How can the following identity for positive integer $n$ be proved? It has been confirmed symbolically by Mathematica:
$$ \sum_{i=1}^n (-1)^{1+i} (2n+1-2i) \binom{2n+1}{i} = 2n + 1 $$
Induction seems quite cumbersome as $n$ appears in three instances on the left hand side. Also the usual binomial sum identities have an upper limit equal to the upper number in the binomial coefficient, whereas here these are $n$ and $2n+1$ respectively.
Background. This identity arose from a series expansion where the functions $U_k(\phi)$ recursively satisfy a differential equation depending on $U_{k-2}(\phi)$. After computing the first few by hand, I was able to see the pattern, and proceeded to prove it by induction. However for odd $k \ge 3$, this involved showing the identity
$$ \sum_{\substack{j=1 \\ \text{$j$ odd}}}^{k-2} (-1)^{(j-1)/2} \frac{2j}{(k-j)!!(k+j)!!} = (-1)^{(k+1)/2} \frac{2k}{(2k)!!}. $$
After rearranging and expanding the double factorials in terms of single factorials, obtained me
$$ \sum_{\substack{j=1 \\ \text{$j$ odd}}}^{k-2} (-1)^{(j+k)/2} \frac{j}{k} \binom{k}{(k-j)/2} = 1. $$
Putting $j = 2i-1$ and $k = 2n+1$ gives
$$ \sum_{i=1}^n (-1)^i (2i-1) \binom{2n+1}{n+1-i} = (-1)^n (2n+1), $$
and writing the sum backwards (i.e. replacing $i$ by $n+1-i$) yields the above.