Prove that `(""^n C_0)/x-(""^n C_1)/(x+1)+(""^n C_2)/(x+2)-.....+(-1)^n(""^n C_n)/(x+n)=(n !)/(x(x+1) . . . (x-n)),` where `n` is any positive integer and `x` is not a negative integer.

Explanation

Let
`f(x) = (n!)/(x(x+1)(x+2)"...."(x+n))`
`= (A_(0))/(x) + (A_(1) )/(x+1) + (A_(2))/(x+2)+"...."+(A_(n))/(x+n)`.
(by partial fractions )
Then `A_(0)=[xf(x)]_(x=0)=(n!)/(1.2.3"...."n) =1=.^(n)C_(0)`
`A_(1) = [(x+1)f(x)]_(x) = - 1`
`= (n!)/((-1){1.2"....."(n-1)})`
`= (-(n!))/((n-1)!) = -.^(n)C_(1)`
`A_(2) = [(x+2)f(x)]_(x=-2)`
`=(n!)/((-2).(-1).1.2"....."(n-2))`
`= (n!)/(2!(n-2)!) = .^(n)C_(2)` and so on
Thus, `(n!)/(x(x+1)(x+2)"....."(x+n))`
`(.^(n)C_(0))/(x) - (.^(n)C_(1))/(x+1) + (.^(n)C_(2))/(x+2) - "....." + (-1)^(n) (.^(n)C_(n))/(x+n)" "(1)`