Binomial Theorem
AdvancedThe binomial theorem expands a binomial power into a polynomial whose coefficients are combination numbers, and the general term formula locks onto a chosen term without expanding everything — the core tool of the extended modules.
- (a + b)ⁿ expands to n + 1 terms, the (r + 1)-th being C(n,r)a^(n−r)b^r
- The coefficients form Pascal's triangle, each being the combination number C(n,r)
- Three steps for a chosen term: write the general term, solve for r from the index condition, substitute into the combination number
文體小知識：數學概念說明文——由逐項相乘的困境引入二項式定理，講解帕斯卡三角形、組合數係數、通項公式與符號處理，是延伸單元二的重要代數工具課。
§Text Passage
1Powers of binomials appear everywhere in algebra: the difference of squares and the cubic formulas all come from them. For powers 2 or 3, multiplying out is still manageable; but a power of 10 or more gives a dozen terms, and step-by-step multiplication becomes slow and error-prone. Mathematicians therefore found the general rule — the binomial theorem: the n-th power of a binomial can be written at once as a sum of n + 1 terms with neatly ordered coefficients, with no repeated multiplication at all.
2The coefficients form Pascal's triangle: each row begins and ends with 1, and every interior number is the sum of the two above it. For example, the row for n = 5 is 1, 5, 10, 10, 5, 1 — exactly the coefficients of the fifth-power expansion; the row for n = 4 is 1, 4, 6, 4, 1. These coefficients are combination numbers: the number of ways to choose r objects from n equals the coefficient of the (r + 1)-th term of the expansion.
3Precisely stated, the binomial theorem says that the n-th power of a binomial equals a sum in which each term is the product of three factors — a combination number, a falling power of the first letter and a rising power of the second letter; the two indices always add up to n. For example, a cubic expansion has four terms with coefficients 1, 3, 3, 1, one index falling from 3 to 0 and the other rising from 0 to 3.
4The general term formula is the key tool: to find a particular term of an expansion, do not write out the whole thing — write the (r + 1)-th term in general form, then use the condition on the index to solve for r. For example, to find the coefficient of the square term in a fifth-power expansion, just take the term with r = 2, substitute the combination number, the coefficient and the indices, and one line of working gives the answer — no need to expand all six terms.
5This topic links closely with counting and bridges both extended modules: in Module 1 the binomial distribution formula mirrors the binomial theorem, while in Module 2 the theorem itself can be proved by mathematical induction. Typical exam tasks ask for a specific coefficient, a constant term, or whether a certain term exists — the entry point is always the general term formula, so practise it until fluent.
1Summary 重點摘要
The binomial theorem expands a binomial power into a polynomial whose coefficients are combination numbers, and the general term formula locks onto a chosen term without expanding everything — the core tool of the extended modules.
2Vocabulary cards Vocabulary
A polynomial with two terms, such as x + 2; the expansion of its powers is the subject of this lesson.
The theorem expanding a binomial to a positive integer power into a sum of n + 1 terms whose coefficients are combination numbers.
The result of writing a power as a sum of terms; the expansion of a binomial to the n-th power has n + 1 terms.
The triangle of expansion coefficients: each row starts and ends with 1, each interior number being the sum of the two above.
The coefficient of each term of an expansion, equal to the corresponding combination number C(n,r).
The number of ways to choose r from n distinct objects without regard to order, written C(n,r).
The general form of the (r + 1)-th term: C(n,r) times a to the (n − r)-th power times b to the r-th power.
The numerical factor in front of the letters of a term; in expansions it is decided by the combination number and the numerical coefficients together.
The term containing no letter (index zero); to find it, set the letter's index in the general term to zero.
The notation using the Greek letter Σ for summing a list of numbers; the binomial theorem can be written compactly with it.
3Step by step Steps
- Step 1: From multiplying out to needing a theoremStart with low powers: the square of a binomial has 3 terms and the cube has 4, both obtainable by the distributive law. For example, (x + 2)² = x² + 4x + 4; multiplying by (x + 2) once more gives (x + 2)³ = x³ + 6x² + 12x + 8. But at the tenth power the expansion has eleven terms, and multiplying out term by term is slow and invites missing terms. The binomial theorem writes all terms at once with one formula, and can even compute just the one term a question asks for — which is exactly why it must be mastered.
- Step 2: Pascal's triangle and the coefficient patternStacking the coefficients of successive powers gives Pascal's triangle: each row starts and ends with 1, and each interior number is the sum of the two above it. For example, the row for n = 4 is 1, 4, 6, 4, 1 — precisely the coefficients of (1 + x)⁴; the row for n = 5 is 1, 5, 10, 10, 5, 1. Note that rows are counted from 0, and row n has n + 1 numbers. The triangle suits small powers — quick and visual; for large powers use combination numbers directly to avoid miscounting rows.
- Step 3: The coefficients are combination numbersThe number at position r (counting from 0) of row n of Pascal's triangle equals the combination number C(n,r), the number of ways to choose r objects from n distinct ones. For example, C(5,2) = 10, matching the middle 10 of row five, while C(5,0) = C(5,5) = 1 match the ends. This is no coincidence: when expanding a binomial power, each term arises from choosing which r of the n factors contribute b and which n − r contribute a, and the number of such choices is exactly the combination number C(n,r).
- Step 4: Stating the binomial theoremThe binomial theorem: for a positive integer n, (a + b)ⁿ equals a sum of terms, each being C(n,r) times a^(n−r) times bʳ, with r running from 0 to n — hence n + 1 terms. For example, expanding (2 + x)³: the terms are C(3,0) × 2³ (= 8), C(3,1) × 2²x (= 12x), C(3,2) × 2x² (= 6x²) and C(3,3) × x³ (= x³), giving 8 + 12x + 6x² + x³. Write terms in increasing r, with one index rising and the other falling — never out of order.
- Step 5: Locking onto a term with the general termThe (r + 1)-th term of the expansion is the general term, T(r + 1) = C(n,r)a^(n−r)b^r. To find a required term, write the general term first, then solve for r from the index condition. For example, for the term independent of x (the constant term) in (x² + 1/x)⁶: the general term carries x to the power (12 − 3r); setting the index to zero gives r = 4, and the term is C(6,4) = 15. Only two lines of working, with no full expansion. The same technique finds any coefficient or tests whether a term exists — a guaranteed feature of extended papers.
- Step 6: Expanding (1 + 2x)⁵ and the x² coefficientTake the classic example: expand (1 + 2x)⁵. Each term is C(5,r) times (2x) to the r-th power: r = 0 gives 1; r = 1 gives C(5,1) × 2x = 10x; r = 2 gives C(5,2) × (2x)² = 10 × 4x² = 40x²; continuing yields 80x³, 80x⁴ and 32x⁵. The coefficient of x² is therefore 40. Take special care: (2x)² is 4x², not 2x² — the numerical coefficient must be squared too, and this is the most common place to lose marks. Check with the ends: the first and last terms must be 1 and 32, i.e. the fifth power of the number 2.
- Step 7: Handling the minus signWhen the binomial contains a minus sign, such as (2x − 3) to a power, read it as (2x + (−3)); then b = −3 and bʳ takes its sign from the parity of r. For example, expanding (x − 2)⁴, the signs run positive, negative, positive, negative, positive; the coefficient of x² is C(4,2) × (−2)² = 6 × 4 = 24. Rule of thumb: with a minus sign, odd powers of b are negative and even powers positive; the two ends of the expansion are always positive — a quick check. Handle sign and size separately: compute the magnitude first, then attach the sign.
5Worked example Example
Example: (a) expand (1 + 2x)⁵ and state the coefficient of x²; (b) use the general term formula to find the constant term in the expansion of (x² + 1/x)⁶.
- (a) The terms are C(5,r)(2x)ʳ for r = 0 to 5, with coefficients 1, 5, 10, 10, 5, 1
- Computing term by term: (1 + 2x)⁵ = 1 + 10x + 40x² + 80x³ + 80x⁴ + 32x⁵
- Hence the coefficient of x² is 40 (from C(5,2) × 2² = 10 × 4)
- (b) Write the general term: T(r + 1) = C(6,r)(x²)^(6−r)(1/x)ʳ = C(6,r)x^(12−3r)
- Set the index 12 − 3r = 0, giving r = 4, and the constant term = C(6,4) = 15
- Answer: (a) (1 + 2x)⁵ = 1 + 10x + 40x² + 80x³ + 80x⁴ + 32x⁵, and the coefficient of x² is 40. (b) The constant term is 15. The key is to write the (r + 1)-th general term first, then solve for r from the index condition; when a numerical coefficient is present, square (or raise) it together with the letter, e.g. (2x)² = 4x².
6Common mistakes Mistakes
7Quick check Quick check
Q1In the binomial theorem, how many terms does the expansion of (a + b)ⁿ have, and what determines each coefficient?
Q2Find the coefficient of x³ in the expansion of (1 + x)⁶.
Q3Expand (2 + x)³.
Q4Find the coefficient of x² in the expansion of (x + 2)⁵.
Q5Why can the general term formula find a chosen term without expanding everything?
8Recap Recap
- (a + b)ⁿ expands to n + 1 terms, the (r + 1)-th being C(n,r)a^(n−r)b^r
- The coefficients form Pascal's triangle, each being the combination number C(n,r)
- Three steps for a chosen term: write the general term, solve for r from the index condition, substitute into the combination number
- Numerical coefficients rise with the index: (2x)² = 4x²; minus signs alternate by the parity of the power
- The general term formula is the core tool for finding specific coefficients and constant terms without full expansion