binomial theorem2

1. Find the constant term in the expansion of [x-(1/x)]^4 – (3x+4)^5
Constant term in [x - (1/x)]^4 is C(4,r)(x^r)(1/x)^(4-r)
So r - (4 - r) = 0 => r = 2
Constant term is C(4,2)(x^2)(-1/x)^2 = 6
Constant term in (3x + 4)^5 = C(5,5)(3x)^0(4^5) = 1024
Overal constant term = 6 - 1024 = -1018
2. Let B(r) be the coefficient of x^r in the expansion of (1+x)^24. If B(r+2) : B(r)=57:7 find the value of r.
B(r) = C(24,r); B(r+2) = C(24,r+2)
B(r + 2) / B(r) = {24! / [(24 - r - 2)!(r + 2)!]} / {24! / [(24 - r)!r!]
= (24 - r)! r! / [(24 - r - 2)!(r + 2)!]
= (24 - r)(24 - r - 1) / [(r + 2)(r + 1)]
= (24 - r)(23 - r) / [(r + 2)(r + 1)]
= (552 - 47r + r^2) / (r^2 + 3r + 2) = 57/7
7(552 - 47r + r^2) = 57(r^2 + 3r + 2)
3864 - 329r + 7r^2 = 57r^2 + 171r + 114
50r^2 + 500r - 3750 = 0
r^2 + 10r - 75 = 0
(r - 5)(r + 15) = 0
r = 5 or r = -15 (rejected)