A.Maths(urgent)－vbd391t的部落格

標題:

A.Maths(urgent)

發問:

1)Let a and b be the roots of x^2-x+f=0,where f くor equal 1/4 Let Sn=a^n+b^n,where n is a positive integer. a)Express S3 and S4 in terms of f b)Shows that a^5-a^4+fa^3= Hence,show that S5-S4+fS3=0 and expressS5 in terms of f c)If S5=11,find the value of f d)Hence,evaluate ((1+√5)/2)^10+((1-√5)/2)^10

最佳解答:

a) a+b = 1, ab = f S3 = a^3+b^3 = (a+b)(a^2 - ab + b^2) = (a+b) [ (a+b)^2 - 3ab ] = 1 - 3f S4 = a^4+b^4 = (a^2+b^2)^2 - 2 a^2 b^2 = [ (a+b)^2 - 2ab ]^2 - 2(ab)^2 = (1 - 2f)^2 - 2f^2 = 2f^2 - 4f + 1 b) a^5-a^4+fa^3 = (a^2-a+f) (a^3) = (0) (a^3) = 0 S5-S4+fS3 = a^5+b^5 - a^4 - b^4 + f a^3 + f b^3 = a^5-a^4+fa^3 + b^5-b^4+f b^3 = 0 + 0 = 0 (同理, b^5-b^4+f b^3 = 0) S5 = S4 - fS3 = 2f^2 - 4f + 1 - f + 3f^2 = 5f^2 - 5f + 1 c) 5f^2 - 5f + 1 = 11 5f^2 - 5f - 10 = 0 f^2 - f - 2 = 0 (f+1)(f-2) = 0 f = -1 or f = 2 (rejected) f = -1 d) a and b be the roots of x^2-x-1=0 (1+√5)/2, (1-√5)/2 are roots of x^2-x-1=0 ((1+√5)/2)^10+((1-√5)/2)^10 = a^10 + b^10 = (a^5 + b^5)^2 - 2 a^5 b^5 = (S5)^2 - 2(ab)^5 = 11^2 - 2(-1)^5 = 123 咁深又咁長, 唔似 A.Maths, 似 pure maths 多 d

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