F.4 A.Math題

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

( 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