📚 C - The binomial theorem | 二项式定理
The binomial theorem provides a powerful way to expand expressions of the form (a + b)n without carrying out lengthy multiplication. It is a cornerstone of algebra, linking combinatorics, Pascal's triangle and polynomial expansions. In IB Mathematics, you are expected to understand the theorem, use summation notation, find particular terms and coefficients, and connect the expansion to combinatorial reasoning.
二项式定理提供了一个强有力的方法，可以在无需进行冗长乘法的情况下展开形如 (a + b)n 的表达式。它是代数中的一块基石，将组合学、帕斯卡三角形和多项式展开联系起来。在 IB 数学中，你需要理解该定理，会使用求和符号，能求出特定项与系数，并将展开式与组合推理建立联系。
1. Starting from the familiar | 从熟悉的情形出发
You already know simple expansions: (a + b)2 = a2 + 2ab + b2 and (a + b)3 = a3 + 3a2b + 3ab2 + b3. These are just the first steps. The coefficients 1, 2, 1 and 1, 3, 3, 1 look familiar – they appear in Pascal's triangle. The binomial theorem generalises this pattern for any positive integer exponent n.
你已经熟悉简单的展开：(a + b)2 = a2 + 2ab + b2 以及 (a + b)3 = a3 + 3a2b + 3ab2 + b3。这只是最初几步。系数 1, 2, 1 和 1, 3, 3, 1 看起来似曾相识——它们出现在帕斯卡三角形中。二项式定理将这一规律推广到任意正整数指数 n。
When we expand (a + b)n by repeatedly multiplying the binomial by itself, each term is formed by choosing either a or b from each factor. The coefficient of an−r br is the number of ways to choose b in r of the n factors, which is exactly the binomial coefficient nCr.
当我们通过反复自乘将 (a + b)n 展开时，每一项都是通过从每个因式中选取 a 或 b 而形成的。an−r br 的系数就是从 n 个因式中选出 r 个 b 的选法数，这恰好就是二项式系数 nCr。
2. The binomial coefficient nCr | 二项式系数 nCr
The notation nCr (also written as C(n, r) or nCr) represents the number of combinations of n items taken r at a time. It is calculated as nCr = n! / [r!(n − r)! st be confident evaluating these coefficients and interpreting them in expansion problems.
记号 nCr（也写作 C(n, r) 或 nCr）表示从 n 个物品中一次取出 r 个的组合数。它的计算公式是 nCr = n! / [r!(n − r)!]。在 IB 课程中，你必须能够熟练地计算这些系数，并在展开问题中予以解释。
For small n, you can simply use Pascal's triangle. For larger n, using the factorial formula is quicker. Remember the property nCr = nCn−r, which explains the symmetry of coefficients in the expansion.
对于较小的 n，你可以直接使用帕斯卡三角形。对于较大的 n，使用阶乘公式会更快。记住这一性质：nCr = nCn−r，这解释了展开式中系数的对称性。
3. Pascal's triangle and its links | 帕斯卡三角形及其联系
Pascal's triangle is constructed by starting with 1 at the top and then each number is the sum of the two numbers directly above it. Row n of the triangle gives the coefficients of (a + b)n. For example, row 4 is 1, 4, 6, 4, 1, matching (a + b)4.
帕斯卡三角形从顶端的 1 开始构建，每一个数都是它正上方两个数之和。三角形的第 n 行给出 (a + b)n 的系数。例如，第 4 行是 1, 4, 6, 4, 1，与 (a + b)4 对应。
The triangle also visualises the identity nCr = n−1Cr−1 + n−1Cr. This is known as Pascal's rule and forms the basis of many proofs and problems in combinatorics. Understanding this relationship can help you quickly generate coefficients without a calculator.
三角形还直观地展示了恒等式 nCr = n−1Cr−1 + n−1Cr。这被称为帕斯卡法则，是组合学中众多证明和问题的基础。理解这一关系可以帮助你无需计算器就能快速生成系数。
4. Statement of the binomial theorem | 二项式定理的表述
The formal theorem states that for any positive integer n,
(a + b)n = nC0 an + nC1 an−1b + nC2 an−2b2 + … + nCr an−rbr + … + nCn bn
该定理的正式表述为：对任意正整数 n，
(a + b)n = nC0 an + nC1 an−1b + nC2 an−2b2 + … + nCr an−rbr + … + nCn bn
Using summation notation, this becomes:
(a + b)n = Σr=0n nCr an−r br
用求和符号表示，则变为：
(a + b)n = Σr=0n nCr an−r br
Here r is the index of summation; when r = 0 we get the first term an, and when r = n we get the last term bn. Make sure you are comfortable reading and writing this compact form, as IB questions frequently expect it.
这里 r 是求和指标；当 r = 0 时得到首项 an，当 r = n 时得到末项 bn。务必熟练读写这一紧凑形式，因为 IB 试题经常要求使用它。
5. The general term | 通项公式
The (r + 1)th term in the expansion of (a + b)n is given by Tr+1 = nCr an−r br is extremely useful when you are asked to find a specific coefficient or the middle term.
(a + b)n 展开式中的第 (r + 1) 项由 Tr+1 = nCr an−r br 给出。请特别注意编号：r 从 0 开始，因此首项对应 r = 0。当你被要求找出某个特定系数或中间项时，这个公式极为有用。
For an expansion like (2x − 3y)5, treat (2x) as 'a' and (−3y) as 'b'. Then the general term becomes 5Cr (2x)5−r (−3y)r. Simplify the powers and signs carefully to obtain the desired coefficient.
对于像 (2x − 3y)5 这样的展开式，将 (2x) 看作 'a'，将 (−3y) 看作 'b'。那么通项就变为 5Cr (2x)5−r (−3y)r。仔细化简幂次和符号，就能得到所需的系数。
6. Finding particular coefficients and constant terms | 求特定系数与常数项
One of the most common IB question types asks for the coefficient of a specific power of x in an expansion. Set up the general term, collect powers of x, and then solve for r. Once r is found, substitute back to get the coefficient. Always check whether the question wants the numerical coefficient or the term including the variable.
IB 常见的题型之一，是要求展开式中 x 的某次幂的系数。写出通项，集中 x 的幂次，然后求解 r。一旦求出 r，代回原式即可得到系数。一定要检查题目要求的是数值系数还是包含变量的项。
For example, in (x + 1/x)8, the general term is 8Cr x8−r (1/x)r = 8Cr x8−2r. Setting 8 − 2r = 0 gives r = 4, so the constant term is 8C4 s approach works equally well for fractional and negative powers of x inside the binomial.
例如，在 (x + 1/x)8 中，通项为 8Cr x8−r (1/x)r = 8Cr x8−2r。令 8 − 2r = 0 得到 r = 4，因此常数项为 8C4 = 70。该方法同样适用于二项式内含有 x 的分数次幂或负次幂的情形。
7. Extensions: (1 + x)n and linkages | 拓展：(1 + x)n 及其关联
A special case used frequently is (1 + x)n = Σr=0n nCr xr is xr. This is simply a = 1 and b = x in the original theorem. It highlights that the coefficient of xr is nCr. This form often appears in calculus, approximations and binomial distribution problems.
一个常用的特例是 (1 + x)n = Σr=0n nCr xr。这不过是在原定理中令 a = 1、b = x。它突显了 xr 的系数就是 nCr。该形式经常出现在微积分、逼近计算以及二项分布问题中。
Furthermore, substituting x = 1 gives Σr=0n nCr = 2n mbinatorial identities. Similarly, substituting x = −1 shows the alternating sum equals 0.
此外，代入 x = 1 可得 Σr=0n nCr = 2n。这一结果，即所有二项式系数之和，是一个有用的检验，也出现在组合恒等式中。类似地，代入 x = −1 则显示交错和等于 0。
8. Proof by mathematical induction | 数学归纳法证明
m can be proved using mathematical induction on n. The base case n = 1 is trivial: (a + b)1 = 1C0 a + 1C1 b = a + b. Assume the theorem holds for n = k, then multiply both sides of (a + b)k expansion by (a + b) to obtain the (k+1) expansion.
二项式定理可以通过对 n 进行数学归纳来证明。基准情形 n = 1 是平凡的：(a + b)1 = 1C0 a + 1C1 b = a + b。假设定理对 n = k 成立，然后将 (a + b)k 的展开式两边同乘 (a + b) 来得到 (k+1) 的展开式。
During the multiplication, the term ak+1 comes from ak × a, and the term bk+1 from bk × b. For the intermediate term ak+1−rbr, the coefficient becomes kCr + kCr−1, which by Pascal's rule equals k+1Cr. Thus the statement is true for all positive integers n.
在乘法过程中，ak+1 项来自 ak × a，bk+1 项来自 bk × b。对于中间的项 ak+1−rbr，其系数变为 kCr + kCr−1，根据帕斯卡法则它等于 k+1Cr。因此该命题对所有正整数 n 都成立。
Understanding this proof deepens your appreciation of the structural link between Pascal's triangle and the theorem, and it is a standard induction question in IB Higher Level papers.
理解该证明能够加深你对帕斯卡三角形与定理之间结构联系的认识，这也是 IB 高水平试卷中标准的归纳法题目。
9. Common pitfalls and tips | 常见陷阱与技巧提示
Many students forget that the expansion has (n + 1) terms, not n terms. Another frequent mistake is misidentifying the variable part: when finding the coefficient, you must be precise about whether the question asks for the coefficient in the expansion of (a + bxp)n or just the constant multiplier.
许多学生忘记展开式有 (n + 1) 项，而不是 n 项。另一个常见错误是错误识别变量部分：在求系数时，你必须精确区分题目问的是 (a + bxp)n 展开式中的系数，还是仅仅指常数因子。
Always write the general term explicitly before solving. If a term appears to have no x – i.e. it is a constant term – set the index of x to zero. Also watch out for negative signs inside the binomial; they affect the sign of the coefficient in alternating fashion.
解题前一定要先明确写出通项。如果某项看起来不含 x——即为常数项——则将 x 的指数设为零。还要留意二项式内部的负号；它们会以交替的方式影响系数的符号。
Use your GDC (calculator) to compute combinations quickly, but always show the substitution steps. IB examiners look for clear method marks, so even if you get the final number slightly wrong, a well-earned method mark can make a big difference.
使用你的图形计算器（GDC）快速计算组合数，但始终要展示代入步骤。IB 阅卷人看重清晰的方法分，因此即使你最终数字稍有差错，扎实的方法分仍会带来显著不同。
10. Applications beyond pure expansion | 超越纯展开的应用
The binomial theorem is not just an algebraic tool; it appears in probability (binomial distribution), in approximations such as (1 + x)n n expanding functions. Recognising these connections is essential for high achievement in IB Mathematics.
二项式定理不只是一个代数工具；它还出现在概率（二项分布）中，出现在诸如小量近似 (1 + x)n ≈ 1 + nx 中，以及在微积分中用于函数展开。认识这些联系对于在 IB 数学中取得高分至关重要。
In the binomial distribution B(n, p), the probability of exactly r successes is nCr pr (1−p)n−r, which is exactly a term in the expansion of [(1−p) + p]n. This shows how the total probability sums to 1.
在二项分布 B(n, p) 中，恰好 r 次成功的概率为 nCr pr (1−p)n−r，这正是 [(1−p) + p]n 展开式中的一项。这表明总概率之和为 1。
These versatile uses explain why the binomial theorem is emphasised throughout the IB syllabus. Practice repeatedly with mixed numbers, fractions and negative signs to build fluency.
这些多用途解释了为什么 IB 教学大纲始终强调二项式定理。反复练习包含混合数、分数和负号的情形，以培养熟练度。
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