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secondary 3 | A Maths
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Date Posted: 4 years ago
Views: 189
(1 - x)(1 + ax)^6
= (1 - x)(1 + 6ax + 6!/2!4! (ax)² + ...)
= (1 - x)(1 + 6ax + 15a²x² + ...)
= 1 + 6ax + 15a²x² - x - 6ax² +...
≈ 1 + (6a - 1)x + (15a² - 6a)x²
Compare coefficients with 1 + bx²
For coefficient of x,
6a - 1 = 0
6a = 1
a = 1/6
For coefficient of x²,
15a² - 6a = b
b = 15(1/6)² - 6(1/6)
= 5/12 - 1
= -7/12
= (1 - x)(1 + 6ax + 6!/2!4! (ax)² + ...)
= (1 - x)(1 + 6ax + 15a²x² + ...)
= 1 + 6ax + 15a²x² - x - 6ax² +...
≈ 1 + (6a - 1)x + (15a² - 6a)x²
Compare coefficients with 1 + bx²
For coefficient of x,
6a - 1 = 0
6a = 1
a = 1/6
For coefficient of x²,
15a² - 6a = b
b = 15(1/6)² - 6(1/6)
= 5/12 - 1
= -7/12
Edited the answer. Please check
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Date Posted:
4 years ago
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