Example 5.2.2: Binomial Expansion | 例题5.2.2：二项展开

📚 Example 5.2.2: Binomial Expansion | 例题5.2.2：二项展开

In this worked example, we will study a typical binomial expansion problem from the AQA A-Level Mathematics syllabus. The problem is to find the coefficient of x³ in the expansion of (2 + 3x)⁶, and then use the expansion to evaluate (2.03)⁶ correct to four significant figures.

在这个工作示例中，我们将学习 AQA A-Level 数学大纲中一个典型的二项展开问题。题目要求找出 (2 + 3x)⁶ 展开式中 x³ 的系数，并利用展开式将 (2.03)⁶ 精确到四位有效数字。

1. Problem Statement | 问题陈述

Write down the coefficient of x³ in the binomial expansion of (2 + 3x)⁶. Hence, or otherwise, evaluate (2.03)⁶ correct to four significant figures.

写出 (2 + 3x)⁶ 的二项展开式中 x³ 的系数。并由此（或以其他方法）将 (2.03)⁶ 精确到四位有效数字。

This question tests both the direct calculation of a binomial coefficient and the application of substitution to estimate a numerical value.

该问题既考查二项系数的直接计算，也考查通过代换估计数值的应用能力。

2. The Binomial Theorem | 二项定理复习

For a positive integer n, the binomial theorem states that (a + b)ⁿ = Σₖ₌₀ⁿ C(n, k) aⁿ⁻ᵏ bᵏ, where C(n, k) = n! / (k!(n−k)!).

对于正整数 n，二项定理指出 (a + b)ⁿ = Σₖ₌₀ⁿ C(n, k) aⁿ⁻ᵏ bᵏ，其中 C(n, k) = n! / (k!(n−k)!)。

In this example, we have a = 2, b = 3x, and n = 6. We can substitute these into the formula and examine each term.

在本例中，a = 2，b = 3x，n = 6。我们将这些值代入公式，并逐一检查各项。

3. General Term | 通项公式

The k-th term in the expansion is given by T₍ₖ₊₁₎ = C(6, k) · 2⁶⁻ᵏ · (3x)ᵏ. This is obtained by fixing the exponent of b as k and the exponent of a as 6 − k.

展开式中的第 (k+1) 项为 T₍ₖ₊₁₎ = C(6, k) · 2⁶⁻ᵏ · (3x)ᵏ。这是通过令 b 的指数为 k、a 的指数为 6 − k 得到的。

To find the coefficient of x³, we need the term where the power of x is exactly 3. Since (3x)ᵏ contains xᵏ, we set k = 3.

为了找到 x³ 的系数，我们需要 x 的幂恰好为 3 的那一项。因为 (3x)ᵏ 中含有 xᵏ，所以我们令 k = 3。

4. Coefficient of x³ | 求 x³ 的系数

Setting k = 3 in the general term gives T₄ = C(6, 3) · 2⁶⁻³ · (3x)³. The coefficient of x³ is therefore C(6, 3) · 2³ · 3³.

在通项中令 k = 3，得到 T₄ = C(6, 3) · 2⁶⁻³ · (3x)³。因此 x³ 的系数为 C(6, 3) · 2³ · 3³。

Now compute: C(6, 3) = 6! / (3!·3!) = 720 / (6·6) = 20. Also 2³ = 8 and 3³ = 27. Therefore the coefficient is 20 × 8 × 27 = 4320.

现在计算：C(6, 3) = 6! / (3!·3!) = 720 / (6·6) = 20。又 2³ = 8，3³ = 27。所以系数为 20 × 8 × 27 = 4320。

Coefficient of x³ = 20 × 8 × 27 = 4320

x³ 的系数 = 20 × 8 × 27 = 4320

5. Full Expansion (First Few Terms) | 展开式（前几项）

Although only the coefficient of x³ was requested, writing out the first few terms helps to verify the result and supports the later approximation.

虽然题目只要求 x³ 的系数，但写出前几项有助于验证结果，并为后面的近似计算作准备。

The expansion of (2 + 3x)⁶ begins as follows:

(2 + 3x)⁶ 的展开式开头如下：

(2 + 3x)⁶ = 64 + 576x + 2160x² + 4320x³ + 4860x⁴ + 2916x⁵ + 729x⁶

Each term is obtained by applying the binomial theorem successively. For instance, T₁ = 2⁶ = 64, T₂ = 6·2⁵·3x = 576x, T₃ = 15·2⁴·9x² = 2160x², and so on.

每一项都是通过依次应用二项定理得到的。例如，T₁ = 2⁶ = 64，T₂ = 6·2⁵·3x = 576x，T₃ = 15·2⁴·9x² = 2160x²，依此类推。

Notice that the coefficient 4320 matches our previous calculation exactly.

注意系数 4320 与我们之前的计算完全一致。

6. Approximation Using x = 0.01 | 用 x = 0.01 近似

We want to evaluate (2.03)⁶. Observe that 2.03 = 2 + 3(0.01), so we can set x = 0.01 in the expansion of (2 + 3x)⁶.

我们要计算 (2.03)⁶。注意到 2.03 = 2 + 3(0.01)，因此我们可以在 (2 + 3x)⁶ 的展开式中令 x = 0.01。

Substituting x = 0.01 into the expansion gives (2.03)⁶ ≈ 64 + 576(0.01) + 2160(0.01)² + 4320(0.01)³ + 4860(0.01)⁴ + 2916(0.01)⁵ + 729(0.01)⁶.

将 x = 0.01 代入展开式，得到 (2.03)⁶ ≈ 64 + 576(0.01) + 2160(0.01)² + 4320(0.01)³ + 4860(0.01)⁴ + 2916(0.01)⁵ + 729(0.01)⁶。

We can now compute each term. Notice that higher powers of 0.01 become very small, but since we require four significant figures, we must include the first few terms carefully.

我们现在可以计算每一项。注意 0.01 的高次幂变得非常小，但由于需要四位有效数字，我们必须仔细处理前几项。

7. Numerical Evaluation | 数值计算

Let us compute each term step by step:

让我们逐步计算每一项：

64
576 × 0.01 = 5.76
2160 × 0.01² = 2160 × 0.0001 = 0.216
4320 × 0.01³ = 4320 × 0.000001 = 0.00432
4860 × 0.01⁴ = 4860 × 0.00000001 = 0.0000486
2916 × 0.01⁵ = 2916 × 10⁻¹⁰ = 0.0000002916
729 × 0.01⁶ = 729 × 10⁻¹² = 0.000000000729

Adding these terms together:

将这些项相加：

64 + 5.76 + 0.216 + 0.00432 + 0.0000486 + 0.0000002916 + 0.000000000729 = 69.980368892329

Now round to four significant figures. The first four significant figures are 6, 9, 9, 8. The next digit is 0, so we do not round up. Thus (2.03)⁶ ≈ 69.98.

现在四舍五入到四位有效数字。前四个有效数字是 6、9、9、8。下一位数字是 0，所以不需要进位。因此 (2.03)⁶ ≈ 69.98。

(2.03)⁶ ≈ 69.98 (4 s.f.)

(2.03)⁶ ≈ 69.98（4位有效数字）

8. Common Mistakes | 常见错误

Students often forget to apply the coefficient C(6, k) when extracting a term. For example, the x³ term in (2 + 3x)⁶ is not simply 2³(3x)³; it must be multiplied by 20.

学生在提取项时经常忘记应用系数 C(6, k)。例如，(2 + 3x)⁶ 中 x³ 项不仅仅是 2³(3x)³，还必须乘以 20。

Another common error is misapplying the power to both 3 and x. Since (3x)³ = 27x³, the factor 27 must be included in the coefficient.

另一个常见错误是没有将幂同时应用到 3 和 x。因为 (3x)³ = 27x³，所以系数中必须包含因子 27。

When approximating, students may stop too early and truncate the series. For four significant figures, the term 0.00432 is still relevant because it affects the first four significant digits. However, the 0.0000486 term is too small to change the rounded result, but it is safe to include it.

在近似计算中，学生可能过早停止并截断级数。对于四位有效数字，0.00432 这个项仍然相关，因为它影响了前四个有效数字。然而，0.0000486 这一项太小，不会改变四舍五入的结果，但保留它是稳妥的做法。

9. Practice Question | 练习题

Try this similar problem by yourself: Find the coefficient of x⁴ in the expansion of (1 − 2x)⁷, and hence evaluate (0.98)⁷ correct to four decimal places.

请你自己尝试一个类似的问题：求 (1 − 2x)⁷ 展开式中 x⁴ 的系数，并由此将 (0.98)⁷ 精确到四位小数。

Hint: Set x = 0.01 to get 0.98 = 1 − 2(0.01). The coefficient of x⁴ is C(7, 4) · 1³ · (−2)⁴ = 35 × 16 = 560.

提示：令 x = 0.01 可得 0.98 = 1 − 2(0.01)。x⁴ 的系数为 C(7, 4) · 1³ · (−2)⁴ = 35 × 16 = 560。

You should find that (0.98)⁷ ≈ 0.8681 when rounded to four decimal places. The full sum is 1 − 0.14 + 0.0084 − 0.00028 + 0.0000056 − … = 0.8681256.

你应该会发现 (0.98)⁷ ≈ 0.8681（四舍五入到四位小数）。完整求和为 1 − 0.14 + 0.0084 − 0.00028 + 0.0000056 − … = 0.8681256。

10. Summary | 总结

In this example, we used the binomial theorem to find the coefficient of x³ in (2 + 3x)⁶, obtaining 4320. We then substituted x = 0.01 to approximate (2.03)⁶ and rounded the result to four significant figures, obtaining 69.98.

在本例中，我们利用二项定理求出了 (2 + 3x)⁶ 中 x³ 的系数，得到 4320。然后我们代入 x = 0.01 来近似 (2.03)⁶，并将结果四舍五入到四位有效数字，得到 69.98。

Key skills tested here include identifying the correct term, computing binomial coefficients accurately, and performing a sensible substitution for numerical approximation. Mastery of these techniques is essential for A-Level mathematics.

这里考查的关键技能包括：确定正确的项、准确计算二项系数，以及进行合理的代换以完成数值近似。熟练掌握这些技巧对于 A-Level 数学至关重要。

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Example 5.2.2: Binomial Expansion | 例题5.2.2：二项展开