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Amaths:)
發問:
In the expansion of (x-3/x)^2．(1+2x)^n ,the constant term is 210.Find the value of n.Suppose that q is a positive integer.a)Ecpand (1+x+px^2)^q in ascending powers of x up to the term x^3.b)If the coefficients of x and x^2 are 6 and 27 respectively,find the values of p and q.Hence find the coefficient of x^3... 顯示更多 In the expansion of (x-3/x)^2．(1+2x)^n ,the constant term is 210.Find the value of n. Suppose that q is a positive integer. a)Ecpand (1+x+px^2)^q in ascending powers of x up to the term x^3. b)If the coefficients of x and x^2 are 6 and 27 respectively,find the values of p and q.Hence find the coefficient of x^3 in the above expansion
最佳解答:
First question: (x - 3xˉ1)2 ( 1+2x)^n = (x2 - 6 + 9xˉ2)[1+ n2x + (nC2)(2x) 2 +...] the constant term: 210 = (-6)(1) + (9xˉ2)[(nC2)(2x) 2] 210 = - 6 + 36[n(n-1)] / 2 0 = 18n2–18n – 216 n = -3 (rejected) or 4 Second question: a)(1+x+px2)^q = 1 + q (x+px2) + (qC2)(x+px2)2 + (q3)(x+px2)^3 + ... = 1 + qx + pqx2 + (qC2)(x2+px^3 + ...) + (qC3)(x^3+...) +... = 1+ qx+[pq+(qC2)] x2 + [(qC2)p+(qC3)] x^3 +... b)coefficients of x = q =6 coefficients of x2 = 27 = [pq+(qC2)] 27 = p(6) + (6C2) 27 = 6p + 15 p = 2 q=6, p=2 From [(qC2)p+(qC3)] x^3, coefficients of x^3 = (qC2)p+(qC3) = (6C2)(2) + (6C3) = 30 + 20 = 50
Amaths:)
發問:
In the expansion of (x-3/x)^2．(1+2x)^n ,the constant term is 210.Find the value of n.Suppose that q is a positive integer.a)Ecpand (1+x+px^2)^q in ascending powers of x up to the term x^3.b)If the coefficients of x and x^2 are 6 and 27 respectively,find the values of p and q.Hence find the coefficient of x^3... 顯示更多 In the expansion of (x-3/x)^2．(1+2x)^n ,the constant term is 210.Find the value of n. Suppose that q is a positive integer. a)Ecpand (1+x+px^2)^q in ascending powers of x up to the term x^3. b)If the coefficients of x and x^2 are 6 and 27 respectively,find the values of p and q.Hence find the coefficient of x^3 in the above expansion
最佳解答:
First question: (x - 3xˉ1)2 ( 1+2x)^n = (x2 - 6 + 9xˉ2)[1+ n2x + (nC2)(2x) 2 +...] the constant term: 210 = (-6)(1) + (9xˉ2)[(nC2)(2x) 2] 210 = - 6 + 36[n(n-1)] / 2 0 = 18n2–18n – 216 n = -3 (rejected) or 4 Second question: a)(1+x+px2)^q = 1 + q (x+px2) + (qC2)(x+px2)2 + (q3)(x+px2)^3 + ... = 1 + qx + pqx2 + (qC2)(x2+px^3 + ...) + (qC3)(x^3+...) +... = 1+ qx+[pq+(qC2)] x2 + [(qC2)p+(qC3)] x^3 +... b)coefficients of x = q =6 coefficients of x2 = 27 = [pq+(qC2)] 27 = p(6) + (6C2) 27 = 6p + 15 p = 2 q=6, p=2 From [(qC2)p+(qC3)] x^3, coefficients of x^3 = (qC2)p+(qC3) = (6C2)(2) + (6C3) = 30 + 20 = 50
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