Notation: $$ \left(\!\! \binom{n}{k}\!\!\right)={n+k-1 \choose k}=\frac{(n+k-1)!}{k!(n-1)!} $$ where $n!$ is the factorial, i.e. $1\cdot 2\cdots n.$
Let $n,N\in \mathbb{Z}_{>0}.$ I'm stuck at trying to write the following sum:
$$\sum_{N=3}^{u}(N-2)\left(\!\!\binom{ n-2 }{ N+1}\!\!\right)$$ in an "acceptable way", perhaps using a generating function or similar. I tried to write it as a product of two formal power series but I don't think it will work. I haven't used combinatorics for a long time, sadly.
$$(1-x)^{-n} = \sum_{k=0}^\infty \left( \left(\!\! \binom{n}{k}\!\!\right) x^{k}\!\! \right) = 1+\left(\!\! \binom{n}{1}\!\!\right)x + \left(\!\! \binom{n}{2}\!\!\right)x^2 + \left(\!\! \binom{n}{3}\!\!\right)x^3 + \sum_{k=4}^\infty \left( \left(\!\! \binom{n}{k}\!\!\right) x^{k}\!\! \right)$$
Rearranging
$$\frac{(1-x)^{-n}-1-nx - \left(\!\! \binom{n}{2}\!\!\right)x^2-\left(\!\! \binom{n}{3}\!\!\right)x^3}{x^3} = \sum_{k=4}^\infty \left( \left(\!\! \binom{n}{k}\!\!\right) x^{k-3}\!\! \right)$$
Differentiating with respect to $x$
$$f(x)=\frac{\partial}{\partial x}\left(\frac{(1-x)^{-n}-1-nx - \left(\!\! \binom{n}{2}\!\!\right)x^2-\left(\!\! \binom{n}{3}\!\!\right)x^3}{x^3}\right) = \sum_{k=4}^{\infty}(k-3)\left(\!\! \binom{n}{k}\!\!\right) x^{k-4} = \sum_{k=0}^{\infty}(k+1)\left(\!\! \binom{n}{k+4}\!\!\right) x^{k}$$
Or simply
$$ \frac{-6 (-1 + (1 - x)^{-n}) + n (4 + 2 (1 - x)^{-1 - n} - x) x + n^2 x^2}{2x^4} = \sum_{k=0}^{\infty}(k+1)\left(\!\! \binom{n}{k+4}\!\!\right) x^{k}$$
What you need is the partial sum generating function of this sequence, which has generating function of $\frac{1}{1-x}f(x)$. So overall, the generating function of the partial sums is
$$ \frac{-6 (-1 + (1 - x)^{-n}) + n (4 + 2 (1 - x)^{-1 - n} - x) x + n^2 x^2}{2x^4(1-x)} = \sum_{u=0}^{\infty}\left(\sum_{k=0}^u(k+1)\left(\!\! \binom{n}{k+4}\!\!\right) \right)x^u = \sum_{u=0}^{\infty}\left(\sum_{k=3}^{u+3}(k-2)\left(\!\! \binom{n}{k+1}\!\!\right)\right)x^u$$