Calculate the binomial sum $ I_n=\sum_{i=0}^n (-1)^i { 2n+1-i \choose i} $ I need any  hint  with calculating  of the sum 
$$
I_n=\sum_{i=0}^n (-1)^i { 2n+1-i \choose i}.
$$
Maple give the strange unsimplified result
$$
I_n={\frac {1/12\,i\sqrt {3} \left( - \left(  \left( 1+i\sqrt {3} \right) 
^{2\,{\it n}+2} \right) ^{2}+16\, \left( -1 \right) ^{2\,{\it n}}
 \left( {2}^{2\,{\it n}} \right) ^{2} \right) }{{2}^{2\,{\it n}}
 \left( 1+i\sqrt {3} \right) ^{2\,{\it n}+2}}},
$$
Сalculation  for  small  $n$ are  as follows $I_1=-1, I_2=0,I_3=1, I_4=-1, \ldots$
and leads to a hypothese: 
$
I_n= -1 \text{ for } n=3k+1, =0, \text{ for } n=3k+2,=1, \text{ for } n=3k. 
$
But  how to prove it?

Generating Functions
Let's compute the generating function of $\displaystyle\sum_{k=0}^n(-1)^k\binom{n-k}{k}$:
$$
\begin{align}
\sum_{n=0}^\infty\sum_{k=0}^n(-1)^k\binom{n-k}{k}x^n
&=\sum_{k=0}^\infty(-1)^kx^k\sum_{n=k}^\infty\binom{n-k}{k}x^{n-k}\\
&=\sum_{k=0}^\infty(-1)^kx^k\frac{x^k}{(1-x)^{k+1}}\\
&=\frac1{1-x}\frac1{1+\frac{x^2}{1-x}}\\
&=\frac1{1-x+x^2}\tag{1}
\end{align}
$$
What the question asks is for the odd terms, which we get by taking the odd part of $(1)$:
$$
\frac12\left(\frac1{1-x+x^2}-\frac1{1+x+x^2}\right)=\frac{x}{1+x^2+x^4}\tag{2}
$$
which implies that $\displaystyle\sum_{k=0}^n(-1)^k\binom{2n+1-k}{k}$ satisfies the recurrence
$$
a_n=-a_{n-1}-a_{n-2}\tag{3}
$$
and starts out: $(1,-1,0,1,-1,\dots)$ and $(3)$ ensures that it will follow this pattern. That is,
$$
\boxed{\displaystyle\bbox[5px]{
\sum_{k=0}^n(-1)^k\binom{2n+1-k}{k}=\left\{\begin{array}{rl}
1&\text{if }n\equiv0\pmod{3}\\
-1&\text{if }n\equiv1\pmod{3}\\
0&\text{if }n\equiv2\pmod{3}\\
\end{array}\right.}}\tag{4}
$$

Solving the Recurrence $\boldsymbol{(3)}$
We can also get a solution by solving the recurrence $(3)$.
Since the roots of $x^2+x+1$ are $\frac{-1\pm i\sqrt3}{2}=e^{\pm i2\pi/3}$ and $a_0=1$ and $a_1=-1$, we get the general solution to be
$$
\begin{align}
a_n
&=\frac{\left(\frac{-1+i\sqrt3}2\right)^{n+1}-\left(\frac{-1-i\sqrt3}2\right)^{n+1}}{i\sqrt3}\\[6pt]
&=\frac{e^{i2\pi(n+1)/3}-e^{-i2\pi(n+1)/3}}{i\sqrt3}\\[4pt]
&=\frac2{\sqrt3}\sin\left(\frac{2\pi(n+1)}{3}\right)\tag{5}
\end{align}
$$
