Prove that $\sum_{k=0}^{\lfloor (n-1)/2 \rfloor} (-1)^k {n+1 \choose k} {2n-2k-1 \choose n} =\frac{ n(n+1)}2 $

We seek to show that

$$\sum_{k=0}^{\lfloor (n-1)/2 \rfloor} (-1)^k {n+1\choose k} {2n-2k-1\choose n} = \frac{1}{2} n (n+1).$$

The LHS is

$$\sum_{k=0}^{\lfloor (n-1)/2 \rfloor} (-1)^k {n+1\choose k} {2n-2k-1\choose n-1-2k} \\ = [z^{n-1}] (1+z)^{2n-1} \sum_{k=0}^{\lfloor (n-1)/2 \rfloor} (-1)^k {n+1\choose k} z^{2k} (1+z)^{-2k}.$$

Now the coefficient extractor $[z^{n-1}]$ combined with the $z^{2k}$ term enforces the range, making for a zero contribution when $2k\gt n-1$ and we may continue with

$$[z^{n-1}] (1+z)^{2n-1} \sum_{k\ge 0} (-1)^k {n+1\choose k} z^{2k} (1+z)^{-2k} \\ = [z^{n-1}] (1+z)^{2n-1} \left(1-\frac{z^2}{(1+z)^2}\right)^{n+1} \\ = [z^{n-1}] \frac{1}{(1+z)^3} (1+2z)^{n+1}.$$

This is

$$\sum_{q=0}^{n-1} (-1)^q {q+2\choose q} {n+1\choose n-1-q} 2^{n-1-q}.$$

Observe that

$${q+2\choose q} {n+1\choose n-1-q} = \frac{(n+1)!}{q!\times 2! \times (n-1-q)!} = {n+1\choose 2} {n-1\choose q}. $$

This yields for the sum

$${n+1\choose 2} \sum_{q=0}^{n-1} (-1)^q {n-1\choose q} 2^{n-1-q} \\ = {n+1\choose 2} (-1+2)^{n-1} = {n+1\choose 2}.$$

We have the claim.

Coefficient of $x^k$ in $(1-x)^{n+1}$ is $(-1)^k$${n+1 \choose k}$

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Coefficient of $$x^{(\frac{n-2k-1}{2})}$$ in $$(1- \sqrt{x})^{-(n+1)}$$ is ${{2n-2k-1} \choose n}$

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In all what we want is coefficient of (multiplying previous 2 series)$$x^{(\frac{n-1}{2})}$$ in $$(1 +\sqrt{x})^{n+1}$$ which is nothing but ${{n+1} \choose 2}$

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