I'm trying to compute the sum $$\sum_{k=\ell+1}^{n}(-1)^{(k-\ell+1)}\binom{k}{\ell}$$ with $\ell \in [2,n-1]$ in an efficient way. I know that this reduces to something like $$ 1 -\sum_{i=0}^{n/2 -1}\binom{n-2i-1}{l-1} $$ in the case that $n-\ell \equiv 0\, mod\,2$, with minor adjustments for the odd case. Therefore, if I had a closed form of the sums $\sum_{i=0}^{n}\binom{2i}{\ell}$ and $\sum_{i=0}^{n}\binom{2i+1}{\ell}$, I could fairly easily construct what I need.
My issue is that WolframAlpha readily produced a solution in terms of $_{3}F_{2}$ with parameters: $$_{3}F_{2}(1,n+2,n+\tfrac{5}{3};-\tfrac{k}{2}+n+2,-\tfrac{k}{2}+n+\tfrac{5}{2};1).$$ In evaluating this, WolframAlpha seems convinced that the hypergeometric series doesn't converge, and in all of my reading the only convergence identity that I could find was related to $\tfrac{d\log (_{3}F_{2})}{dz}$.
In effect, this leaves me with two questions:
(1) Does $_{3}F_{2}$ converge in this case? If so, to what?
(2) Is there an easier or more effective way to approach this summation?
Maple writes your sum as
$$ -{2}^{-1-l}+1- \left( -1 \right) ^{n-l}{n+1\choose l} {\mbox{$_2$F$_1$}(1,n+2;\,n-l+2;\,-1)} $$
However, the series for this hypergeometric function doesn't converge at $z=-1$ either. Nevertheless, the function has an analytic continuation to a neighbourhood of $-1$ and there are ways to compute it. For example, $g(z)={\mbox{$_2$F$_1$}(1,n+2;\,n-l+2;\,z)}$ satisfies the hypergeometric differential equation $$ (z^2-z) g'' + ((n+4) z + l - n - 2) g' + (n+2) g = 0, \; g(0) = 1, \; g'(0) = \frac{n+2}{n-l+2} $$ which can be numerically integrated from $z=0$ to $z=-1$.