Exercise 7.1
Expand (1 – 2x)⁵.
Answer
Using (a+b)ⁿ = Σ ⁿCᵣ aⁿ⁻ʳ bʳ with a=1, b=–2x, n=5:
= ⁵C₀(1)⁵ + ⁵C₁(1)⁴(–2x) + ⁵C₂(1)³(–2x)² + ⁵C₃(1)²(–2x)³ + ⁵C₄(1)(–2x)⁴ + ⁵C₅(–2x)⁵
= 1 + 5(–2x) + 10(4x²) + 10(–8x³) + 5(16x⁴) + (–32x⁵)
= 1 – 10x + 40x² – 80x³ + 80x⁴ – 32x⁵
Expand (x/2 – 2/x)⁵.
Answer
a = x/2, b = –2/x, n = 5:
T(r+1) = ⁵Cᵣ(x/2)⁵⁻ʳ(–2/x)ʳ = ⁵Cᵣ · x⁵⁻ʳ/2⁵⁻ʳ · (–1)ʳ · 2ʳ/xʳ = ⁵Cᵣ (–1)ʳ 2^(2r–5) x^(5–2r)
r=0: ⁵C₀ · x⁵/32 = x⁵/32
r=1: ⁵C₁(–1)·2⁻³·x³ = –5x³/8
r=2: ⁵C₂·2⁻¹·x = 10x/2 = 5x
r=3: ⁵C₃(–1)·2¹·x⁻¹ = –10·2/x = –20/x
r=4: ⁵C₄·2³·x⁻³ = 5·8/x³ = 40/x³
r=5: ⁵C₅(–1)·2⁵·x⁻⁵ = –32/x⁵
(x/2 – 2/x)⁵ = x⁵/32 – 5x³/8 + 5x – 20/x + 40/x³ – 32/x⁵
Expand (2x – 3)⁶.
Answer
a=2x, b=–3, n=6:
= ⁶C₀(2x)⁶ + ⁶C₁(2x)⁵(–3) + ⁶C₂(2x)⁴(–3)² + ⁶C₃(2x)³(–3)³ + ⁶C₄(2x)²(–3)⁴ + ⁶C₅(2x)(–3)⁵ + ⁶C₆(–3)⁶
= 64x⁶ + 6(32x⁵)(–3) + 15(16x⁴)(9) + 20(8x³)(–27) + 15(4x²)(81) + 6(2x)(–243) + 729
= 64x⁶ – 576x⁵ + 2160x⁴ – 4320x³ + 4860x² – 2916x + 729
Expand (x/3 + 1/x)⁵.
Answer
a = x/3, b = 1/x, n = 5:
= ⁵C₀(x/3)⁵ + ⁵C₁(x/3)⁴(1/x) + ⁵C₂(x/3)³(1/x²) + ⁵C₃(x/3)²(1/x³) + ⁵C₄(x/3)(1/x⁴) + ⁵C₅(1/x⁵)
= x⁵/243 + 5·x⁴/(81x) + 10·x³/(27x²) + 10·x²/(9x³) + 5·x/(3x⁴) + 1/x⁵
= x⁵/243 + 5x³/81 + 10x/27 + 10/(9x) + 5/(3x³) + 1/x⁵
Expand (x + 1/x)⁶.
Answer
a = x, b = 1/x, n = 6:
= x⁶ + 6x⁵·(1/x) + 15x⁴·(1/x²) + 20x³·(1/x³) + 15x²·(1/x⁴) + 6x·(1/x⁵) + 1/x⁶
= x⁶ + 6x⁴ + 15x² + 20 + 15/x² + 6/x⁴ + 1/x⁶
Evaluate (96)³ using Binomial Theorem.
Answer
96 = 100 – 4. (100–4)³ = 100³ – 3·100²·4 + 3·100·16 – 64
= 1000000 – 120000 + 4800 – 64 = 884736
Evaluate (102)⁵ using Binomial Theorem.
Answer
102 = 100 + 2. (100+2)⁵:
= 100⁵ + 5·100⁴·2 + 10·100³·4 + 10·100²·8 + 5·100·16 + 32
= 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32
= 11040808032
Evaluate (101)⁴ using Binomial Theorem.
Answer
101 = 100 + 1. (100+1)⁴ = 100⁴ + 4·100³ + 6·100² + 4·100 + 1
= 100000000 + 4000000 + 60000 + 400 + 1 = 104060401
Evaluate (99)⁵ using Binomial Theorem.
Answer
99 = 100–1. (100–1)⁵ = 100⁵ – 5·100⁴ + 10·100³ – 10·100² + 5·100 – 1
= 10000000000 – 500000000 + 10000000 – 100000 + 500 – 1 = 9509900499
Which is larger: (1.1)¹⁰⁰⁰⁰ or 1000?
Answer
(1.1)¹⁰⁰⁰⁰ = (1 + 0.1)¹⁰⁰⁰⁰
Using binomial expansion, the 4th term alone: ¹⁰⁰⁰⁰C₃(0.1)³ = (10000 × 9999 × 9998)/(6) × 0.001
≈ 1.67 × 10¹¹ × 0.001 = 1.67 × 10⁸, which is much greater than 1000.
∴ (1.1)¹⁰⁰⁰⁰ > 1000
Find (a+b)⁴ – (a–b)⁴. Hence evaluate (√3+√2)⁴ – (√3–√2)⁴.
Answer
(a+b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴
(a–b)⁴ = a⁴ – 4a³b + 6a²b² – 4ab³ + b⁴
Difference = 8a³b + 8ab³ = 8ab(a² + b²)
For a=√3, b=√2: 8·√3·√2·(3+2) = 8√6·5 = 40√6
Find (x+1)⁶ + (x–1)⁶. Hence evaluate (√2+1)⁶ + (√2–1)⁶.
Answer
(x+1)⁶ + (x–1)⁶ = 2[⁶C₀x⁶ + ⁶C₂x⁴ + ⁶C₄x² + ⁶C₆]
= 2[x⁶ + 15x⁴ + 15x² + 1]
For x = √2: 2[8 + 15·4 + 15·2 + 1] = 2[8 + 60 + 30 + 1] = 2 × 99 = 198
Show that 9ⁿ⁺¹ – 8n – 9 is divisible by 64 for all positive integers n.
Answer
9ⁿ⁺¹ = (1+8)ⁿ⁺¹ = ∑ ⁽ⁿ⁺¹⁾Cᵣ 8ʳ
= 1 + (n+1)·8 + ⁽ⁿ⁺¹⁾C₂·64 + higher terms with 64 as factor
= 1 + 8n + 8 + 64k = 8n + 9 + 64k
9ⁿ⁺¹ – 8n – 9 = 64k, divisible by 64 ∎
Prove that ∑ᵣ₌₀ⁿ ⁿCᵣ · 3ʳ = 4ⁿ.
Answer
∑ᵣ₌₀ⁿ ⁿCᵣ · 3ʳ = ∑ᵣ₌₀ⁿ ⁿCᵣ · 1ⁿ⁻ʳ · 3ʳ = (1+3)ⁿ = 4ⁿ ∎
Miscellaneous Exercise
If a and b are distinct integers, prove that a–b is a factor of aⁿ–bⁿ.
Answer
Write aⁿ = ((a–b)+b)ⁿ = ∑ⁿCᵣ(a–b)ʳ bⁿ⁻ʳ
= bⁿ + ∑ᵣ₌₁ⁿ ⁿCᵣ(a–b)ʳ bⁿ⁻ʳ
aⁿ – bⁿ = (a–b)[∑ᵣ₌₁ⁿ ⁿCᵣ(a–b)ʳ⁻¹ bⁿ⁻ʳ]
Thus (a–b) is a factor of aⁿ–bⁿ ∎
Evaluate (√3+√2)⁶ – (√3–√2)⁶.
Answer
(a+b)⁶ – (a–b)⁶ = 2[⁶C₁a⁵b + ⁶C₃a³b³ + ⁶C₅ab⁵]
= 2[6a⁵b + 20a³b³ + 6ab⁵] = 4ab[3a⁴ + 10a²b² + 3b⁴]
For a=√3, b=√2: 4·√6·[3·9 + 10·6 + 3·4] = 4√6·[27+60+12] = 4√6·99 = 396√6
Find value of (a² + √(a²–1))⁴ + (a² – √(a²–1))⁴.
Answer
Let x = a², y = √(a²–1). (x+y)⁴ + (x–y)⁴ = 2[x⁴ + 6x²y² + y⁴]
= 2[(a²)⁴ + 6(a²)²(a²–1) + (a²–1)²]
= 2[a⁸ + 6a⁶ – 6a⁴ + a⁴ – 2a² + 1]
= 2a⁸ + 12a⁶ – 10a⁴ – 4a² + 2
Find approximation of (0.99)⁵ using first three terms.
Answer
0.99 = 1 – 0.01
(1–0.01)⁵ ≈ 1 – 5(0.01) + 10(0.01)² = 1 – 0.05 + 0.001 = 0.951
Expand (1 + x/2 – 2/x)⁴, x ≠ 0.
Answer
Write as ((1–2/x) + x/2)⁴. Let a = 1–2/x, b = x/2.
= ⁴C₀a⁴ + ⁴C₁a³b + ⁴C₂a²b² + ⁴C₃ab³ + ⁴C₄b⁴
= (1–2/x)⁴ + 4(1–2/x)³(x/2) + 6(1–2/x)²(x/2)² + 4(1–2/x)(x/2)³ + (x/2)⁴
Expanding each term carefully:
= 1 – 8/x + 24/x² – 32/x³ + 16/x⁴ + 4(1–6/x+12/x²–8/x³)(x/2) + 6(1–4/x+4/x²)(x²/4) + 4(1–2/x)(x³/8) + x⁴/16
= 1 – 8/x + 24/x² – 32/x³ + 16/x⁴ + 2x – 12 + 48/x – 16/x² + (3x²/2 – 6x + 6) + (x³/2 – x²) + x⁴/16
= x⁴/16 + x³/2 + x²/2 – 4x – 11 + 40/x + 8/x² – 32/x³ + 16/x⁴
Find expansion of (3x² – 2ax + 3a²)³.
Answer
Write as ((3x²+3a²) – 2ax)³ or use direct expansion. Let A = 3x²–2ax, B = 3a²:
(A+B)³ = A³ + 3A²B + 3AB² + B³
A³ = (3x²–2ax)³ = 27x⁶ – 3·9x⁴·2ax + 3·3x²·4a²x² – 8a³x³ = 27x⁶ – 54ax⁵ + 36a²x⁴ – 8a³x³
3A²B = 3(9x⁴–12ax³+4a²x²)(3a²) = 9a²(9x⁴–12ax³+4a²x²) = 81a²x⁴–108a³x³+36a⁴x²
3AB² = 3(3x²–2ax)(9a⁴) = 27a⁴(3x²–2ax) = 81a⁴x²–54a⁵x
B³ = 27a⁶
Sum: 27x⁶ – 54ax⁵ + (36a²+81a²)x⁴ + (–8a³–108a³)x³ + (36a⁴+81a⁴)x² – 54a⁵x + 27a⁶
= 27x⁶ – 54ax⁵ + 117a²x⁴ – 116a³x³ + 117a⁴x² – 54a⁵x + 27a⁶
Important Extra Questions
Find the middle term in the expansion of (x/3 + 9y)¹⁰.
Answer
n=10 (even), middle term = T(n/2+1) = T₆.
T₆ = ¹⁰C₅(x/3)⁵(9y)⁵ = 252 · x⁵/243 · 59049y⁵ = 252 × 243 × x⁵y⁵ = 61236 x⁵y⁵
Find the general term in (x² – y)⁶ and the term independent of y.
Answer
T(r+1) = ⁶Cᵣ(x²)⁶⁻ʳ(–y)ʳ = ⁶Cᵣ(–1)ʳ x^(12–2r) yʳ
For term independent of x: 12–2r = 0
⇒ r = 6. T₇ = ⁶C₆(–1)⁶y⁶ = y⁶. (Not independent of y, just x)
Find the coefficient of x⁵ in (x+3)⁸.
Answer
T(r+1) = ⁸Cᵣ x^(8–r) · 3ʳ. For x⁵: 8–r=5
⇒ r=3.
Coefficient = ⁸C₃ · 3³ = 56 × 27 = 1512
Using binomial theorem, prove that 6ⁿ – 5n always leaves remainder 1 when divided by 25.
Answer
6ⁿ = (1+5)ⁿ = 1 + 5n + ⁿC₂·25 + higher terms of 25
6ⁿ – 5n = 1 + 25(ⁿC₂ + ⁿC₃·5 + ...) = 1 + 25k
So 6ⁿ – 5n = 25k + 1, leaving remainder 1 when divided by 25 ∎
Find the term independent of x in (√x – 1/√x)¹⁰.
Answer
T(r+1) = ¹⁰Cᵣ(√x)^(10–r)(–1/√x)ʳ = ¹⁰Cᵣ(–1)ʳ x^((10–r)/2 – r/2) = ¹⁰Cᵣ(–1)ʳ x^(5–r)
For term independent of x: 5–r=0 → r=5.
T₆ = ¹⁰C₅(–1)⁵ = –252