簡易特殊級數求解，需要過程?

f(x) = x/[(x+2)(x+3)(x+4)]
      = A/(x+2) + B/(x+3) + C/(x+4)
∴ x = A(x+3)(x+4) + B(x+2)(x+4) + C(x+2)(x+3)
分別以 x = -2, -3, -4 代入, 得
    -2 = A(1)(2)
    -3 = B(-1)(1)
    -4 = C(-2)(-1)
∴ A = -1, B = 3, C = -2
f(x) = -1/(x+2) + 3/(x+3) - 2/(x+4)

Σ_[r=1～n] r/[(r+2)(r+3)(r+4)]
    = Σ_[r=1～n]  { -1/(r+2) + 3/(r+3) - 2/(r+4) }
    = Σ_[r=1～n] [-1/(r+2) + 1/(r+3)] 
           +  2 Σ_[r=1～n] [1/(r+3) - 1/(r+4)] 
   = [-1/(1+2) + 1/(n+3)] + 2[1/(1+3) - 1/(n+4)]
   = -n/[3(n+3)] + 2n/[4(n+4)]
   = n [-2(n+4)+3(n+3)]/[6(n+3)(n+4)]
   = n(n+1)/[6(n+3)(n+4)]