📚 Expanding (1 + x)ⁿ | 展开 (1 + x)ⁿ
The binomial expansion transforms an expression of the form (1 + x)ⁿ into a sum of powers of x with predictable coefficients. Mastering this topic is vital for A-Level Edexcel Mathematics, whether n is a positive integer, a fraction, or a negative number.
二项式展开将形如 (1 + x)ⁿ 的表达式转化为带有可预测系数的 x 幂次之和。无论 n 是正整数、分数还是负数，掌握此专题对 A-Level Edexcel 数学至关重要。
1. What Is a Binomial? | 什么是二项式？
A binomial is an algebraic expression containing exactly two terms, such as (1 + x), (a + b), or (2x – 3y). The process of expanding a binomial raised to a power n means expressing it as a sum of individual monomials, each consisting of a coefficient multiplied by a power of the variable(s).
二项式是恰好包含两项的代数表达式，例如 (1 + x)、(a + b) 或 (2x – 3y)。对一个二项式进行 n 次幂展开，是指将它表示成一组单项式的和，其中每一项由一个系数乘以变量的某次幂构成。
When n is a positive integer, the expansion yields a finite polynomial of degree n. For example, expanding (1 + x)² gives 1 + 2x + x², and (1 + x)³ gives 1 + 3x + 3x² + x³.
当 n 为正整数时，展开得到一个 n 次有限多项式。例如，展开 (1 + x)² 得 1 + 2x + x²，展开 (1 + x)³ 得 1 + 3x + 3x² + x³。
2. Pascal's Triangle and Binomial Coefficients | 帕斯卡三角与二项式系数
s of the expanded terms for small positive integer powers can be read directly from Pascal' 1 2 1; n = 3 gives 1 3 3 1, and so on. Every entry is the sum of the two entries above it.
小正整数次幂展开后的项系数可以直接从帕斯卡三角读得。三角形的每一行对应幂次 n，顶上由 n = 0 开始：1；n = 1 得 1 1；n = 2 得 1 2 1；n = 3 得 1 3 3 1，依此类推。每个数值等于上方两个数值之和。
These numbers are also known as binomial coefficients, often written as nCr or C(n, r), and they count the number of ways to choose r items from a set of n items. The binomial expansion can thus be expressed systematically without drawing Pascal’s Triangle for large n.
这些数字也称为二项式系数，常写作 nCr 或 C(n, r)，它们表示从 n 个元素中选取 r 个元素的组合数。因此二项式展开可以系统化表示，而不必为大 n 绘制帕斯卡三角。
3. The Binomial Theorem for Positive Integer n | 正整数指数 n 的二项式定理
For any positive integer n, the expansion of (1 + x)ⁿ is given by the Binomial Theorem:
(1 + x)ⁿ = 1 + nC1 x + nC2 x² + ... + nCr xʳ + ... + nCn xⁿ
对于任意正整数 n，(1 + x)ⁿ 的展开由二项式定理给出：
(1 + x)ⁿ = 1 + nC1 x + nC2 x² + ... + nCr xʳ + ... + nCn xⁿ
Here, nCr is the binomial coefficient, calculated as nCr = n! / [r!(n – r)!], where n! (n factorial) is the product of all positive integers up to n. The term involving xʳ is the general term of the expansion.
此处 nCr 是二项式系数，计算公式为 nCr = n! / [r!(n – r)!]，其中 n!（n 阶乘）是直至 n 的所有正整数的乘积。含 xʳ 的那一项就是展开的通项。
4. General Term and the (r+1)th Term | 通项与第 r+1 项
In the binomial expansion of (1 + x)ⁿ, the power of x increases by 1 each time. If we label the terms starting with the constant term (x⁰) as the first term, then the coefficient of xʳ is nCr. Therefore, the (r+1)th term is:
Tr+1 = nCr · xʳ
在 (1 + x)ⁿ 的二项展开中，x 的幂次每次增加 1。若我们将常数项（x⁰）计为第一项，那么 xʳ 的系数就是 nCr。因此第 r+1 项为：
Tr+1 = nCr · xʳ
t expanding the whole series. For instance, the coefficient of x⁵ in (1 + x)⁸ is 8C5 = 56.
该公式对于在不展开整个级数的情况下求特定项极其有用。例如，(1 + x)⁸ 中 x⁵ 的系数为 8C5 = 56。
5. Expanding (1 + x)ⁿ Using Factorial Form | 使用阶乘形式展开 (1 + x)ⁿ
Substituting the factorial definition of nCr gives a fully explicit form of the expansion:
(1 + x)ⁿ = 1 + n·x + [n(n – 1)/2!]x² + [n(n – 1)(n – 2)/3!]x³ + ... + xⁿ
将 nCr 的阶乘定义代入，得到完全显式的展开形式：
(1 + x)ⁿ = 1 + n·x + [n(n – 1)/2!]x² + [n(n – 1)(n – 2)/3!]x³ + ... + xⁿ
This representation highlights the pattern of descending factors in the numerator and factorial denominators. It is also the form that extends naturally to non-integer and negative n.
这种表示突显了分子中递减因子与阶乘分母的规律，也正是这种形式可以自然地推广到非整数和负指数 n。
6. Worked Examples (Positive Integer n) | 正整数指数例题
Example 1: Expand (1 + x)⁴. Using Pascal's Triangle or the formula, we obtain:
(1 + x)⁴ = 1 + 4x + 6x² + 4x³ + x⁴.
示例 1：展开 (1 + x)⁴。使用帕斯卡三角或公式得：
(1 + x)⁴ = 1 + 4x + 6x² + 4x³ + x⁴。
Example 2: Find the coefficient of x³ in the expansion of (1 + x)¹⁰.
The required coefficient is 10C3 = 10!/(3!7!) = (10×9×8)/(3×2×1) = 120.
示例 2：求 (1 + x)¹⁰ 展开中 x³ 的系数。
所求系数为 10C3 = 10!/(3!7!) = (10×9×8)/(3×2×1) = 120。
Example 3: Expand (1 + 2x)³ by treating it as (1 + u)³ with u = 2x, then replace and simplify:
= 1 + 3(2x) + 3(2x)² + (2x)³ = 1 + 6x + 12x² + 8x³.
示例 3：将 (1 + 2x)³ 视为 (1 + u)³，其中 u = 2x，然后替换并化简：
= 1 + 3(2x) + 3(2x)² + (2x)³ = 1 + 6x + 12x² + 8x³。
7. Extending to Rational and Negative Indices | 扩展到有理指数和负指数
When n is not a positive integer (e.g., a fraction like ½ or a negative number like –2), the binomial expansion becomes an infinite series. The formula is:
(1 + x)ⁿ = 1 + n·x + [n(n – 1)/2!]x² + [n(n – 1)(n – 2)/3!]x³ + ...
当 n 不是正整数（例如分数 ½ 或负数 –2）时，二项展开成为一个无穷级数。公式为：
(1 + x)ⁿ = 1 + n·x + [n(n – 1)/2!]x² + [n(n – 1)(n – 2)/3!]x³ + ...
Because n is not a positive integer, the factors (n – + 1) will never become zero, so the series continues forever. For instance, (1 + x)⁻¹ = 1 – x + x² – x³ + ... (a geometric series), and (1 + x)¹⁄² = 1 + (1/2)x – (1/8)x² + (1/16)x³ – ... .
由于 n 不是正整数，因子 (n – r + 1) 永远不会变为零，因此级数无穷延续。例如，(1 + x)⁻¹ = 1 – x + x² – x³ + ...（一个几何级数），而 (1 + x)¹⁄² = 1 + (1/2)x – (1/8)x² + (1/16)x³ – ... 。
It is essential to use the factorial form here, because nCr is defined only for non‑negative integers r ≤ n. The factorial‑based formula works universally for any real n.
此处必须使用阶乘形式，因为 nCr 仅在 r ≤ n 且为非负整数时有定义。基于阶乘的公式对任何实数 n 均适用。
8. Range of Validity |x| < 1 | 有效性范围 |x| < 1
ial expansion to be valid (convergent), the absolute value of x must be less than 1, i.e., |x| < 1, or equivalently –1 < x < 1. If |x| diverges and no longer represents the function accurately unless n is a positive integer.
无穷二项展开成立（收敛）的条件是 x 的绝对值必须小于 1，即 |x| < 1，或等价地 –1 < x < 1。若 |x| ≥ 1，则级数发散，除 n 为正整数外，不再精确表示该函数。
en using the expansion for approximations. For example, the expansion of (1 + x)⁻¹ gives 1 – x + x² – ..., but substituting x = 2 would yield a divergent series that does not equal 1/3.
当利用展开进行近似计算时，这一条件至关重要。例如，(1 + x)⁻¹ 的展开是 1 – x + x² – ...，但代入 x = 2 会得到一个发散级数，它不等于 1/3。
9. Adapting (a + b)ⁿ to the (1 + x) Form | 将 (a + b)ⁿ 改写为 (1 + x) 形式
Many exam problems feature expressions like (a + b)ⁿ or (p + qx)ⁿ. To apply the standard binomial expansion, factor out the first term to create a '1' inside the bracket:
(a + b)ⁿ = aⁿ (1 + b/a)ⁿ
许多考题会给出形如 (a + b)ⁿ 或 (p + qx)ⁿ 的表达式。要应用标准二项式展开，可提取第一项，在括号内构造出 “1”：
(a + b)ⁿ = aⁿ (1 + b/a)ⁿ
x = b/a and expand (1 + x)ⁿ, then multiply each term by aⁿ. The validity condition becomes |b/a| < 1, i.e., |b| < |a|, unless n is a positive integer.
令 x = b/a，展开 (1 + x)ⁿ，然后将每一项乘以 aⁿ。除非 n 为正整数，否则有效性条件变为 |b/a| < 1，即 |b| < |a|。
For example, to expand (4 + 3x)⁻¹, write it as 4⁻¹ (1 + ¾x)⁻¹ = ¼ [1 – ¾x + (¾x)² – ...], valid when |¾x| < 1, i.e., |x| < 4/3.
例如，要展开 (4 + 3x)⁻¹，可改写为 4⁻¹ (1 + ¾x)⁻¹ = ¼ [1 – ¾x + (¾x)² – ...]，当 |¾x| < 1，即 |x| < 4/3 时有效。
10. Approximations Using Binomial Expansion | 利用二项式展开进行近似
he higher-power terms become negligible. We can approximate (1 + x)ⁿ by truncating the series after a few terms:
(1 + x)ⁿ ≈ 1 + n x + [n(n – 1)/2] x²
当 x 非常小时（例如 x = 0.01），高次项可以忽略不计。我们可以通过只取级数的前几项来近似 (1 + x)ⁿ：
(1 + x)ⁿ ≈ 1 + n x + [n(n – 1)/2] x²
For instance, to estimate √1.02, use (1 + 0.02)¹⁄² ≈ 1 + ½(0.02) + (½)(–½)/2 (0.02)² = 1 + 0.01 – 0.00005 = 1.00995. This method is widely used in physics and engineering.
例如，要估算 √1.02，使用 (1 + 0.02)¹⁄² ≈ 1 + ½(0.02) + (½)(–½)/2 (0.02)² = 1 + 0.01 – 0.00005 = 1.00995。此法在物理和工程中应用广泛。
11. Common Pitfalls to Avoid | 需避免的常见错误
Forgetting the validity condition: For non‑integer n, always state |x| < 1 or the equivalent condition after factoring. An expansion without a validity statement may lose marks.
忘记有效性条件：对于非整数 n，务必给出 |x| < 1 或提取因子后的等价条件。缺少有效性说明的展开可能会失分。
Misapplying nCr for non‑integer n: Using the nCr button on a calculator for fractional or negative n produces an error. Always use the factorial‑based formula n(n – 1)(n – 2).../(r!) for these cases.
对非整数 n 误用 nCr：对于分数或负数 n，使用计算器上的 nCr 键会出错。此类情况务必使用基于阶乘的公式 n(n – 1)(n – 2).../(r!)。
Sign errors: When n is negative, the alternating signs must be handled carefully. For (1 – x)ⁿ, replace x with (–x) in the expansion directly to avoid mistakes: (1 – x)ⁿ = 1 – n x + [n(n – 1)/2!]x² – ... .
符号错误：当 n 为负数时，必须仔细处理交替符号。对于 (1 – x)ⁿ，直接在展开中将 x 替换为 (–x) 可避免出错：(1 – x)ⁿ = 1 – n x + [n(n – 1)/2!]x² – ... 。
Truncating too early without checking: In approximation, if x is not small enough, truncating after the linear term may give a poor estimate. Always judge the size of x.
过早截断而未检查：在近似计算中，若 x 不够小，仅保留线性项可能给出较差的估计。务必判断 x 的大小。
12. Summary and Key Takeaways | 总结与核心要点
The expansion of (1 + x)ⁿ is a cornerstone of A-Level algebra. For a positive integer n, it produces a finite polynomial with coefficients given by nCr or Pascal's Triangle. The general term Tr+1 = nCr xʳ allows speedy extraction of any coefficient.
(1 + x)ⁿ 的展开是 A-Level 代数的基石。当 n 为正整数时，它生成一个有限多项式，其系数由 nCr 或帕斯卡三角给出。通项 Tr+1 = nCr xʳ 能快速提取任意系数。
For rational or negative n, the expansion becomes an infinite series, valid exclusively for |x| < 1. Expressions of the form (a + b)ⁿ must be rewritten as aⁿ (1 + b/a)ⁿ to use the standard formula. Approximations for small x are a practical application, but always note the validity region.
对于有理指数或负指数 n，展开成为无限级数，仅在 |x| < 1 时有效。形如 (a + b)ⁿ 的表达式必须改写为 aⁿ (1 + b/a)ⁿ 才能套用标准公式。对小 x 进行近似是一种实际应用，但务必指明有效范围。
Mastery of these techniques, together with careful handling of signs and factorial expressions, will equip you to tackle any binomial expansion question in the Edexcel A-Level Mathematics exam.
掌握这些技巧，加之谨慎处理符号与阶乘表达式，将使你能够应对 Edexcel A-Level 数学考试中的任何二项式展开问题。
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