮快速求出这些系数。理解帕斯卡三角与 nCr 之间的联系至关重要，这使你能处理任意正整数指数 n 的展开，而无需画出巨大的三角形。
4. The Binomial Theorem Formula | 二项式定理公式
The binomial theorem states that for any positive integer n:
(a + b)n = ⁿC₀ an b⁰ + ⁿC₁ an-1 b¹ + ⁿC₂ an-2 b² + … + ⁿCn a⁰ bn
二项式定理表明，对于任意正整数 n：
(a + b)n = ⁿC₀ an b⁰ + ⁿC₁ an-1 b¹ + ⁿC₂ an-2 b² + … + ⁿCn a⁰ bn
Notice how the powers of a decrease from n to 0, while the powers of b increase from 0 to n. The sum of the exponents in each term is always n. The coefficient of the general term is given by ⁿCr, where r is the index of b (and also the term number minus 1).
注意 a 的指数从 n 递减至 0，而 b 的指数从 0 递增至 n。每一项中两个指数之和始终为 n。一般项的系数由 ⁿCr 给出，其中 r 是 b 的指数（也是项数减 1）。
5. Expanding (a + b)n Step by Step | 逐步展开 (a + b)n
To expand (a + b)n, follow these steps: 1) Write down the binomial coefficients for that power using Pascal’s triangle or nCr. 2) Write the first term as an. 3) Decrease the power of a by 1 and increase the power of b by 1 in each subsequent term. 4) Multiply each term by the corresponding coefficient. 5) Simplify if there are any additional factors inside the bracket.
要展开 (a + b)n，遵循以下步骤：1) 利用帕斯卡三角或 nCr 写出该次幂的二项式系数。2) 写出第一项 an。3) 在后续每一项中，将 a 的指数减 1，b 的指数加 1。4) 每一项乘上对应的系数。5) 如果括号内还有其他因子，则化简。
For example, to expand (x + 2)3: coefficients are 1, 3, 3, 1. The terms are x3, x2(2), x(2)2, (2)3. Multiply and simplify: x3 + 6x2 + 12x + 8. Always check that the number of terms is n + 1.
例如，展开 (x + 2)3：系数为 1, 3, 3, 1。各项为 x3, x2(2), x(2)2, (2)3。相乘并化简得：x3 + 6x2 + 12x + 8。始终检查项数是否为 n + 1。
6. Finding Specific Terms in an Expansion | 寻找展开式中的特定项
A common IGCSE exam question asks you to find a particular term, such as the term in xk, or the constant term, without expanding the whole expression. Use the general term formula: the (r + 1)th term is Tr+1 = ⁿCr an-r br. Identify the values of n, a, b, and the required power of x, then solve for r.
IGCSE 考试中常见的一种题目是要求你找出某一特定项，例如含 xk 的项，或常数项，而不必展开整个表达式。使用通项公式：第 (r + 1) 项为 Tr+1 = ⁿCr an-r br。确定 n, a, b 的值以及所需 x 的指数，然后解出 r。
For instance, to find the term in x3 in the expansion of (2x – 1)5, set n = 5, a = 2x, b = -1. General term: ⁵Cr (2x)5-r (-1)r = ⁵Cr 25-r x5-r (-1)r. We need the power of x to be 3, so 5 – r = 3 → r = 2. Substitute r = 2: ⁵C₂ (2)3 x3 (-1)2 = 10 × 8 × 1 × x3 = 80x3.
例如，求 (2x – 1)5 展开式中含 x3 的项，设 n = 5, a = 2x, b = -1。通项： ⁵Cr (2x)5-r (-1)r = ⁵Cr 25-r x5-r (-1)r。我们需要 x 的指数为 3，所以 5 – r = 3 → r = 2。代入 r = 2：⁵C₂ (2)3 x3 (-1)2 = 10 × 8 × 1 × x3 = 80x3。
7. Using the nCr Button on Your Calculator | 使用计算器上的 nCr 按钮
Most scientific calculators have an nCr function, often accessed by pressing the 'nCr' button (sometimes above the division key) or via a combinations menu. To calculate ⁵C₂, you would usually type 5 nCr 2. In exam conditions, using the calculator reduces errors, but you must still be able to write the expansion correctly. Practice typing nCr for various values until you are quick and confident.
大多数科学计算器都有 nCr 功能，通常通过按 “nCr” 按钮（有时在除号键上方）或组合数菜单访问。要计算 ⁵C₂，你通常输入 5 nCr 2。在考试环境下，使用计算器能减少错误，但你仍需能正确写出展开式。多练习输入不同的 nCr 值，直到快速熟练。
s 1. For the first term r = 0, so ⁵C₀ = 1. If you get an unexpected value, recheck your r.
始终核实你输入的 n 和 r 的值。记住 n 是二项式被乘的次幂，r 是项的位置序号减 1。第一项 r = 0，所以 ⁵C₀ = 1。如果你得到意外的数值，重新检查 r。
8. Common Mistakes to Avoid | 常见错误要避免
• Forgetting to apply the power to both the coefficient and the variable, e.g., (2x)3 = 8x3, not 2x3.
• 忘记将幂同时应用到系数和变量上，例如 (2x)3 = 8x3，而不是 2x3。
• Missing negative signs: when b is negative, the signs of terms alternate. Always include brackets around negative b when substituting.
• 遗漏负号：当 b 为负时，各项的符号交替。代入时始终将负的 b 用括号括起来。
• erm corresponds to r = 0, second term r = 1, etc. Many marks are lost by being off by one.
• 混淆项数与 r：第一项对应 r = 0，第二项 r = 1，依此类推。许多失分都是因为错位了 1。
• Expanding too many terms unnecessarily; if only one term is requested, use the general term formula to save time.
• 不必要地展开太多项；如果只要求某一项，则用通项公式以节省时间。
9. Worked Example 1: (2x + 3)4 | 例题 1： (2x + 3)4
Expand and simplify (2x + 3)4 rem, n = 4. The coefficients are ⁴C₀ = 1, ⁴C₁ = 4, ⁴C₂ = 6, ⁴C₃ = 4, ⁴C₄ = 1. Write terms:
展开并化简 (2x + 3)4。使用二项式定理，n = 4。系数为 ⁴C₀ = 1, ⁴C₁ = 4, ⁴C₂ = 6, ⁴C₃ = 4, ⁴C₄ = 1。写出各项：
T₁ = 1·(2x)4·3⁰ = 16x4
T₂ = 4·(2x)3·3¹ = 4·8x3·3 = 96x3
T₃ = 6·(2x)2·3² = 6·4x2·9 = 216x2
T₄ = 4·(2x)1·3³ = 4·2x·27 = 216x
T₅ = 1·(2x)0·3⁴ = 81
So, (2x + 3)4 = 16x4 + 96x3 + 216x2 + 216x + 81. This matches the pattern that powers of x decrease and coefficients are from Pascal’s triangle multiplied by the powers of 2 and 3.
因此，(2x + 3)4 = 16x4 + 96x3 + 216x2 + 216x + 81。这符合 x 指数递减、系数来自帕斯卡三角并乘以 2 和 3 的幂的规律。
10. Worked Example 2: Find the Coefficient of x³ in (1 – 2x)⁵ | 例题 2：求 (1 – 2x)⁵ 中 x³ 的系数
Here a = 1, b = –2x, n = 5. General term: Tr+1 = ⁵Cr (1)5-r (–2x)r = ⁵Cr (–2)r xr. We need the term with x³, so set r = 3. Then T4 = ⁵C₃ (–2)3 x³ = 10 × (–8) x³ = –80x³. The coefficient is –80. Notice that r matched the power of x because a = 1 simplified the expression.
这里 a = 1, b = –2x, n = 5。通项：Tr+1 = ⁵Cr (1)5-r (–2x)r = ⁵Cr (–2)r xr。我们需要含 x³ 的项，所以令 r = 3。那么 T4 = ⁵C₃ (–2)3 x³ = 10 × (–8) x³ = –80x³。系数为 –80。注意到 r 正好与 x 的指数一致，因为 a = 1 简化了表达式。
11. Special Cases: (1 + x)ⁿ | 特殊情况： (1 + x)ⁿ
When the first term is 1, the expansion simplifies significantly: (1 + x)n = 1 + nx + [n(n–1)/2!] x2 + [n(n–1)(n–2)/3!] x3 + … This is often used in approximations and is valid for any positive integer n. In IGCSE, you may see this form and be asked to use it or contrast it with the general expansion.
当第一项为 1 时，展开式大大简化：(1 + x)n = 1 + nx + [n(n–1)/2!] x2 + [n(n–1)(n–2)/3!] x3 + … 这个形式常用于近似计算，并对任意正整数 n 成立。在 IGCSE 中，你可能会见到这种形式并被要求使用它，或将其与一般展开式进行对比。
For example, to find the first four terms of (1 + 2x)6, you can apply the formula: 1 + 6(2x) + [6×5/2!] (2x)2 + [6×5×4/3!] (2x)3 = 1 + 12x + 15 × 4x2 + 20 × 8x3 = 1 + 12x + 60x2 + 160x3. Always check whether it is simpler to use the standard binomial theorem or this special case.
例如，求 (1 + 2x)6 的前四项，你可以应用公式：1 + 6(2x) + [6×5/2!] (2x)2 + [6×5×4/3!] (2x)3 = 1 + 12x + 15 × 4x2 + 20 × 8x3 = 1 + 12x + 60x2 + 160x3。始终判断是用标准二项式定理还是这种特殊情况更简便。
12. Exam Tips and Practice | 考试技巧与练习
• Write down the general term formula at the start of any ‘find a specific term’ question: Tr+1 = ⁿCr an-r br. This earns method marks even if the arithmetic is wrong.
• 在任何“求特定项”的题目开头，写下通项公式： Tr+1 = ⁿCr an-r br。即便计算有误，这也能为你赢得方法分。
• Practice with different combinations of a and b, especially negative b and coefficients other than 1. Watch out for signs and powers.
• 多练习 a 和 b 的不同组合，特别是 b 为负数以及系数不为 1 的情形。注意符号和幂次。
• If you are asked to find the constant term (term independent of x), set the power of x in the general term to zero and solve for r. Then substitute back to find the coefficient.
• 如果要求找常数项（与 x 无关的项），将通项中 x 的指数设为零，解出 r。然后代回求出系数。
• Always write your final answer with terms in descending powers of the variable unless otherwise instructed. Keep it neat and systematic.
• 除非另有要求，始终按变量指数递减的顺序写出最终答案。保持整洁和条理。
• Use past papers and mark schemes to familiarise yourself with the exact phrasing and expected layout of AQA binomial expansion questions.
• 使用历年真题和评分标准，熟悉 AQA 二项式展开题目的确切措辞和期望的书写格式。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图