For any even number $2n$, can the expression $\displaystyle\sum_{k=0}^n \frac{n!}{(n-k)!} \cdot (2n-1-k)!$ be simplified to $\dfrac{(2n)!} n \text{?}$
If Yes, then how? If No then what can it be simplified to?
For any even number $2n$, can the expression $\displaystyle\sum_{k=0}^n \frac{n!}{(n-k)!} \cdot (2n-1-k)!$ be simplified to $\dfrac{(2n)!} n \text{?}$
If Yes, then how? If No then what can it be simplified to?
Copyright © 2021 JogjaFile Inc.
Use the hockey stick identity \begin{eqnarray*} \sum_{k=0}^{n} \binom{2n-1-k}{n-1} = \binom{2n}{n}. \end{eqnarray*} Your result follows with a little rearrangement.