Prove that

$$x^{(n)} = x(x+1)(x+2)\cdots(x+n-1) = \sum_{k=0}^n \left[n\atop k\right]x^k$$

Or

$$ \sum_{k=0}^n \left[n\atop k\right]x^k = \frac{\Gamma(x+n)}{\Gamma(x)}$$

$$\textit{proof}$$

By induction for \( n=0\) we have

$$ \left[0\atop 0\right] = 1$$

Assume it is true for \( n \) then for \( n+1 \)

$$ \sum_{k=0}^{n+1} \left[n+1\atop k\right]x^k =\sum_{k=0}^{n} \left[n+1\atop k\right]x^k +x^{n+1} =n\sum_{k=0}^{n} \left[n\atop k\right]x^k + \sum_{k=1}^{n} \left[n\atop k-1\right]x^k +x^{n+1} $$

Simplify and use the inductive step

$$ \sum_{k=0}^{n+1} \left[n+1\atop k\right]x^k =n\sum_{k=0}^{n} \left[n\atop k\right]x^k +x \sum_{k=0}^{n} \left[n\atop k\right]x^k = \frac{(n+x)\Gamma(x+n)}{\Gamma(x)} = \frac{\Gamma(n+1+x)}{\Gamma(x)}$$

### Like this:

Like Loading...

*Related*