📚 Exercise 6C.3: Mastering Binomial Expansion Coefficients | 练习6C.3：掌握二项式展开系数
Exercise 6C.3 in the IB Mathematics course focuses on a core algebraic skill: applying the binomial theorem to find specific coefficients and constant terms in polynomial expansions. These problems test your ability to manipulate the general term formula, manage powers accurately, and think strategically about the index of the variable. Whether you are aiming for a high score on Paper 1 or Paper 2, mastering these techniques will save you time and prevent algebraic errors.
IB 数学课程中的练习 6C.3 聚焦于一项核心代数技能：利用二项式定理求出多项式展开中的特定系数和常数项。此类问题旨在考察你是否能熟练运用通项公式、准确处理指数，并围绕变量的指数进行策略性思考。无论你的目标是卷面高分还是强化代数基本功，掌握这些技巧都能帮你节省时间并避免计算失误。
1. Understanding the Binomial Theorem | 理解二项式定理
The binomial theorem states that for any positive integer n, (a + b)ⁿ = Σ [C(n, r) aⁿ⁻ʳ bʳ] from r = 0 to n. The symbol C(n, r) or "n choose r" (a + b)ⁿ, the powers of a decrease from n to 0 while the powers of b increase from 0 to n.
二项式定理指出，对任意正整数 n，有 (a + b)ⁿ = Σ [C(n, r) aⁿ⁻ʳ bʳ]（r 从 0 到 n）。记号 C(n, r)（或 “n 选 r”）表示二项式系数，即从 n 个元素中不计顺序地选取 r 个的方法数。在展开 (a + b)ⁿ 时，a 的指数从 n 递减至 0，而 b 的指数则从 0 递增至 n。
C(5,1)x⁴·2¹ + C(5,2)x³·2² + C(5,3)x²·2³ + C(5,4)x¹·2⁴ + C(5,5)x⁰·2⁵. This expansion yields a polynomial where each term is a product of a binomial coefficient, a power of x, and a power of 2.
例如， + 2)⁵ = C(5,0)x⁵·2⁰ + C(5,1)x⁴·2¹ + C(5,2)x³·2² + C(5,3)x²·2³ + C(5,4)x¹·2⁴ + C(5,5)x⁰·2⁵。该展开式得到一个多项式，其中每一项均由二项式系数、x 的幂以及 2 的幂相乘得到。
2. The General Term Tᵣ₊₁ | 通项 Tᵣ₊₁
or solving Exercise 6C.3 is the general term, often denoted Tᵣ₊₁. For (a + b)ⁿ, the (r+1)ᵗʰ term is given by Tᵣ₊₁ = C(n, r) · aⁿ⁻ʳ · bʳ f the variable x and set up an equation to find the required coefficient or constant term.
解决练习 6C.3 最得力的工具是通项，常记作 Tᵣ₊₁。对 (a + b)ⁿ，第 (r+1) 项为 Tᵣ₊₁ = C(n, r) · aⁿ⁻ʳ · bʳ，其中 r 依次取 0,1,2,...,n。通过专注于某一项，你能够分离出变量 x 的指数，并由此建立方程去求出所需的系数或常数项。
For example, in (2x – 3)⁶, we set a = 2x, b = –3, and n = 6. Then Tᵣ₊₁ = C(6, r) · (2x)⁶⁻ʳ · (–3)ʳ otice that both the numerical coefficient and the sign of b must be raised to the power r.
例如，在 (2x – 3)⁶ 中，令 a = 2x，b = –3，n = 6。于是 Tᵣ₊₁ = C(6, r) · (2x)⁶⁻ʳ · (–3)ʳ。请留意，数值系数以及 b 的符号都要做 r 次乘方。
3. Finding a Specified Coefficient | 求指定项系数
A typical question asks: Find the coefficient of x³ in the expansion of (2x – 3)⁶. First, write the general term: Tᵣ₊₁ = C(6, r) · (2x)⁶⁻ʳ · (–3)ʳ. The power of x is 6 – r. Set this equal to 3 → 6 – r = 3 → r = 3.
一道典型题会问：求 (2x – 3)⁶ 的展开式中 x³ 的系数。首先写出通项：Tᵣ₊₁ = C(6, r) · (2x)⁶⁻ʳ · (–3)ʳ。x 的指数为 6 – r，令其等于 3，得 6 – r = 3，即 r = 3。
Then substitute r = 3: T₄ = C(6, 3) · (2x)³ · (–3)³ = 20 · 8x³ · (–27). The coefficient is 20 × 8 × (–27) = –4320. Always remember to include the sign arising from negative b.
然后代入 r = 3：T₄ = C(6, 3) · (2x)³ · (–3)³ = 20 · 8x³ · (–27)。系数为 20 × 8 × (–27) = –4320。永远记得带上由负的 b 所产生的符号。
4. Constant Term Problems | 常数项问题
1/x)⁸. Here a = x², b = 1/x = x⁻¹, and n = 8. The general term is Tᵣ₊₁ = C(8, r) · (x²)⁸⁻ʳ · (x⁻¹)ʳ = C(8, r) · x¹⁶⁻²ʳ · x⁻ʳ = C(8, r) · x¹⁶⁻³ʳ.
要求常数项，需令 x 的指数为零。考虑 (x² + 1/x)⁸，这里 a = x²，b = 1/x = x⁻¹，n = 8。通项为 Tᵣ₊₁ = C(8, r) · (x²)⁸⁻ʳ · (x⁻¹)ʳ = C(8, r) · x¹⁶⁻²ʳ · x⁻ʳ = C(8, r) · x¹⁶⁻³ʳ。
Set the exponent 16 – er, there is no constant term in this expansion. This teaches an important lesson: always check whether the index r is a non-negative integer; if not, the requested term does not exist.
令指数 16 – 3r = 0，解得 r = 16/3。由于 r 必须为整数，故该展开式中不存在常数项。这一点教给我们一个重要教训：务必检验指数 r 是否为非负整数；若不是，则所求的项就不存在。
5. Solving for Unknown Powers | 求解未知指数
in the expansion of (1 + 2x)ⁿ is 40, find n. Write the general term: Tᵣ₊₁ = C(n, r) · 1ⁿ⁻ʳ · (2x)ʳ e term containing x² corresponds to r = 2. So the coefficient is C(n, 2) · 2² = 4 C(n, 2).
IB 试题中常问：已知 (1 + 2x)ⁿ 的展开式中 x² 的系数为 40，求 n。写出通项：Tᵣ₊₁ = C(n, r) · 1ⁿ⁻ʳ · (2x)ʳ。含 x² 的项对应 r = 2，系数为 C(n, 2) · 2² = 4 C(n, 2)。
Set 4 C(n, 2) = 40 → C(n, 2) = 10. Since C(n, 2) = n(n – 1)/2, we have n(n – 1)/2 = 10 → n² – n – 20 = 0 → (n – 5)(n + 4) = 0. n must be positive, so n = 5.
令 4 C(n, 2) = 40，得 C(n, 2) = 10。因为 C(n, 2) = n(n – 1)/2，于是 n(n – 1)/2 = 10 → n² – n – 20 = 0 → (n – 5)(n + 4) = 0。n 必须为正整数，故 n = 5。
6. When the Expression Has Coefficients | 当表达式含有系数时
2)⁴, do not simply take the coefficient as 3. The general term is Tᵣ₊₁ = C(4, r) · (3x)⁴⁻ʳ · 2ʳ. The coefficient of x⁴⁻ʳ comes from 3⁴⁻ʳ, and the constant part from 2ʳ. Both must be multiplied by the binomial coefficient.
一个常见错误是忘记把括号内的系数进行相应次方运算。对于 (3x + 2)⁴，不要简单地把系数当作 3。通项为 Tᵣ₊₁ = C(4, r) · (3x)⁴⁻ʳ · 2ʳ。x⁴⁻ʳ 的系数来自 3⁴⁻ʳ，常数部分则来自 2ʳ，两者均须乘以二项式系数。
For instance, to find the coefficient of x², set 4 – = 2. Then the term is C(4, 2) · (3x)² · 2² = 6 · 9x² · 4 = 216 x². The coefficient is 216.
例如，要求 x² 的系数，令 4 – r = 2，得 r = 2。该项为 C(4, 2) · (3x)² · 2² = 6 · 9x² · 4 = 216 x²，系数为 216。
7. Combining Binomial Expansions | 组合二项式展开
Some Exercise 6C.3 problems require you to find a coefficient in the product of two binomial expansions, such as (x + 1)²(2x – 1)³. Here, you can expand each factor completely if the powers are small, or you can select terms strategically. To find the coefficient of x³, you must consider which terms from the first expansion multiply with which terms from the second to yield an x³ overall.
部分练习 6C.3 的题目要求你求出两个二项式展开式乘积中的某个系数，例如 (x + 1)²(2x – 1)³。此时，若指数较小可将其各自完全展开，或策略性地挑选项的组合。为求 x³ 的系数，你需要考虑第一个展开式中的哪些项与第二个展开式中的哪些项相乘后可得到总的 x³ 次幂。
Expand (x+1)² = x² + 2x + 1. For (2x – 1)³, use the general term: C(3, r)(2x)³⁻ʳ(–1)ʳ , or the 2x term with the x² term, or the constant 1 with the x³ term. Sum the contributions carefully.
将 (x+1)² 展开得 x² + 2x + 1。对于 (2x – 1)³，使用通项：C(3, r)(2x)³⁻ʳ(–1)ʳ。第二个展开式中 x 的可能次幂为 x³, x², x¹, x⁰。为得到整体的 x³，可将第一个因式的 x² 项与第二个因式的 x¹ 项相乘，或将 2x 项与 x² 项相乘，或将常数 1 与 x³ 项相乘。仔细地将这些贡献相加即可。
8. Common Pitfalls and How to Avoid Them | 常见陷阱及避免方法
ying a and b. Always write the binomial in standard form (a + b)ⁿ. If the expression is (1 – 3x)⁵, take b = (–3x), not 3x. Another trap is forgetting that the general term’s power of a is n – xtract the power of x; (4) set the equation; (5) solve for r; (6) compute the coefficient.
一个常见错误是弄错 a 和 b。永远把二项式写成标准形式 (a + b)ⁿ。比如表达式为 (1 – 3x)⁵，应设 b = (–3x) 而非 3x。另一个陷阱是忘记通项中 a 的指数是 n – r 而不是 r。准备一份检查清单：(1) 确定 a, b, n；(2) 写出 Tᵣ₊₁；(3) 提炼 x 的指数；(4) 建立方程；(5) 解出 r；(6) 计算系数。
) as a simple fraction; always calculate binomial coefficients correctly using C(n, r) = n! / [r! (n – r)! ascal’s triangle. Double-check the final sign, especially when b is negative and r is odd.
另有一些学生有时会把 C(n, r) 当作简单的分数来处理；务必使用 C(n, r) = n! / [r! (n – r)!] 或帕斯卡三角形正确地计算二项式系数。最后还要反复检查符号，尤其当 b 为负且 r 为奇数时。
9. Worked Examples from Exercise 6C.3 | 练习 6C.3 中的典型例题
Example 1: Find the term independent of x in (x/2 – 2/x²)⁹. Set a = x/2, b = –2/x² = –2x⁻², n = 9. Then Tᵣ₊₁ = C(9, r) · (x/2)⁹⁻ʳ · (–2x⁻²)ʳ = C(9, r) · x⁹⁻ʳ / 2⁹⁻ʳ · (–2)ʳ · x⁻²ʳ. The power of x is 9 – r – 2r = 9 – 3r. Set 9 – 3r = 0 → r = 3. Substitute: T₄ = C(9,3) · (x/2)⁶ · (–2x⁻²)³ = 84 · x⁶ / 2⁶ · (–8) · x⁻⁶ = 84 · (x⁶/64) · (–8) · x⁻⁶ = 84 · (–8)/64 = –10.5. The constant term is –10.5.
例 1：求 (x/2 – 2/x²)⁹ 中与 x 无关的项。设 a = x/2，b = –2/x² = –2x⁻²，n = 9。则 Tᵣ₊₁ = C(9, r) · (x/2)⁹⁻ʳ · (–2x⁻²)ʳ = C(9, r) · x⁹⁻ʳ / 2⁹⁻ʳ · (–2)ʳ · x⁻²ʳ。x 的指数为 9 – r – 2r = 9 – 3r。令 9 – 3r = 0，得 r = 3。代入：T₄ = C(9,3) · (x/2)⁶ · (–2x⁻²)³ = 84 · x⁶ / 2⁶ · (–8) · x⁻⁶ = 84 · (x⁶/64) · (–8) · x⁻⁶ = 84 · (–8)/64 = –10.5。常数项为 –10.5。
is 8 and the coefficient of x² is 30. Find p and n. The general term: Tᵣ₊₁ = C(n, r) (px)ʳ. Coefficient of x: r=1 → C(n,1)p = np = 8. Coefficient of x²: r=2 → C(n,2)p² = [n(n–1)/2] p² = 30. Substitute p = 8/n: [n(n–1)/2] (64/n²) = 30 → (n–1)·32 / n = 30 → 32(n–1) = 30n → 32n – 32 = 30n → 2n = 32 → n = 16. Then p = 8/16 = 0.5. Thus p = 0.5, n = 16.
例 2：在 (1 + px)ⁿ 的展开式中，x 的系数为 8，x² 的系数为 30。求 p 和 n。通项：Tᵣ₊₁ = C(n, r) (px)ʳ。x 系数：r=1 → C(n,1)p = np = 8。x² 系数：r=2 → C(n,2)p² = [n(n–1)/2] p² = 30。代入 p = 8/n： [n(n–1)/2] (64/n²) = 30 → (n–1)·32 / n = 30 → 32(n–1) = 30n → 32n – 32 = 30n → 2n = 32 → n = 16。于是 p = 8/16 = 0.5。故 p = 0.5，n = 16。
10. Extension: Using the Binomial Theorem for Proofs | 扩展：用二项式定理进行证明
hen x = 1. By setting x = 1, the expansion gives 2ⁿ = sum of all coefficients. Setting x = – g these reveals properties. While not always required, such insights deepen understanding.
练习 6C.3 还可能引入简单恒等式，例如证明 (1 + x)ⁿ 在 x = 1 时偶次幂系数之和等于奇次幂系数之和。令 x = 1，展开式给出 2ⁿ 等于所有系数之和；令 x = –1 得到交替和为零。将两者相加、相减便揭示了系数的性质。虽然不常要求，但这些见解可加深理解。
Another useful extension is identifying the greatest coefficient in a binomial expansion. By comparing successive terms, you can find which term is maximal—a skill that appears in some challenge sets.
另一项有用的扩展是确定二项式展开中的最大系数。通过比较相邻项，可以找出哪一项系数最大——这一技巧在某些挑战题组中会出现。
11. Practice Strategy | 练习策略
k calculations by testing a small case with Pascal’s triangle if possible. Finally, attempt the mixed problems requiring combination of expansions to simulate exam conditions.
在完成练习 6C.3 的过程中，先从直接求系数的题目入手以建立信心，将通项作为公式化的锚点。然后逐步进阶到常数项问题和未知指数问题。每道题完成后，验证 r 是否为整数且满足 0 ≤ r ≤ n。如有可能，可用帕斯卡三角形测试一个小规模情形来核对计算。最后尝试那些需要组合展开的混合题，以模拟考试情境。
ial coefficient values. Frequent practice with negative b is essential, as sign errors are among the most costly.
此外，准备一张公式卡，上面记有通项公式、(1 ± x)ⁿ 的标准展开式以及常见二项式系数值。经常练习负值 b 的题型至关重要，毕竟符号错误是最“昂贵”的失分项之一。
12. Summary and Key Takeaways | 总结与关键要点
ise 6C.3 boils down to three skills: (1) correctly setting up the general term Tᵣ₊₁ = C(n,
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