2
$\begingroup$

How did i prove

$$\frac{\binom{n}{0}}{x}-\frac{\binom{n}{1}}{x+1}+\frac{\binom{n}{2}}{x+2}-\cdots \cdots +(-1)^n\frac{\binom{n}{n}}{x+n}=\frac{n!}{x(x+1)(x+2)\cdots (x+n)}$$

what i try

$$\sum^{n}_{r=0}(-1)^r\frac{\binom{n}{r}}{x+r}=\int^{1}_{0}\sum^{n}_{r=0}(-1)^r\binom{n}{r}t^{x+r-1}dt$$

$$\begin{align}=\int^{1}_{0}t^{x-1}\sum^{n}_{r=0}(-1)^r\binom{n}{r}t^{r}dt =\int^{1}_{0}t^{x-1}(1-t)^ndt\end{align}$$

$\endgroup$
  • 4
    $\begingroup$ Continuing further $$\int_0^1 t^{x-1} \left(\sum_{r=0}^n (-1)^r \binom nr t^r \right) dt=\int_0^1 t^{x-1}(1-t)^n dt =B(x,n+1)$$ $\endgroup$ – Rohan Shinde Feb 17 at 6:09
  • 2
    $\begingroup$ And now you can finish by using the usual relation between the Beta and Gamma functions and the usual recursion for the Gamma function. $\endgroup$ – Ian Feb 17 at 6:17
  • $\begingroup$ thanks Diagamma got it $\endgroup$ – jacky Feb 17 at 6:30
4
$\begingroup$

A possible way is as follows:

Multiplying by $\prod_{i=0}^n(x+i)$ gives

$$\sum_{k=0}^n(-1)^k\binom{n}{k}\prod_{\stackrel{i=0}{i\neq k}}^n(x+i) = n!$$

On the left side is a polynomial $P(x)$ of degree $n$. So we only need to check the identity at $n+1$ points. A good choice for that are the zeros of $\prod_{i=0}^n(x+i)$:

$$P(-k) = (-1)^k \binom{n}{k}\underbrace{\prod_{\stackrel{i=0}{i\neq k}}^n(i-k)}_{=(-1)^k\cdot k!\cdot (n-k)!} = n!$$

So, we are done.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.