a maths

(1+2x-3x^2)^n=1+ax+bx^2+terms involving higher power of x

A) express a and b in terms of n

(1+2x-3x^2)^n

=(1+(2x-3x^2)) ^n

=1+ nC1 (2x-3x^2) + nC2 (2x-3x^2)^2 + ... 去到lei 度夠, be cos juz 去到x^2

=1 + nC1 (2x-3x^2) + nC2 (4x^2 + ... ) + ... 大至小拆 去到x^2

=1 + n (2x-3x^2) + (0.5)(n(n-1))(4x^2 + ... ) + ...

=1 + 2nx - 3nx^2 + 2x^2 (n^2 -n) + ...

=1+ 2nx + (-3n+2n^2 -2n) x^2 +...

=1 +2nx + (2n^2 -5n) x^2 + ...

i.e a=2n b=2n^2 -5n


B)if b=63,find the value of n

b=63

2n^2 - 5n = 63 by A)

2n^2 - 5n -63 = 0

(n-7) (2n+9) = 0

i.e n=7 or n=-4.5 (rej.)

a=nC1x2=2n

, b=nx(-3)+nC2x(2^2)=-3n+2n!/(n-2)!