close
標題:
f4 maths~~please help me
發問:
1. If the coefficients of x^11 and x^12 are the same in the expansion of (2 + ax)^15 , where a not equal 0, find the value of a .2. If the coefficient of x^3 is triple that of x in the expansion of (1- x/2)^n , where n is a positive integer, find the value of n .3.If the coefficients of x^2 in the expansion... 顯示更多 1. If the coefficients of x^11 and x^12 are the same in the expansion of (2 + ax)^15 , where a not equal 0, find the value of a . 2. If the coefficient of x^3 is triple that of x in the expansion of (1- x/2)^n , where n is a positive integer, find the value of n . 3.If the coefficients of x^2 in the expansion of (3 + x)(2 - ax)^3 is 30, find the value of a . 4.Using the binomial theorem, find the remainder when 8^88 is divided by each of the following integers. a) 7 b) 49 5. Find the constant term in the expansion of (1 + 2/x)^3 (3 + x)^6 .
(1) In terms of a, coeff of x11 = 15C11 a11 = 1365 a11 Coeff of x12 = 15C12 a12 = 455 a12 Thus, 1365 a11 = 455 a12 a = 3 (2) In terms of n, coeff of x3 = - nC3 /8 Coeff of x = - n/2 Thus, nC3 /8 = 3n/2 n(n - 1)(n - 2)/6 = 12n (n - 1)(n - 2) = 72 n2 - 3n - 70 = 0 (n - 10)(n + 3) = 0 n = 10 or - 3 (rejected) (3) (3 + x)(2 - ax)3 = (3 + x)(8 - 12ax + 6a2x2 - ...) Coefficient of x2 = - 12a + 18a2 = 30 3a2 - 2a - 5 = 0 (3a - 5)(a + 1) = 0 a = 5/3 or -1 (4) 888 = (1 + 7)88 = (1 + 88 x 7 + higher powers of 7) So when it is divided by 7, remainder = 1 When it is divided by 49, remainder will be the same as when 1 + 88 x 7 is divided by 49, i.e. Remainder of 617/49 = 29 (5) (1 + 2/x)3 (3 + x)6 = (1 + 6/x + 12/x2 + 8/x3) (729 + 1458x + 1215x2 + 540x3 + ...) Constant term = 729 + 6 x 1458 + 12 x 1215 + 8 x 540 = 28377
其他解答:
f4 maths~~please help me
發問:
1. If the coefficients of x^11 and x^12 are the same in the expansion of (2 + ax)^15 , where a not equal 0, find the value of a .2. If the coefficient of x^3 is triple that of x in the expansion of (1- x/2)^n , where n is a positive integer, find the value of n .3.If the coefficients of x^2 in the expansion... 顯示更多 1. If the coefficients of x^11 and x^12 are the same in the expansion of (2 + ax)^15 , where a not equal 0, find the value of a . 2. If the coefficient of x^3 is triple that of x in the expansion of (1- x/2)^n , where n is a positive integer, find the value of n . 3.If the coefficients of x^2 in the expansion of (3 + x)(2 - ax)^3 is 30, find the value of a . 4.Using the binomial theorem, find the remainder when 8^88 is divided by each of the following integers. a) 7 b) 49 5. Find the constant term in the expansion of (1 + 2/x)^3 (3 + x)^6 .
此文章來自奇摩知識+如有不便請留言告知
最佳解答:(1) In terms of a, coeff of x11 = 15C11 a11 = 1365 a11 Coeff of x12 = 15C12 a12 = 455 a12 Thus, 1365 a11 = 455 a12 a = 3 (2) In terms of n, coeff of x3 = - nC3 /8 Coeff of x = - n/2 Thus, nC3 /8 = 3n/2 n(n - 1)(n - 2)/6 = 12n (n - 1)(n - 2) = 72 n2 - 3n - 70 = 0 (n - 10)(n + 3) = 0 n = 10 or - 3 (rejected) (3) (3 + x)(2 - ax)3 = (3 + x)(8 - 12ax + 6a2x2 - ...) Coefficient of x2 = - 12a + 18a2 = 30 3a2 - 2a - 5 = 0 (3a - 5)(a + 1) = 0 a = 5/3 or -1 (4) 888 = (1 + 7)88 = (1 + 88 x 7 + higher powers of 7) So when it is divided by 7, remainder = 1 When it is divided by 49, remainder will be the same as when 1 + 88 x 7 is divided by 49, i.e. Remainder of 617/49 = 29 (5) (1 + 2/x)3 (3 + x)6 = (1 + 6/x + 12/x2 + 8/x3) (729 + 1458x + 1215x2 + 540x3 + ...) Constant term = 729 + 6 x 1458 + 12 x 1215 + 8 x 540 = 28377
其他解答:
文章標籤
全站熱搜
留言列表
