Binomial Expansion of (1+x)^n | 二项式展开 (1+x)^n

📚 Binomial Expansion of (1+x)^n | 二项式展开 (1+x)^n

The binomial expansion is one of the most powerful tools in algebra. It allows us to expand expressions of the form (1+x)^n without repeatedly multiplying by hand, and it forms the backbone of many exam questions in algebra, probability, and calculus.

二项式展开是代数中最强大的工具之一。它让我们无需反复手算乘法，就能展开形如(1+x)^n的表达式，同时也是代数、概率和微积分中许多考试题目的基础。

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1. Understanding the Binomial Theorem | 理解二项式定理

The binomial theorem states that for any positive integer n, the expansion of (1+x)^n is given by a sum of terms involving binomial coefficients. Each term has the form C(n, r) x^r, where r ranges from 0 to n.

二项式定理指出：对于任意正整数 n，(1+x)^n 的展开式由若干包含二项式系数的项组成。每一项的形式为 C(n, r) x^r，其中 r 从 0 取到 n。

(1+x)^n = C(n,0) + C(n,1)x + C(n,2)x² + … + C(n,n)xⁿ

The binomial coefficient C(n, r) is also written as ⁿCᵣ or (n choose r), and it counts how many ways to choose r items from n items. It is calculated using factorials.

二项式系数 C(n, r) 也写作 ⁿCᵣ 或组合数 C(n, r)，它表示从 n 个物品中选取 r 个物品的方法数，通过阶乘计算。

C(n, r) = n! / (r! (n – r)!)

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2. Pascal’s Triangle | 帕斯卡三角形

Pascal’s triangle provides a simple visual way to find binomial coefficients. Each row of the triangle gives the coefficients for (1+x)^n, starting with n = 0 at the top.

帕斯卡三角形提供了一种直观地寻找二项式系数的简单方法。三角形中每一行都对应 (1+x)^n 的系数，从顶部的 n = 0 开始。

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

Each number is the sum of the two numbers directly above it. For example, in row n=4, the coefficient 6 comes from 3+3 in the row above. This pattern continues indefinitely.

每个数字都是其正上方两个数字之和。例如，在 n=4 这一行中，系数 6 来自上一行的 3+3。这个规律可以无限延续。

Using Pascal’s triangle, we can quickly write out expansions for small n without computing factorials.

利用帕斯卡三角形，我们可以快速写出较小 n 的展开式，而无需计算阶乘。

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3. The General Term | 通项公式

In an expansion of (1+x)^n, the term containing x^r is called the (r+1)th term, because counting starts from r = 0. This general term is extremely useful in exams.

在 (1+x)^n 的展开式中，含有 x^r 的项被称为第 (r+1) 项，因为计数从 r = 0 开始。这个通项公式在考试中极为有用。

T_{r+1} = C(n, r) x^r

For example, the 5th term of (1+x)^10 corresponds to r = 4, so it is C(10, 4) x⁴ = 210x⁴. Notice that the power of x is always one less than the term number.

例如，(1+x)^10 的第 5 项对应 r = 4，因此为 C(10, 4) x⁴ = 210x⁴。注意 x 的幂总是比项数少 1。

When the expression is (1 + ax)^n instead of (1+x)^n, each x is replaced by ax, so the general term becomes C(n, r) (ax)^r = C(n, r) a^r x^r.

当表达式是 (1 + ax)^n 而不是 (1+x)^n 时，每个 x 都被替换为 ax，因此通项变为 C(n, r) (ax)^r = C(n, r) a^r x^r。

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4. Expansion for Positive Integer n | 正整数 n 的展开

When n is a positive integer, the expansion has exactly n+1 terms. The powers of x increase from 0 to n, and the coefficients follow the symmetric pattern of Pascal’s triangle.

当 n 是正整数时，展开式恰好有 n+1 项。x 的幂从 0 增加到 n，系数遵循帕斯卡三角形的对称模式。

• For n=2: (1+x)² = 1 + 2x + x²

  当 n=2 时：(1+x)² = 1 + 2x + x²

• For n=3: (1+x)³ = 1 + 3x + 3x² + x³

  当 n=3 时：(1+x)³ = 1 + 3x + 3x² + x³

• For n=4: (1+x)⁴ = 1 + 4x + 6x² + 4x³ + x⁴

  当 n=4 时：(1+x)⁴ = 1 + 4x + 6x² + 4x³ + x⁴

Notice that the coefficients are the same forward and backward. This symmetry comes from the identity C(n, r) = C(n, n-r).

注意系数前后对称。这种对称性来自于恒等式 C(n, r) = C(n, n – r)。

Also, the sum of all coefficients in the expansion of (1+x)^n is found by setting x = 1, giving 2ⁿ. This is a useful shortcut for checking answers.

此外，将 x = 1 代入展开式，所有系数之和为 2ⁿ。这是一个检查答案的有用技巧。

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5. Binomial Coefficients and Factorials | 二项式系数与阶乘

To calculate binomial coefficients without Pascal’s triangle, we use the factorial formula. For example, C(6, 2) = 6! / (2! 4!) = 720 / (2 × 24) = 15.

为了在没有帕斯卡三角形时计算二项式系数，我们使用阶乘公式。例如，C(6, 2) = 6! / (2! 4!) = 720 / (2 × 24) = 15。

Many students find it easier to use the shortcut form: C(n, r) = n(n-1)(n-2)…(n-r+1) / r!. This avoids writing out large factorials.

许多学生发现使用简化形式更容易：C(n, r) = n(n-1)(n-2)…(n-r+1) / r!。这样可以避免写出很大的阶乘。

For example, C(10, 3) = (10 × 9 × 8) / (3 × 2 × 1) = 720 / 6 = 120. This method is especially fast when r is small.

例如，C(10, 3) = (10 × 9 × 8) / (3 × 2 × 1) = 720 / 6 = 120。当 r 较小时，这种方法特别快。

Remember that C(n, 0) = 1 and C(n, 1) = n for every positive integer n. These are the first two coefficients in every expansion.

记住对于任意正整数 n，C(n, 0) = 1 且 C(n, 1) = n。这是每个展开式的前两项系数。

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6. Expanding (1 + ax)^n | 展开 (1 + ax)^n

A common exam question asks for the expansion of (1 + ax)^n. The key is to treat ax as a single unit and apply the same binomial formula.

一个常见的考试题是要求展开 (1 + ax)^n。关键是将 ax 视为一个整体，并应用相同的二项式公式。

(1 + ax)^n = 1 + n(ax) + C(n,2)(ax)² + C(n,3)(ax)³ + …

Simplifying each term gives powers of a as well as powers of x. For instance, when n = 5 and a = 2:

化简每一项会同时得到 a 的幂和 x 的幂。例如，当 n = 5 且 a = 2 时：

(1 + 2x)⁵ = 1 + 10x + 40x² + 80x³ + 80x⁴ + 32x⁵

Notice how the coefficients involve powers of 2: 2, 4, 8, 16, 32 multiplied by the binomial coefficients 1, 5, 10, 10, 5, 1.

注意系数如何包含 2 的幂：2、4、8、16、32 分别乘以二项式系数 1、5、10、10、5、1。

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7. Finding a Specific Coefficient | 求特定项的系数

To find the coefficient of x^k in (1+x)^n, simply set r = k in the general term. For example, the coefficient of x³ in (1+x)^8 is C(8, 3) = 56.

要求 (1+x)^n 中 x^k 的系数，只需在通项中令 r = k。例如，(1+x)^8 中 x³ 的系数是 C(8, 3) = 56。

When the bracket is (1 + ax)^n, the coefficient of x^k becomes C(n, k) a^k. This is because the term is C(n, k)(ax)^k.

当括号是 (1 + ax)^n 时，x^k 的系数变为 C(n, k) a^k。这是因为该项为 C(n, k)(ax)^k。

For example, in the expansion of (1 + 3x)^7, the coefficient of x⁴ is C(7, 4) × 3⁴ = 35 × 81 = 2835.

例如，在 (1 + 3x)^7 的展开式中，x⁴ 的系数是 C(7, 4) × 3⁴ = 35 × 81 = 2835。

Always remember to include the power of a when it is not 1. A common mistake is forgetting to raise a to the correct power.

当 a 不为 1 时，务必记得包含 a 的幂。一个常见错误是忘记将 a 提升到正确的次数。

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8. The Independent Term | 常数项

The independent term in an expansion is the term that does not contain x, meaning the power of x is zero. In (1+x)^n, the independent term is always 1, from r = 0.

展开式中的常数项是那些不含 x 的项，即 x 的幂为零。在 (1+x)^n 中，常数项始终为 1，来自 r = 0。

However, for more complex expressions like (1 + ax)^n × (1 + bx)^m, finding the constant term requires considering combinations of terms whose x powers cancel out.

然而，对于更复杂的表达式，如 (1 + ax)^n × (1 + bx)^m，求常数项需要考虑 x 幂相互抵消的项的组合。

This type of question often appears in advanced algebra exams, requiring careful tracking of exponents across multiple brackets.

这类问题经常出现在高级代数考试中，需要仔细追踪多个括号之间的指数变化。

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9. Binomial Expansion for Negative or Fractional n | 负指数或分数指数的二项式展开

When n is not a positive integer, the binomial expansion becomes an infinite series. The formula uses the generalised binomial coefficient, defined for any real n.

当 n 不是正整数时，二项式展开成为一个无穷级数。公式使用广义二项式系数，该系数对任意实数 n 都有定义。

(1+x)^n = 1 + nx + n(n-1)x²/2! + n(n-1)(n-2)x³/3! + …

This series is valid only when |x| < 1 for most fractional or negative n. This condition is called the interval of convergence.

对于大多数分数或负指数 n，这个级数仅在 |x| < 1 时有效。这个条件被称为收敛区间。

For example, (1+x)^(-1) = 1 – x + x² – x³ + … for |x| < 1. This is the well-known geometric series.

例如，(1+x)^(-1) = 1 – x + x² – x³ + …，其中 |x| < 1。这就是著名的等比级数。

Also, (1+x)^(1/2) = 1 + (1/2)x – (1/8)x² + (1/16)x³ – … which can be used to approximate square roots.

此外，(1+x)^(1/2) = 1 + (1/2)x – (1/8)x² + (1/16)x³ – …，可用于近似计算平方根。

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10. Using Binomial Expansion for Approximations | 用二项式展开做近似计算

One practical application is approximating values like (1.01)¹⁰. By writing 1.01 = 1 + 0.01, we can use the first few terms of the binomial expansion to get a very close estimate.

一个实际应用是近似计算像 (1.01)¹⁰ 这样的值。将 1.01 写成 1 + 0.01，我们可以使用二项式展开的前几项得到一个非常接近的估计值。

(1.01)¹⁰ = 1 + 10(0.01) + C(10,2)(0.01)² + C(10,3)(0.01)³ + …

Computing the first three terms gives 1 + 0.1 + 0.0045 = 1.1045. Adding the next term C(10,3)(0.01)³ = 120 × 0.000001 = 0.00012 gives 1.10462, which is extremely close to the true value 1.10462.

计算前三项得到 1 + 0.1 + 0.0045 = 1.1045。再加上下一项 C(10,3)(0.01)³ = 120 × 0.000001 = 0.00012，得到 1.10462，这与真实值 1.10462 极为接近。

This technique is particularly useful when calculators are not allowed, or when only a few decimal places of accuracy are needed.

当不允许使用计算器，或只需要几位小数精度时，这种技巧特别有用。

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11. Common Mistakes and How to Avoid Them | 常见错误及避免方法

One common mistake is forgetting that the expansion of (1+x)^n has n+1 terms, not n terms. Another is incorrectly applying the general term by confusing r with the term number.

一个常见错误是忘记 (1+x)^n 的展开式有 n+1 项，而不是 n 项。另一个错误是混淆 r 与项数，从而错误应用通项公式。

• Always start counting from r = 0, so the first term is C(n,0)x⁰ = 1.

  始终从 r = 0 开始计数，因此第一项是 C(n,0)x⁰ = 1。

• When expanding (1 + ax)^n, do not forget to raise a to the power r.

  展开 (1 + ax)^n 时，不要忘记将 a 提升到 r 次幂。

• For fractional n, check that |x| < 1 before using the infinite series.

  对于分数 n，使用无穷级数前检查 |x| < 1。

• Use Pascal’s triangle or the factorial formula to double-check small coefficients.

  使用帕斯卡三角形或阶乘公式来复核较小的系数。

By practising these patterns, you can avoid careless errors and solve binomial expansion questions quickly and confidently.

通过练习这些模式，你可以避免粗心错误，并迅速而自信地解决二项式展开问题。

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12. Exam-style Practice Questions | 考试风格练习题

Here are three typical exam questions to test your understanding. Try to solve them before checking the results.

以下是三道典型考试题，用于测试你的理解。请在查看答案前先自己尝试解答。

• 1. Find the coefficient of x⁵ in (1 + 2x)¹².

  1. 求 (1 + 2x)¹² 中 x⁵ 的系数。

• 2. Expand (1 + x/2)⁶ up to the term in x³.

  2. 展开 (1 + x/2)⁶ 至 x³ 项。

• 3. Use the binomial expansion to estimate (0.98)⁸ correct to 4 decimal places.

  3. 使用二项式展开估算 (0.98)⁸，精确到 4 位小数。

For question 1, the general term is C(12, r)(2x)^r, so we set r = 5: C(12, 5) × 2⁵ = 792 × 32 = 25344.

对于第 1 题，通项为 C(12, r)(2x)^r，令 r = 5：C(12, 5) × 2⁵ = 792 × 32 = 25344。

For question 2, write (1 + x/2)⁶ = 1 + 6(x/2) + 15(x/2)² + 20(x/2)³ = 1 + 3x + (15/4)x² + (5/2)x³.

对于第 2 题，写出 (1 + x/2)⁶ = 1 + 6(x/2) + 15(x/2)² + 20(x/2)³ = 1 + 3x + (15/4)x² + (5/2)x³。

For question 3, write 0.98 = 1 – 0.02, so (1 – 0.02)⁸ = 1 – 8(0.02) + 28(0.02)² – 56(0.02)³ + … = 1 – 0.16 + 0.0112 – 0.000448 + … = 0.8508 correct to 4 decimal places.

对于第 3 题，将 0.98 写成 1 – 0.02，因此 (1 – 0.02)⁸ = 1 – 8(0.02) + 28(0.02)² – 56(0.02)³ + … = 1 – 0.16 + 0.0112 – 0.000448 + … = 0.8508，精确到 4 位小数。

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