Expanding (a + b)ⁿ
Multiplying (a + b) by itself again and again soon becomes long. The binomial theorem gives the answer straight away.
For a positive integer n, (a + b)ⁿ = ⁿC₀aⁿ + ⁿC₁aⁿ⁻¹b + ⁿC₂aⁿ⁻²b² + … + ⁿCₙbⁿ. It has n + 1 terms; the powers of a go down and the powers of b go up, and the sum of the powers is n in every term. The coefficients form Pascal's triangle.
Formula: (a + b)ⁿ = Σ ⁿCᵣ aⁿ⁻ʳ bʳ
In the figure: Each number is the sum of the two above it. Row n gives the coefficients of (a + b)ⁿ.
Rule 1: Expand — Take coefficients from Pascal's triangle or ⁿCᵣ.
- Easy: (x + y)⁴ = x⁴ + 4x³y + 6x²y² + 4xy³ + y⁴.
- Medium: (x + 2)³ = x³ + 6x² + 12x + 8.
- Exam-type: (2x − 1)⁴ = 16x⁴ − 32x³ + 24x² − 8x + 1.
Rule 2: Number and sign of terms — Alternate signs when b is negative.
- Easy: (a − b)ⁿ has terms with alternating signs: +, −, +, − …
- Medium: (x + 1)⁷ has 8 terms.
- Exam-type: The sum of the coefficients of (1 + x)ⁿ is 2ⁿ (put x = 1).
Replace b by −b to expand (a − b)ⁿ.
Forgetting that (a + b)ⁿ has n + 1 terms, not n.
Pascal: each number is the sum of the two above.
Practice questions
- 1Expand (x + 1)⁴.
- 2Expand (a − b)³.
- 3How many terms are in the expansion of (x + 3)⁹?
- 4Expand (x + 2)⁴.
- 5Expand (1 − 2x)³.
- 6Find the sum of the coefficients in (1 + x)⁶.
- 7Expand (√2 + 1)⁴ + (√2 − 1)⁴.
- 8Evaluate (2 + √3)³ + (2 − √3)³.
Show answers (try them yourself first)
- 1x⁴ + 4x³ + 6x² + 4x + 1
- 2a³ − 3a²b + 3ab² − b³
- 310
- 4x⁴ + 8x³ + 24x² + 32x + 16
- 51 − 6x + 12x² − 8x³
- 664
- 7The odd-power terms cancel and the even ones double: 2(4 + 12 + 1) = 34
- 8Odd terms cancel: 2(8 + 3·2·3) = 52
General term
To find the 4th term of a huge expansion you do not want to expand everything. A single formula gives any term.
The (r + 1)th term of (a + b)ⁿ is Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ. Note r starts from 0: the first term has r = 0.
Formula: Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ
Rule 1: Find a term — Put r = (term number − 1).
- Easy: 4th term of (x + 2)⁶: r = 3, T₄ = ⁶C₃ x³ 2³ = 160x³.
- Medium: 5th term of (x + 1)⁸: r = 4, T₅ = ⁸C₄ x⁴ = 70x⁴.
- Exam-type: 3rd term of (2x − 1)⁵: r = 2, T₃ = ⁵C₂ (2x)³ (−1)² = 80x³.
Rule 2: Find the term containing x^k — Collect the power of x in Tᵣ₊₁ and solve for r.
- Easy: Coefficient of x³ in (1 + x)⁸: ⁸C₃ = 56.
- Medium: Coefficient of x² in (2 + x)⁵: r = 2, ⁵C₂ 2³ = 80.
- Exam-type: Term independent of x in (x + 1/x)⁶: x⁶⁻ʳ x⁻ʳ = x⁶⁻²ʳ, so r = 3, term = ⁶C₃ = 20.
The r + 1 in the subscript is why the 4th term uses r = 3.
Using r = 4 for the 4th term.
Term number minus 1 = r.
Practice questions
- 1Find the 3rd term of (x + y)⁷.
- 2Find the 5th term of (1 + x)⁹.
- 3Find the coefficient of x⁴ in (1 + x)¹⁰.
- 4Find the 4th term of (2x + 1)⁶.
- 5Find the coefficient of x³ in (3 + x)⁶.
- 6Find the term independent of x in (x² + 1/x)⁶.
- 7Find the coefficient of x² in (x − 1/x)⁸.
- 8The 3rd term of (1 + x)ⁿ has coefficient 28. Find n.
Show answers (try them yourself first)
- 1⁷C₂ x⁵y² = 21x⁵y²
- 2⁹C₄ x⁴ = 126x⁴
- 3210
- 4⁶C₃ (2x)³ = 160x³
- 5⁶C₃ 3³ = 540
- 6x¹²⁻²ʳ⁻ʳ = x¹²⁻³ʳ, r = 4: ⁶C₄ = 15
- 7x⁸⁻²ʳ = x², r = 3: ⁸C₃ (−1)³ = -56
- 8ⁿC₂ = 28 gives n(n − 1) = 56, so n = 8
Middle term
In a row of beads, the middle bead is easy to point to: it has equal beads on both sides.
If n is even, (a + b)ⁿ has one middle term, the (n/2 + 1)th. If n is odd, it has two middle terms, the ((n + 1)/2)th and ((n + 3)/2)th.
Rule 1: n even — One middle term T(n/2 + 1).
- Easy: (x + 1)⁶: middle term is T₄ = ⁶C₃ x³ = 20x³.
- Medium: (a + b)⁸: T₅ = ⁸C₄ a⁴b⁴ = 70a⁴b⁴.
- Exam-type: (2x + 1)⁴: T₃ = ⁴C₂ (2x)² = 24x².
Rule 2: n odd — Two middle terms.
- Easy: (x + y)⁵: T₃ and T₄ = 10x³y² and 10x²y³.
- Medium: (1 + x)⁷: T₄ and T₅ = 35x³ and 35x⁴.
- Exam-type: (a + b)⁹: T₅ and T₆.
The largest coefficient in (1 + x)ⁿ belongs to the middle term(s).
Giving only one middle term for odd n.
Even → one, Odd → two.
Practice questions
- 1How many middle terms does (x + y)¹⁰ have?
- 2How many middle terms does (x + y)¹¹ have?
- 3Find the middle term of (1 + x)⁴.
- 4Find the middle term of (x + 2)⁶.
- 5Find the middle terms of (1 + 2x)⁵.
- 6Find the middle term of (a − b)⁶.
- 7Find the middle term of (x + 1/x)⁸.
- 8In (1 + x)ⁿ the middle term is the 4th. Find n.
Show answers (try them yourself first)
- 1One, the 6th term.
- 2Two, the 6th and 7th terms.
- 3T₃ = 6x²
- 4T₄ = ⁶C₃ x³ 2³ = 160x³
- 5T₃ = ⁵C₂ (2x)² = 40x² and T₄ = ⁵C₃ (2x)³ = 80x³
- 6T₄ = ⁶C₃ a³ (−b)³ = −20a³b³
- 7T₅ = ⁸C₄ x⁴ x⁻⁴ = 70
- 8n/2 + 1 = 4 gives n = 6
Applications
Without a calculator, how big is 1.02⁵? The binomial theorem gives a very accurate estimate.
Use the binomial theorem to find approximate values, remainders and sums of coefficients, by writing a number as (1 + small).
Rule 1: Approximate values — Write the number as 1 + small and keep the first few terms.
- Easy: 1.02⁵ = (1 + 0.02)⁵ ≈ 1 + 5(0.02) + 10(0.02)² = 1.1040 (exact 1.10408).
- Medium: 0.99⁴ ≈ 1 − 4(0.01) + 6(0.0001) = 0.9606 (exact 0.960596).
- Exam-type: 101³ = (100 + 1)³ = 1000000 + 30000 + 300 + 1 = 1030301.
Rule 2: Divisibility — Write the number as (multiple + 1) or (multiple − 1).
- Easy: 11⁹ − 1 = (10 + 1)⁹ − 1 is divisible by 10.
- Medium: 9ⁿ − 1 = (8 + 1)ⁿ − 1 is divisible by 8.
- Exam-type: 6ⁿ − 5n has remainder 1 when divided by 25.
Sum of coefficients of (a + bx)ⁿ: put x = 1.
Using too few terms when the accuracy asked is high.
Small number → keep few terms.
Practice questions
- 1Find 99³ using (100 − 1)³.
- 2Estimate 1.01⁶ up to 4 decimal places using 3 terms.
- 3Find 102⁴ using (100 + 2)⁴.
- 4Find the sum of the coefficients in (2x + 3)⁴.
- 5Show that 7ⁿ − 1 is divisible by 6.
- 6Find the remainder when 8⁵ is divided by 7.
- 7Find (1.1)⁴ exactly using the theorem.
- 8Find the sum of coefficients of x² and x³ in (1 + x)⁷.
Show answers (try them yourself first)
- 1970299
- 21 + 6(0.01) + 15(0.0001) = 1.0615
- 3108243216
- 4625
- 57ⁿ = (6 + 1)ⁿ = 6·(…) + 1, so 7ⁿ − 1 is a multiple of 6
- 68⁵ = (7 + 1)⁵ leaves remainder 1; check: 32768 mod 7 = 1
- 71 + 0.4 + 0.06 + 0.004 + 0.0001 = 1.4641
- 8⁷C₂ + ⁷C₃ = 21 + 35 = 56