📚 The Binomial Expansion | 二项式展开
The binomial expansion is a fundamental algebraic tool that allows us to write expressions of the form (a+b)n as a sum of terms involving powers of a and b. For A-Level Edexcel Mathematics, understanding how to expand binomials for positive integer exponents, as well as for fractional and negative exponents, is essential. This topic bridges algebraic manipulation, combinatorial reasoning, and infinite series, and it frequently appears in both pure mathematics and applied contexts such as approximations and probability.
二项式展开是一个基础代数工具，它使我们能够将形如 (a+b)n 的表达式写成包含 a 和 b 的幂次项之和。对于 Edexcel A-Level 数学，掌握正整数指数以及分数指数和负指数下的二项式展开至关重要。这一主题连接了代数运算、组合推理和无穷级数，并在近似计算和概率等应用情境中频繁出现。
1. Pascal's Triangle and Combinatorial Coefficients | 帕斯卡三角形与组合系数
Before introducing the general formula, it helps to recall Pascal's triangle. For a small positive integer n, the coefficients of the expansion of (a+b)n are the numbers in the (n+1)-th row of Pascal's triangle. For example, (a+b)3 itten as ³C₀, ³C₁, ³C₂, ³C₃. The binomial coefficient C(n,r) counts the number of ways to choose r items from n, and is given by C(n,r) = n! / [r! (n−r)! rovides a systematic way to find any coefficient without drawing the triangle for large n.
在引入一般公式之前，回顾帕斯卡三角形很有帮助。对于一个较小的正整数 n，(a+b)n 展开式的系数就是帕斯卡三角形第 (n+1) 行的数字。例如，(a+b)³ 展开得到 1·a³ + 3·a²b + 3·ab² + 1·b³。系数 1, 3, 3, 1 对应组合数 C(3,0), C(3,1), C(3,2), C(3,3)，也常写作 ³C₀, ³C₁, ³C₂, ³C₃。二项式系数 C(n,r) 表示从 n 个物品中选取 r 个的方式数，其公式为 C(n,r) = n! / [r! (n−r)!]。这提供了一种系统的方法，无需画出三角形便可求得较大的 n 下的任意系数。
2. The Binomial Theorem for Positive Integer n | 正整数 n 的二项式定理
For a positive integer n, the binomial theorem states that (a+b)n = C(n,0)an + C(n,1)an−1b + C(n,2)an−2b² + ... + C(n,n)bn. More compactly, using summation notation:
对于正整数 n，二项式定理表明 (a+b)n = C(n,0)an + C(n,1)an−1b + C(n,2)an−2b² + ... + C(n,n)bn。用求和符号可更简洁地表示为：
(a+b)n = Σ (r=0 to n) C(n,r) an−r br
ify a=1, b=2x, and n=5, then systematically apply the theorem. The expansion quickly generates terms such as 1 + 10x + 40x² + 80x³ + 80x⁴ + 32x⁵.
每一项由一个二项式系数、a 的递减幂和 b 的递增幂组成。在展开诸如 (1+2x)⁵ 的表达式时，确定 a=1, b=2x, n=5，然后系统地应用该定理。展开将迅速生成 1 + 10x + 40x² + 80x³ + 80x⁴ + 32x⁵ 这样的项。
3. Finding Specific Coefficients and Terms | 求特定系数和指定项
Exam questions often ask for the coefficient of a particular power of x, or for a constant term, without expanding the whole expression. The general term (the (r+1)-th term) is given by Tr+1 = C(n,r) an−r br. By substituting the parts of the binomial and equating the power of x to the required value, you can solve for r and then compute the corresponding coefficient.
考试题经常要求在不完全展开的前提下，求出某个特定 x 的幂的系数或常数项。一般项（第 r+1 项）由 Tr+1 = C(n,r) an−r br 给出。将二项式的各部分代入，并令 x 的指数等于所要求的值，便可解出 r，进而计算相应的系数。
pendent of x in (3x + 2/x)⁸, set a=3x, b=2/x, n=8. The power of x in the general term is 38−r·(2r)·x8−r · x−r = constant × x8−2r independent of x, set 8−2r = 0, giving r=4. The coefficient is then C(8,4)·3⁴·2⁴ = 90720.
例如，要求 (3x + 2/x)⁸ 中与 x 无关的项，设 a=3x, b=2/x, n=8。一般项中 x 的幂次为 38−r·(2r)·x8−r · x−r = 常数 × x8−2r。为了使该项不含 x，令 8−2r = 0，得到 r=4。其系数便是 C(8,4)·3⁴·2⁴ = 90720。
4. Extending to Rational and Negative Powers | 扩展到分数指数与负指数
When the exponent n is not a positive integer (e.g., n is a fraction or a negative number), the finite binomial expansion is replaced by an infinite series. The standard form to expand is (1+x)n, where |x|<1. The series is:
当指数 n 不是正整数时（例如 n 为分数或负数），有限项的二项式展开便被一个无穷级数所替代。展开的标准形式为 (1+x)n，且要求 |x|<1。其级数为：
(1+x)n = 1 + nx + [n(n−1)/2!] x² + [n(n−1)(n−2)/3!] x³ + ...
This formula works for any rational n and gives an infinite series that converges when |x|<1. For example, (1+x)−1 = 1 − x + x² − x³ + ... and √1+x = (1+x)1/2 = 1 + (1/2)x − (1/8)x² + (1/16)x³ − ... .
该公式适用于任何有理数 n，并给出一个在 |x|<1 时收敛的无穷级数。例如，(1+x)−1 = 1 − x + x² − x³ + ... ，而 √1+x = (1+x)1/2 = 1 + (1/2)x − (1/8)x² + (1/16)x³ − ... 。
5. The Infinite Series Form and Factorial Coefficients | 无穷级数形式与阶乘系数
The coefficients in the expansion of (1+x)n non-integer n are no longer simple C(n,r) but rather products like n(n−1)(n−2)...(n−r+1)/r! helps in writing the series without memorising: the r-th term (r≥1) is [n(n−1)...(n−r+1)/r!] xr. The factorial denominator r! grows quickly, and for fractional n, the series is infinite because the product in the numerator never becomes zero.
对于非整数 n，(1+x)n 展开式中的系数不再是简单的 C(n,r)，而是诸如 n(n−1)(n−2)...(n−r+1)/r! 的乘积。我们常用广义二项式系数来书写。理解这一规律有助于无需死记硬背地写出级数：第 r 项（r≥1）为 [n(n−1)...(n−r+1)/r!] xr。分母 r! 增长很快，而对于分数 n，级数之所以是无穷的，是因为分子中的乘积永远不会变为零。
For (1+2x)1/3, the first four terms are 1 + (1/3)(2x) + (1/3)(−2/3)/2! (2x)² + (1/3)(−2/3)(−5/3)/3! (2x)³ + ..., simplifying to 1 + (2/3)x − (4/9)x² + (40/81)x³ + ... . Care with algebraic simplification is crucial here.
对于 (1+2x)1/3，前四项为 1 + (1/3)(2x) + (1/3)(−2/3)/2! (2x)² + (1/3)(−2/3)(−5/3)/3! (2x)³ + ...，化简后得 1 + (2/3)x − (4/9)x² + (40/81)x³ + ... 。此处代数化简务必小心。
6. Validity and Convergence Conditions | 有效性与收敛条件
The infinite binomial expansion (1+x)n is only valid (converges) when |x| < 1. If a binomial is given in the form (a+bx)n with a ≠ 1, we must first rewrite it as an (1 + (b/a)x)n the bracket takes the form (1+X) with X = (b/a)x. The condition for validity then becomes |(b/a)x| < 1, i.e., |x| < |a/b|. Overlooking this validity range is a common source of mark loss in exams.
无穷二项式展开 (1+x)n 仅在 |x| < 1 时有效（收敛）。如果二项式以 (a+bx)n 的形式给出，且 a ≠ 1，我们必须先将其改写为 an (1 + (b/a)x)n，从而使括号具有 (1+X) 的形式，其中 X = (b/a)x。有效性的条件随之变为 |(b/a)x| < 1，即 |x| < |a/b|。忽视这一有效性范围是考试中常见的丢分原因。
For example, expanding (4+3x)1/2 directly is incorrect; we must write it as 41/2 [1 + (3/4)x]1/2 = 2 (1 + 0.75x)1/2. The expansion is then valid for |0.75x| < 1 ⇒ |x| < 4/3. Always state the validity interval alongside the series.
例如，直接展开 (4+3x)1/2 是不正确的；我们必须写成 41/2 [1 + (3/4)x]1/2 = 2 (1 + 0.75x)1/2。该展开在 |0.75x| < 1 ⇒ |x| < 4/3 时有效。请务必在给出级数的同时注明有效区间。
7. Adapting the Form (a+bx)n for Rational n | 对有理数指数调整 (a+bx)n 的形式
To handle (a+bx)n when n is not a positive integer, follow these steps: factor out an to get an[1+(b/a)x]n; then apply the standard expansion to [1+(b/a)x]n; finally multiply each term by an. It is often easier to find the first few terms only, as the series is infinite. Remember that the coefficient of xr will involve an multiplied by the r-th series term of the (1+X)n expansion.
当 n 不是正整数时，要处理 (a+bx)n ，遵循以下步骤：提取因子 an 得到 an[1+(b/a)x]n；然后对 [1+(b/a)x]n 应用标准展开式；最后用 an 乘以每一项。通常只需找出前几项，因为级数是无限的。要记得 xr 的系数将包含 an 与 (1+X)n 展开式第 r 项之积。
Example: Expand (2−5x)−2 in ascending powers of x up to x³. First, rewrite as 2−2[1 − (5/2)x]−2 = (1/4)[1 − (5/2)x]−2. Using n=−2, X = −(5/2)x, the expansion is (1/4)[1 + (−2)(−5/2)x + (−2)(−3)/2! (−5/2)²x² + (−2)(−3)(−4)/3! 5/2)³x³ + ...] = (1/4)[1 + 5x + (75/4)x² + (125/2)x³ + ...]. Simplify each term carefully.
示例：将 (2−5x)−2 按 x 的升幂展开至 x³。首先改写为 2−2[1 − (5/2)x]−2 = (1/4)[1 − (5/2)x]−2。利用 n=−2, X = −(5/2)x，展开得 (1/4)[1 + (−2)(−5/2)x + (−2)(−3)/2! (−5/2)²x² + (−2)(−3)(−4)/3! (−5/2)³x³ + ...] = (1/4)[1 + 5x + (75/4)x² + (125/2)x³ + ...]。请仔细化简每一项。
8. Partial Fractions and Binomial Expansion | 部分分式与二项式展开
A common Edexcel question combines algebraic fractions with binomial expansion. A rational function like (4+3x)/[(1−x)(1+2x)] can be split into partial fractions first, yielding A/(1−x) + B/(1+2x). Each fraction can then be written as A(1−x)−1 + B(1+2x)−1 and expanded separately using the binomial series. This technique allows the expansion of complex rational expressions as a single power series, valid for the intersection of the individual validity intervals.
一个常见的 Edexcel 题型将代数分式与二项式展开相结合。像 (4+3x)/[(1−x)(1+2x)] 这样的有理函数可以先拆分成部分分式，得到 A/(1−x) + B/(1+2x)。接着每个分式可写为 A(1−x)−1 + B(1+2x)−1 的形式，并分别用二项级数展开。这一技巧可将复杂的有理表达式展开为单一的幂级数，其有效区间为各个单独区间之交集。
After finding A and B, ensure each term is expressed in the form (1+X)n. For (1−x)−1, n=−1, X=−x, valid for |x|<1; for (1+2x)−1, n=−1, X=2x, valid for |2x|<1 ⇒ |x|<1/2. The combined expansion is valid for |x|<1/2. Adding the series term by term gives the required coefficients.
求出 A 和 B 后，确保每一项都写成 (1+X)n 的形式。对于 (1−x)−1，n=−1, X=−x，在 |x|<1 时有效；对于 (1+2x)−1，n=−1, X=2x，在 |2x|<1 ⇒ |x|<1/2 时有效。合并展开在 |x|<1/2 时有效。逐项相加即可得到所需的系数。
9. Using Expansions for Approximations | 利用展开进行近似计算
If x is small, the terms of a binomial expansion quickly decrease in magnitude. This allows us to approximate functions. For example, (1+x)1/2 ≈ 1 + (1/2)x − (1/8)x² for very small x. To estimate √26, write √26 = (25+1)1/2 = 5(1+0.04)1/2. Using the first three terms of the expansion for (1+0.04)1/2 gives an approximation for √26. The error decreases as more terms are taken.
如果 x 很小，二项展开式的各项数值会迅速减小。这使得我们可以对函数进行近似。例如，当 x 非常小时，(1+x)1/2 ≈ 1 + (1/2)x − (1/8)x²。要估算 √26，可写 √26 = (25+1)1/2 = 5(1+0.04)1/2。利用 (1+0.04)1/2 展开式的前三项即可得到 √26 的近似值。取的项数越多，误差越小。
In practice, you may be asked to find an approximation to a rational number or a root, and state the degree of accuracy. Always compare the next term to judge the possible error. For √26 with the first three terms, the next term is positive and roughly 0.00002, so the approximation 5.09902 is accurate to several decimal places.
实际考题可能要求计算一个有理数或方根的近似值，并说明精确度。始终比较下一项以判断可能的误差。对于 √26 取前三项的情况，下一项为正，约为 0.00002，因此近似值 5.09902 在小数点后多位是准确的。
10. Expansion of Products and Powers | 乘积与乘幂的展开
Sometimes you need the expansion of a product like (1+3x)⁴(1−2x)⁻² up to x³. One approach is to expand each factor separately as far as required, then multiply the series, collecting like terms. Because both are polynomial-like series, term-by-term multiplication is manageable up to a certain power. This avoids trying to find a single binomial form.
有时你需要展开像 (1+3x)⁴(1−2x)⁻² 这样的乘积至 x³。一种方法是分别将每个因子展开到所需的项数，然后将级数相乘并合并同类项。由于两者都是类似多项式的级数，逐项相乘到某个幂次是可行的。这避免了尝试寻找单一的二项式形式。
d keeping terms up to x³ yields 1 + 16x + 114x² + 492x³ + ... . Always check whether an alternative route via partial fractions or careful factorisation is simpler.
例如，(1+3x)⁴ = 1 + 12x + 54x² + 108x³ + ...，而利用 n=−2 得 (1−2x)⁻² = 1 + 4x + 12x² + 32x³ + ...。相乘并保留至 x³ 项得到 1 + 16x + 114x² + 492x³ + ... 。始终要检查是否通过部分分式或巧妙的因式分解能使过程更简单。
11. Identifying the Range of Validity for Composite Expansions | 确定复合展开的有效范围
When combining several binomial expansions, the overall validity interval is the intersection of the individual intervals. If one factor requires |x| < 1 and another requires |x| < 1/3, the combined expansion is only guaranteed to hold for |x| <
Published by TutorHao | A-Level 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