Q:

Simplify given expression about Narayana numbers

Let $N(n,r)=\frac{1}{n}\binom{n}{r}\binom{n}{r-1}$ be Narayana numbers. We have $\binom{n}{r-1}N(n,r+1)=\binom{n}{r+1}N(n,r)$.
How  can we simplify following expression:
$$\sum\limits_{i=1}^{r} N(n-1, i)-\sum\limits_{i=1}^{r}N( n-2,i).$$

A:

Given the function defined (for integers $\,n>0\,$ and $\,k>0$) by
$$ N(n,k) :=\frac{1}{n}\binom{n}{k}\binom{n}{k-1}, \tag{1} $$
the function defined (for integers $\,n>0\,$ and $\,k>0$) by
$$ D(n,i) := N(n-1, i) - N( n-2,i), \tag{2} $$
and the function defined by
$$ S(n,k) := \sum_{i=1}^k D(n,i), \tag{3} $$
then this function does not have a simpler expression as far as
I know and I doubt it is possible to simplify it. You might be
able to express it with hypergeometric functions, but that
is almost exactly the same as the original summation.

