標題:

F.4 A.Math題

發問:

Given ( 1+ 2x - 3x^2)^n = 1+ ax + bx^2 + terms involving higher powers of x , where n is a positive integer . a. Express a and b in terms of n. b.If b = 63 , find the value of n.

最佳解答:

Given (1+2x-3x^2)^n = 1+ ax +bx^2 + terms involving higher powers of x , where n is a positive integer. (a) Express a and b in terms of n. (1+2x-3x^2)^n =1+nC1(2x-3x^2)+nC2(2x-3x^2)^2+... =1+n(2x-3x^2)+[n(n-1)/2](4x^2+...)+... =1+2nx-3nx^2+2n(n-1)x^2+... =1+2nx+[2n^2-2n-3n]x^2+... =1+2nx+(2n^2-5n)x^2+... so a=2n,b=2n^2-5n (b) if b= 63 , find the value of n. 2n^2-5n=63 2n^2-5n-63=0 (n-7)(2n+9)=0 n=7 or n= -9 (rejected) so n=7

其他解答:

( 1+ 2x - 3x^2)^n =[ 1+ (2x - 3x^2)]^n =1 + n(2x - 3x^2) + [n(n-1)/2]*[2x - 3x^2]^2 + ... =1+ 2nx - 3nx^2 + [n(n-1)/2] *[4x^2 - 12x^3 + 9x^4] +... =1+ 2nx - 3nx^2 + [n(n-1)/2] *(4x^2) +... =1+ 2nx - 3nx^2 + 2n(n-1)x^2 + ... =1+ 2nx + (2n^2 - 2n -3n)x^2 + ... =1+ 2nx + (2n^2 - 5n)x^2 + ... a. a=2n b= 2n^2 - 5n b. 63= 2n^2 - 5n (2n+9)(n-7) =0 n=7 or n = -9/2(rejected) ANS: n =7

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