標題:
發問:
1. given (1+3x)^4(1-2x)^5=1+ax+bx^2+ higher power of x , find the values of the constants a and b.2. IN the expansion of (1+3x)^2 (1+x)^n , where n is a positive integer , the coefficient of x is 10.(A) Find the calue of n(B) Find the coefficient of x^23(A) Expand(1-2x)^3... 顯示更多 1. given (1+3x)^4(1-2x)^5=1+ax+bx^2+ higher power of x , find the values of the constants a and b. 2. IN the expansion of (1+3x)^2 (1+x)^n , where n is a positive integer , the coefficient of x is 10. (A) Find the calue of n (B) Find the coefficient of x^2 3(A) Expand(1-2x)^3 and ( 1+1/x)^5 (B) Find, in the expansive of (1-2x)^3 (1+1/x)^5 , 1-6x+12x^2-8x^3 , 1+5/x+10/x^2+10/x^3+5/x^4+1/x^5 4.Given (x^2+1/x)^5-(x^2-1/X)^5=ax^7+bx+c/x^5 , find th calues of a, b and c. HENCE, evaluate (2+1/√2)^5-(2-1/√2)^5 5.Expand(1+x)^n (1-2x)^4 in ascending powers of x up to the term x^2, where n is a positive integer. IF the coefficient of x^2 is 54 , find the coefficient of x.
最佳解答:
1. (1 + 3x)^4(1 - 2x)^5 = [1 + 4(3x) + 6(3x)^2 + ...][1 + 5(-2x) + 10(-2x)^2 + ...] = (1 + 12x + 54x^2 + ...)(1 - 10x + 40x^2 + ...) = 1 + 12x + 54x^2 + ... - 10x - 120x^2 - ... + 40 x^2 + ... = 1 + 2x - 26x^2 + ... Hence a = 2, b = -26 2. (A) (1 + 3x)^2(1 + x)^n = (1 + 6x + 9x^2)(1 + nx + ...) = 1 + 6x + 9x^2 + nx + ... The coefficient of x term is 6+n = 10 => n = 4 (B) (1 + 3x)^2(1 + x)^4 = (1 + 6x + 9x^2)(1 + 4x + 6x^2 + ...) = 1 + 6x + 9x^2 + 4x + 24x^2 + ... + 6x^2 + ... = 1 + 10x + 39x^2 + ... The coefficient of x^2 is 39 3(A) (1 - 2x)^3 = 1 + 3(-2x) + 3(-2x)^2 + (-2x)^3 = 1 - 6x + 12x^2 - 8x^3 (1 + 1/x)^5 = 1 + 5(1/x) + 10(1/x)^2 + 10(1/x)^3 + 5(1/x)^4 + (1/x)^5 = 1 + 5/x + 10/x^2 + 10/x^3 + 5/x^4 + 1/x^5 (B) Find, in the expansive of (1-2x)^3 (1+1/x)^5 , 1-6x+12x^2-8x^3 , 1+5/x+10/x^2+10/x^3+5/x^4+1/x^5 ??? 4.(x^2 + 1/x)^5 - (x^2 - 1/x)^5 = ax^7 + bx + c/x^5 (x^2 + 1/x)^5 = (x^2)^5 + 5(x^2)^4(1/x) + 10(x^2)^3(1/x)^2 + 10(x^2)^2(1/x)^3 + 5(x^2)(1/x)^4 + (1/x)^5 = x^10 + 5x^7 + 10x^5 + 10x + 5/x^2 + 1/x^5 (x^2 - 1/x)^5 = x^10 - 5x^7 + 10x^5 - 10x + 5/x^2 - 1/x^5 (x^2 + 1/x)^5 - (x^2 - 1/x)^5 = 10x^7 + 20x + 2/x^5 So a = 10, b = 20, c = 2 Put x = √2 (2 + 1/√2)^5 - (2 - 1/√2)^5 = 10(√2)^7 + 20(√2) + 2/(√2)^5 = 80√2 + 20√2 + √2 /4 = (100.25)√2 = 141.77 5. (1 + x)^n(1 - 2x)^4 = [1 + nx + n(n-1)/2 x^2 + ...][1 + 4(-2x) + 6(-2x)^2 + ...] = [1 + nx + n(n-1)/2 x^2 + ...](1 - 8x + 24x^2 + ...) = 1 + nx + n(n-1)/2 x^2 + ... - 8x - 8nx^2 + ... + 24x^2 + ... = 1 + (n-8)x + [n(n-1)/2 - 8n + 24]x^2 + ... = 1 + (n-8)x + (n^2 -17n +48)/2 x^2 + ... Since the coefficient of x^2 is 54 (n^2 - 17n + 48)/2 = 54 n^2 - 17n + 48 = 108 n^2 - 17n - 60 = 0 (n - 20)(n + 3) = 0 n = 20 or n = -3 (rejected) Coefficient of x is 20 - 8 = 12
其他解答:
這裏可以幫到你 http://actionte.subo.cn35.com/yahoo.com.hk/hk/auction/178987536|||||(B) Find, in the expansive of (1-2x)^3 (1+1/x)^5 , 1-6x+12x^2-8x^3 , 1+5/x+10/x^2+10/x^3+5/x^4+1/x^5 ??? 應該要expand曬佢= =我見到頭痛= =xD:P
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F.4 BINOMIAL THEOREM question發問:
1. given (1+3x)^4(1-2x)^5=1+ax+bx^2+ higher power of x , find the values of the constants a and b.2. IN the expansion of (1+3x)^2 (1+x)^n , where n is a positive integer , the coefficient of x is 10.(A) Find the calue of n(B) Find the coefficient of x^23(A) Expand(1-2x)^3... 顯示更多 1. given (1+3x)^4(1-2x)^5=1+ax+bx^2+ higher power of x , find the values of the constants a and b. 2. IN the expansion of (1+3x)^2 (1+x)^n , where n is a positive integer , the coefficient of x is 10. (A) Find the calue of n (B) Find the coefficient of x^2 3(A) Expand(1-2x)^3 and ( 1+1/x)^5 (B) Find, in the expansive of (1-2x)^3 (1+1/x)^5 , 1-6x+12x^2-8x^3 , 1+5/x+10/x^2+10/x^3+5/x^4+1/x^5 4.Given (x^2+1/x)^5-(x^2-1/X)^5=ax^7+bx+c/x^5 , find th calues of a, b and c. HENCE, evaluate (2+1/√2)^5-(2-1/√2)^5 5.Expand(1+x)^n (1-2x)^4 in ascending powers of x up to the term x^2, where n is a positive integer. IF the coefficient of x^2 is 54 , find the coefficient of x.
最佳解答:
1. (1 + 3x)^4(1 - 2x)^5 = [1 + 4(3x) + 6(3x)^2 + ...][1 + 5(-2x) + 10(-2x)^2 + ...] = (1 + 12x + 54x^2 + ...)(1 - 10x + 40x^2 + ...) = 1 + 12x + 54x^2 + ... - 10x - 120x^2 - ... + 40 x^2 + ... = 1 + 2x - 26x^2 + ... Hence a = 2, b = -26 2. (A) (1 + 3x)^2(1 + x)^n = (1 + 6x + 9x^2)(1 + nx + ...) = 1 + 6x + 9x^2 + nx + ... The coefficient of x term is 6+n = 10 => n = 4 (B) (1 + 3x)^2(1 + x)^4 = (1 + 6x + 9x^2)(1 + 4x + 6x^2 + ...) = 1 + 6x + 9x^2 + 4x + 24x^2 + ... + 6x^2 + ... = 1 + 10x + 39x^2 + ... The coefficient of x^2 is 39 3(A) (1 - 2x)^3 = 1 + 3(-2x) + 3(-2x)^2 + (-2x)^3 = 1 - 6x + 12x^2 - 8x^3 (1 + 1/x)^5 = 1 + 5(1/x) + 10(1/x)^2 + 10(1/x)^3 + 5(1/x)^4 + (1/x)^5 = 1 + 5/x + 10/x^2 + 10/x^3 + 5/x^4 + 1/x^5 (B) Find, in the expansive of (1-2x)^3 (1+1/x)^5 , 1-6x+12x^2-8x^3 , 1+5/x+10/x^2+10/x^3+5/x^4+1/x^5 ??? 4.(x^2 + 1/x)^5 - (x^2 - 1/x)^5 = ax^7 + bx + c/x^5 (x^2 + 1/x)^5 = (x^2)^5 + 5(x^2)^4(1/x) + 10(x^2)^3(1/x)^2 + 10(x^2)^2(1/x)^3 + 5(x^2)(1/x)^4 + (1/x)^5 = x^10 + 5x^7 + 10x^5 + 10x + 5/x^2 + 1/x^5 (x^2 - 1/x)^5 = x^10 - 5x^7 + 10x^5 - 10x + 5/x^2 - 1/x^5 (x^2 + 1/x)^5 - (x^2 - 1/x)^5 = 10x^7 + 20x + 2/x^5 So a = 10, b = 20, c = 2 Put x = √2 (2 + 1/√2)^5 - (2 - 1/√2)^5 = 10(√2)^7 + 20(√2) + 2/(√2)^5 = 80√2 + 20√2 + √2 /4 = (100.25)√2 = 141.77 5. (1 + x)^n(1 - 2x)^4 = [1 + nx + n(n-1)/2 x^2 + ...][1 + 4(-2x) + 6(-2x)^2 + ...] = [1 + nx + n(n-1)/2 x^2 + ...](1 - 8x + 24x^2 + ...) = 1 + nx + n(n-1)/2 x^2 + ... - 8x - 8nx^2 + ... + 24x^2 + ... = 1 + (n-8)x + [n(n-1)/2 - 8n + 24]x^2 + ... = 1 + (n-8)x + (n^2 -17n +48)/2 x^2 + ... Since the coefficient of x^2 is 54 (n^2 - 17n + 48)/2 = 54 n^2 - 17n + 48 = 108 n^2 - 17n - 60 = 0 (n - 20)(n + 3) = 0 n = 20 or n = -3 (rejected) Coefficient of x is 20 - 8 = 12
其他解答:
這裏可以幫到你 http://actionte.subo.cn35.com/yahoo.com.hk/hk/auction/178987536|||||(B) Find, in the expansive of (1-2x)^3 (1+1/x)^5 , 1-6x+12x^2-8x^3 , 1+5/x+10/x^2+10/x^3+5/x^4+1/x^5 ??? 應該要expand曬佢= =我見到頭痛= =xD:P
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