a) Prove $\sum_{k=0}^{n} s(n,k) = n!$, for all $n\in \mathbb{N}$.
So my idea was to start with induction over $n\\$.
- Base case: $n=1$
$\sum_{k=0}^{1}s(1,k)=s(1,0)+s(1,1) = 0+1=1 =1!$
- Inductive hypothesis. Assume true for $\sum_{k=0}^{n} s(n,k) = n!$
- Inductive step. Check the n+1 case. We get
$\sum_{k=0}^{n+1}s(n+1,k)=\sum_{k=0}^{n}s(n,k)+$
Could someone help me with the last step. I can not seem to understand how to split the last sum.
b) Prove $s(n,n-2) =\frac{1}{24}n(n-1)(n-2)(3n-1)$ for all $n\in\mathbb{N},n\geq2.$
I have tried proving b) by induction ( just like the problem below): (https://math.stackexchange.com/users/438651/thesmallprint), Proving $S(n,n-2) = \frac{n(n-1)(n-2)(3n-5)}{24}.$, URL (version: 2018-06-06): https://math.stackexchange.com/q/2810239
The problem is, that what I get in the Inductive step is: $s(n+1,n-1)=\frac{n(n-1)(3n^2+5n-10)}{24}$. Sadly I can not seem to factorize $3n^2+5n-10$. Some help with that would be appreciated.
Recall that the Stirling numbers of the first kind satisfy the recurrence relation $s(n+1,k)=ns(n,k)+s(n-1,k)$. ($s(n,0)=0$)
The first part follow easily by induction.
For the second part,again induction and use \begin{eqnarray*} s(n,n-1)&=& \frac{n(n-1)}{2} \\ s(n+1,n-1)&=& n \, s(n,n-1)+ s(n,n-2). \end{eqnarray*}