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?

Start by restating the problem: we seek to evaluate
$$\sum_{q=0}^n (-1)^q {2n+1-q\choose q}
= \sum_{q=0}^n (-1)^q {2n+1-q\choose 2n+1-2q}.$$
Introduce the integral representation
$${2n+1-q\choose 2n+1-2q}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n+1-q}}{z^{2n+2-2q}} \; dz.$$
This gives the following for the sum
(note that the integral correctly represents the fact of the binomial coefficient being zero for $q>n$, so we may extend the sum to infinity):
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n+1}}{z^{2n+2}}
\sum_{q\ge 0} (-1)^q \frac{z^{2q}}{(1+z)^q} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n+1}}{z^{2n+2}}
\frac{1}{1+z^2/(1+z)} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n+2}}{z^{2n+2}}
\frac{1}{1+z+z^2} \; dz.$$
Extracting coefficients we find
$$\sum_{q=0}^{2n+1} {2n+2\choose 2n+1-q} 
[z^q] \frac{1}{1+z+z^2}
= \sum_{q=0}^{2n+1} {2n+2\choose q+1} 
[z^q] \frac{1}{1+z+z^2}.$$
Introduce the generating function
$$Q(w) = \sum_{n\ge 0} w^{2n+1} 
\sum_{q=0}^{2n+1} {2n+2\choose q+1} 
[z^q] \frac{1}{1+z+z^2},$$
which is
$$\sum_{q\ge 0} w^q [z^q] \frac{1}{1+z+z^2}
\sum_{2n+1\ge q} {2n+2\choose q+1} w^{2n+1-q}
\\ = \sum_{q\ge 0} w^q [z^q] \frac{1}{1+z+z^2}
\sum_{p\ge 0} {p+q+1\choose q+1} w^p
\\ = \sum_{q\ge 0} w^q [z^q] \frac{1}{1+z+z^2}
\frac{1}{(1-w)^{q+2}}
\\ = \frac{1}{(1-w)^2} \sum_{q\ge 0} 
\left(\frac{w}{1-w}\right)^q [z^q] \frac{1}{1+z+z^2}.$$
What we  have here is an annihilated coefficient extractor  and the
sum simplifies to
$$Q(w) = \frac{1}{(1-w)^2}
\frac{1}{1+w/(1-w) + w^2/(1-w)^2}
\\ = \frac{1}{(1-w)^2+w(1-w) + w^2}
= \frac{1}{1-w+w^2}.$$
This is the ordinary generating function of a sequence with recurrence
$$a_{n+2} = a_{n+1} - a_n.$$
Since $a_0 = 1$ and $a_1 = 1$ we obtain
$$1, 1, 0, -1, -1, 0, 1, 1, 0, -1, -1, 0\ldots$$
thereby proving periodicity with period six.
Selecting  the  odd-index  values  preserves periodicity and we get the sequence
$$1, -1, 0, 1, -1, 0, \ldots$$
Post Scriptum.
If we introduce $$\rho_{1,2} = \frac{1\pm i\sqrt{3}}{2}
\quad\text{and set}\quad
c_{1,2} = \frac{3\pm i\sqrt{3}}{6}$$
we have the closed form
$$[w^n] \frac{1}{1-w+w^2}
= c_1 \rho_1^{-n} + c_2 \rho_2^{-n}$$
from which $[w^{2n+1}] Q(w)$ may be extracted.

The technique of annihilated coefficient extractors (ACE) is also employed at this
MSE link I and this MSE link II.

A trace as to when this method appeared on MSE and by whom starts at this
MSE link.

