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Harder Partial Fractions

發問:

It is known that for any positive integer n, (n+1)!/[x(x+1)(x+2)...(x+n+1)] = summation(r=0->n+1) [(-1)^r C(n+1,r)/(x+r)] , where C(n,r) is the binomial coefficient. Show that summation(r=0->n) [(-1)^r C(n,r)/(n+r)(n+r+1)] = 1/[(2n+1) C(2n,n+1) ]

最佳解答:

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)]

其他解答:

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