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?

Since $ \dbinom{n}{r} = 0 $ for $ r > n $, we can rewrite the sum as   
$$ \text{S} = \sum_{r=0}^{\infty} (-1)^r \dbinom{2n+1-r}{r} $$
From the Binomial Theorem, we see that the sum is the coefficient of $x^n$ in  
$$f(x) = x^n (1-x)^{2n+1} + x^{n-1} (1-x)^{2n} + \ldots $$    
$$ = \dfrac{(1-x)^{n+1}}{x^2-x+1} $$     
Note that,  
$$ \sum_{n=0}^{\infty}{{U}_{r}(a) {x}^{r}} = \dfrac{1}{x^2 -2ax+1} $$    
where $ U_{r} (x) $ is the Chebyshev Polynomial of the second kind.    
Putting $a = \dfrac{1}{2}$, we see that,    
$$ f(x) = (1-x)^{n+1} \sum_{n=0}^{\infty}{{U}_{r} \left( \dfrac{1}{2} \right) {x}^{r}} \quad (*)  $$     
To calculate the coefficient of $x^n$ in $ (*) $, we see that coefficient of $x^{n-r}$ for fixed $r$ in $(1-x)^{n+1}$ is $ (-1)^{n-r} \dbinom{n+1}{n-r} $, so coefficient of $x^n$ is,  
$$ \sum_{r=0}^{n} (-1)^{n-r} \dbinom{n+1}{n-r} U_{r} \left(\dfrac{1}{2}\right) $$    
Substituting $ r \mapsto r-1 $, we have,   
$$ \text{S} = \sum_{r=1}^{n+1} (-1)^{n+1-r} \dbinom{n+1}{n+1-r} U_{r-1} \left(\dfrac{1}{2}\right) $$   
$$ = (-1)^{n+1} \sum_{r=1}^{n+1} (-1)^{r} \dbinom{n+1}{r} U_{r-1} \left(\dfrac{1}{2}\right) \quad \quad \left( \because \dbinom{n}{n-r} = \dbinom{n}{r} \right) $$      
$$ = (-1)^{n+1} \left( \dfrac{2}{\sqrt{3}} \right) \sum_{r=1}^{n+1} (-1)^{r} \dbinom{n+1}{r} \sin \left( \dfrac{r \pi}{3} \right) \quad \quad \left( \because U_{r-1} \left(\dfrac{1}{2}\right) = \sin \left( \dfrac{r \pi}{3} \right) \right) $$   
$$ (-1)^{n+1} \left( \dfrac{2}{\sqrt{3}} \right) \sum_{r=0}^{n+1} (-1)^{r} \dbinom{n+1}{r} \sin \left( \dfrac{r \pi}{3} \right) \quad \quad \left( \because \sin 0 = 0 \right) $$   
Now, $$\displaystyle \sum_{r=0}^{n+1} (-1)^{r} \dbinom{n+1}{r} \sin \left( \dfrac{r \pi}{3} \right) = \Im \left( \sum_{r=0}^{n+1} (-1)^{r} \dbinom{n+1}{r} e^{\frac{i r \pi}{3}} \right) = (-1)^{n+1} \sin \left( \dfrac{2 (n+1) \pi}{3} \right) $$  
$$ \implies \text{S} = \dfrac{2}{\sqrt{3}} \sin \left( \dfrac{2 (n+1) \pi}{3} \right) \quad \square $$     
