Prove the following:
$$\sum\limits_{k=0}^{r}|s(n,k)| = n! - \sum\limits_{k=0}^{n} - |s(n,k+r+1)|$$
Workings:
Proof:
Base case: $n = 0, r = 0$
$s(0,0) = 1$
$0! - s(0,0+0+1) = 1 - 0 = 1$
Base case holds
Induction hypothesis:
Suppose $\sum\limits_{k=0}^{r} |s(n,k)| = n! - \sum\limits_{k=0}^{n} - |s(n,k+r+1)|$ holds for some $n$.
Then for $n+1$:
$\sum\limits_{k=0}^{r} |s(n+1,k)| = \sum\limits_{k=0}^{r} |s(n,k-1)| + n|s(n,k)|$
I don't know what to do next and I believe what I did so far is incorrect.
Any help will be appreciated.
Assuming you didn't mean the double negative on the RHS:
$$\sum\limits_{k=0}^{r}|s(n,k)| + \sum\limits_{k=0}^{n}|s(n,k+r+1)|$$
$$= \sum\limits_{k=0}^{r}|s(n,k)| + \sum\limits_{k={r+1}}^{n+r+1}|s(n,k)|$$
$$= \sum\limits_{k=0}^{r}|s(n,k)| + \sum\limits_{k={r+1}}^{n}|s(n,k)|$$
$$ \text{ (since } s(n,k) = 0 \text{ for }k > n)$$
$$= \sum\limits_{k=0}^{n}|s(n,k)| $$
$$= n!$$