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secondary 3 | A Maths
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General formula for Binomial Theorem :
(a + b)ⁿ = aⁿ + nC1 aⁿ-¹ b + nC2 aⁿ-² b² + ... + bⁿ
So we can apply this to (1 + x - ax²)ⁿ by letting the a = 1 and b = (x - ax²)
(1 + x - ax²)ⁿ = 1ⁿ + nC1 1¹-¹ (x - ax²) + nC2 1ⁿ-² (x - ax²)² + ... + (x - ax²)ⁿ
We only need the terms that have powers of x and x² in them so it will be sufficient to expand the first three terms.
For nCr we can use the formula n!/(n-r)!r! or the other one in the formula sheet. Also recall that nC1 = n
1 to the power of any real number will always be equal to 1.
So,
(1 + x - ax²)ⁿ = 1ⁿ + nC1 1¹-¹ (x - ax²) + nC2 1ⁿ-² (x - ax²)² + ... + (x - ax²)ⁿ
= 1 + n(x - ax²) + n!/(n-2)!2! (x² - 2x(ax²) + (ax²)² ) + ...
= 1 + nx - anx² + n(n-1)/2 (x² - 2ax³ + a²x⁴) + ...
= 1 + nx - anx² + ½n(n-1) x² + ...
= 1 + nx + n(½(n-1) - a) x² + ...
Comparing this with 1 + 5x - 5x²,
①nx = 5x → n = 5
②n(½(n-1) - a) = -5
Since n = 5,
5(½(5-1) - a) = -5
5(2 - a) = -5
2 - a = -1
a = 2 + 1 = 3
So n = 5, a = 3 and the expression is actually (1 + x - 3x²)^5
(a + b)ⁿ = aⁿ + nC1 aⁿ-¹ b + nC2 aⁿ-² b² + ... + bⁿ
So we can apply this to (1 + x - ax²)ⁿ by letting the a = 1 and b = (x - ax²)
(1 + x - ax²)ⁿ = 1ⁿ + nC1 1¹-¹ (x - ax²) + nC2 1ⁿ-² (x - ax²)² + ... + (x - ax²)ⁿ
We only need the terms that have powers of x and x² in them so it will be sufficient to expand the first three terms.
For nCr we can use the formula n!/(n-r)!r! or the other one in the formula sheet. Also recall that nC1 = n
1 to the power of any real number will always be equal to 1.
So,
(1 + x - ax²)ⁿ = 1ⁿ + nC1 1¹-¹ (x - ax²) + nC2 1ⁿ-² (x - ax²)² + ... + (x - ax²)ⁿ
= 1 + n(x - ax²) + n!/(n-2)!2! (x² - 2x(ax²) + (ax²)² ) + ...
= 1 + nx - anx² + n(n-1)/2 (x² - 2ax³ + a²x⁴) + ...
= 1 + nx - anx² + ½n(n-1) x² + ...
= 1 + nx + n(½(n-1) - a) x² + ...
Comparing this with 1 + 5x - 5x²,
①nx = 5x → n = 5
②n(½(n-1) - a) = -5
Since n = 5,
5(½(5-1) - a) = -5
5(2 - a) = -5
2 - a = -1
a = 2 + 1 = 3
So n = 5, a = 3 and the expression is actually (1 + x - 3x²)^5
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