06/24/2010, 07:56 PM
Hi, I discovered it myself 
To derive it, use the rule that the multiplication of two power series is the discrete convolution of their coefficients and that \( (1 - x)^{-1} = \sum_{n=0}^{\infty}{x^n} \).
Now try to multiply a formal power series with \( (1 - x)^{-1} = \sum_{n=0}^{\infty}{x^n} \).
Repeat it k times and use the formula of \( (1 - x)^{-k} \) to derive the formula.
It however reminds Cauchy's formula for repeated integration (with all umbral calculus variations).
To derive it, use the rule that the multiplication of two power series is the discrete convolution of their coefficients and that \( (1 - x)^{-1} = \sum_{n=0}^{\infty}{x^n} \).
Now try to multiply a formal power series with \( (1 - x)^{-1} = \sum_{n=0}^{\infty}{x^n} \).
Repeat it k times and use the formula of \( (1 - x)^{-k} \) to derive the formula.
It however reminds Cauchy's formula for repeated integration (with all umbral calculus variations).