yffuhxy的部落格

標題:

F.4 AMaths Binomial Theorem

發問:

(a) If n is a positive integer, expand (1+qx-3x^2)^n in assending powers of x up to the term x^4. (b) If the coefficients of x and x^2 are -12 and 42 respectively, (i) find q and n (ii) evaluate the coefficient of x^4 in the expansion. (b)(i) & (b)(ii) both are have two set answer. 更新: (b)(ii) is wrong.

• l z+i l + l z - i l = 3 -@1@
• 問what's going on 用法(有情境)
• what is X-man-

 

此文章來自奇摩知識+如有不便請留言告知

最佳解答:

(a) (1+qx-3x^2)^n =[1+x(q-3x)]^n = 1+nx(q-3x) + n(n-1)/2 x^2(q-3x)^2 + n(n-1)(n-2)/6 x^3(q-3x)^3 + n(n-1)(n-2)(n-3)/24 x^4(q-3x)^4 + ... = 1+nqx - 3nx^2 +n(n-1)/2 (q^2x^2 - 6qx^3 + 9x^4) + n(n-1)(n-2)/6 (q^3x^3 -9q^2x^4+...) + n(n-1)(n-2)(n-3)/24 q^4x^4 + ... = 1+nqx + [n(n-1)q^2/2 - 3n]x^2 + [n(n-1)(n-2)q^3/6 -3n(n-1)q]x^3 + [9n(n-1)/2 - 2n(n-1)(n-2)q^2/3 + n(n-1)(n-2)(n-3)q^4/24]x^4 + ... (b) (i) nq = -12 and n(n-1)q^2/2 - 3n = 42 q = -12/n and n(n-1)(144/n^2)/2 3n = 42 72-72/n - 3n = 42 n^2 - 10n + 24 = 0 (n-4)(n-6) = 0 n = 4 or 6 q = -3 or -2 When n = 4, q = -3 and when n = 6 q = -2 (ii) When n = 4, q = -3, coefficient of x^4 is 9(4)(3)/2 - 2(4)(3)(2)(9)/3+4(3)(2)(1)(81)/24 = -9 When n = 6, q = -2, coefficient of x^4 is 9(6)(5)/2-2(6)(5)(4)(4)/3 + (6)(5)(4)(3)(16)/24 = 55

其他解答: