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1Exercise 8.1
Expand (2x + 3y)⁴ using the binomial theorem.
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(a+b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴
With a = 2x, b = 3y:
= (2x)⁴ + 4(2x)³(3y) + 6(2x)²(3y)² + 4(2x)(3y)³ + (3y)⁴
= 16x⁴ + 4(8x³)(3y) + 6(4x²)(9y²) + 4(2x)(27y³) + 81y⁴
= 16x⁴ + 96x³y + 216x²y² + 216xy³ + 81y⁴
2Exercise 8.1
Find the coefficient of x⁵ in the expansion of (1 + 2x)⁷.
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General term: T(r+1) = 7Cr (1)7-r (2x)r = 7Cr 2r xr
For x⁵: r = 5
Coefficient = 7C5 × 2⁵ = 21 × 32 = 672
3Exercise 8.1
Find the term independent of x in the expansion of (2x² - 1/x)⁹.
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General term: T(r+1) = 9Cr (2x²)9-r (-1/x)r
= 9Cr 29-r (-1)r x18-2r · x-r
= 9Cr 29-r (-1)r x18-3r
Independent of x: 18 - 3r = 0 ⇒ r = 6
T(7) = 9C6 2³ (-1)⁶ = 84 × 8 × 1 = 672
4Exercise 8.1
Using binomial theorem, evaluate (0.99)⁵ correct to four decimal places.
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(0.99)⁵ = (1 - 0.01)⁵
= 1 - 5(0.01) + 10(0.01)² - 10(0.01)³ + 5(0.01)⁴ - (0.01)⁵
= 1 - 0.05 + 0.001 - 0.00001 + 0.00000005 - ...
= 0.95099
≈ 0.9510
5Exercise 8.1
Prove that nC0 + nC1 + nC2 + ... + nCn = 2ⁿ.
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Consider (1 + x)ⁿ = nC0 + nC1 x + nC2 x² + ... + nCn xⁿ.
Put x = 1:
(1 + 1)ⁿ = nC0 + nC1 + nC2 + ... + nCn
2ⁿ = nC0 + nC1 + nC2 + ... + nCn. Hence proved.