航空百度

標題:

A.MATHS urgent~

發問:

Let a,b be the roots of x^2-x-1=0, where a>b. For any integern,let f(n)= 1/√5(a^n-b^n0. a) Find f(1) and f(2). b) Prove that f(n+2)=f(n)+f(n+1) c) Prove, by mathematical induction, that f(n0 is an integer for all poositive integer n.

• 關於番禺o既名勝,景點(老人家去的) (超急)
• 第一次上門補習應該準備d咩--
• A question about equilibrium

此文章來自奇摩知識+如有不便請留言告知

最佳解答:

(a) x^2-x-1=0 a+b = 1, ab = -1 a-b = [(a+b)^2 - 4ab]^0.5 = (1^2-4(-1))^0.5 = √5 f(1) = 1/√5(a^1-b^1). = 1//√5(√5) = 1 f(2) = 1/√5(a^2-b^2) = 1/√5(a-b)(a+b) = 1/√5(√5)(1) = 1 (b) f(n+2) = 1/√5[a^(n+2)-b^(n+2)] = 1/√5[(a+b-b)a^(n+1) -(a+b-a)b^(n+1)] = 1/√5[(1-b)a^(n+1)-(1-a)b^(n+1)] = 1/√5[(a-ab)a^n - (b - ab) b^n] = 1/√5[(a+1)a^n- (b+1)b^n] = 1/√5(a^n-b^n) + 1/√5[(a^(n+1)-b^(n+1)] = f(n) + f(n+1) (c) f(n) = 1/√5(a^n-b^n) For n = 1, f(1) = 1 (from part(a) result) Assume n = k and n = k-1 are true f(k) = 1/√5(a^k-b^k) is an integer f(k-1) = 1/√5[a^(k-1)-b^(k-1)] is an integer when n = k+1 f(k+1) = 1 /√5[a^(k+1)-b^(k+1)] = f(k) + f(k-1) (from part (b) result) = integer. By M.I., f(n) is an integer is true for all positive integer n.

其他解答:

你有D野打錯左..