中四附加數

1
(1+2x-3x^2)^n=1+ax+bx^2+ terms involving higher powers of x
(1+x(2-3x))^n
=1+nx(2-3x)+nC2[x(2-3x)]^2+...
=1+2nx-3nx^2+nC2[4x^2]+...
=1+2nx+(4nC2-3n)x^2+...
so
a=2n
b=4*nC2-3n
b)b=63
4n(n-1)/2-3n=63
2n(n-1)-3n=63
2n^2-5n-63=0
(2n+9)(n-7)=0
n=-9/2 or n=7
so n=7
2
(a)
(1+4x+x^2)^n=1+ax+bx^2+other terms involving higher powers of x
(1+4x+x^2)^n
= [1+x(4+x)]^n
=1+nx(4+x)+nC2[x(4+x)]^2+other terms involving higher powers of x
=1+4nx+nx^2+16(nC2)x^2+other terms involving higher powers of x
=1+4nx+[n+16(nC2)]x^2+other terms involving higher powers of x
So
a=4n, b=n+16(nC2)
(b)
a=20
4n=20
n=5
sub into b=n+16(nC2)
b=5+16(5C2)=5+160=165