Harder Partial Fractions

Firstly 1 / (x + r)(x + r + 1) = 1 / (x + r) - 1 / (x + r + 1) : you should be able to prove this.
Σ(r = 0 - >n) [( - 1)^r C(n,r)/(x + r)(x + r + 1)]
= Σ(r = 0 - >n) [( - 1)^r C(n,r)][1/(x + r) - 1/(x + r + 1)]
= Σ(r = 0 - >n) [( - 1)^r C(n,r)]/(x + r) - Σ(r = 0 - >n) [( - 1)^r C(n,r)]/(x + r + 1)] ... (1)
For the second sum, let k = r + 1, then it becomes
Σ(k = 1 - >n + 1) [( - 1)^(k - 1) C(n,k - 1)]/(x + k)], renaming k back to r,
= Σ(r = 1 - >n + 1) [( - 1)^(r - 1) C(n,r - 1)]/(x + r)]
(1) becomes
Σ(r = 0 - >n) [( - 1)^r C(n,r)]/(x + r) - Σ(r = 1 - >n + 1) [( - 1)^(r - 1) C(n,r - 1)]/(x + r)]
= Σ(r = 1 - >n) [( - 1)^r C(n,r)]/(x + r) - Σ(r = 1 - >n) [( - 1)^(r - 1) C(n,r - 1)]/(x + r)] + C(n,0)/x - ( - 1)^nC(n,n)/(x + n + 1)
= Σ(r = 1 - >n) [( - 1)^r/(x + r)][C(n,r) + C(n,r - 1)] + 1/x + ( - 1)^(n + 1)/(x + n + 1)
Note that C(n,r) + C(n,r - 1) = C(n + 1,r)
= Σ(r = 1 - >n) [( - 1)^r/(x + r)][C(n + 1,r)] + 1/x + ( - 1)^(n + 1)/(x + n + 1)
= Σ(r = 0 - >n + 1) [( - 1)^r/(x + r)][C(n + 1,r)]
= (n + 1)!/[x(x + 1)(x + 2)...(x + n + 1)]
Put x = n, the sum becomes
(n + 1)!/[n(n + 1)(n + 2)...(n + n + 1)]
= (n + 1)!(n - 1)! / (2n + 1)!
= (n + 1)!(n - 1)! / [(2n)!(2n + 1)]
= 1 / [(2n + 1)C(2n,n + 1)]

這裏可以幫到你
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