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?
 A: 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
$$
\begin{align}
a_n
&=\frac{\left(\frac{-1+i\sqrt3}2\right)^{n+1}-\left(\frac{-1-i\sqrt3}2\right)^{n+1}}{i\sqrt3}\\[6pt]
&=\frac{e^{i2\pi(n+1)/3}-e^{-i2\pi(n+1)/3}}{i\sqrt3}\\[4pt]
&=\frac2{\sqrt3}\sin\left(\frac{2\pi(n+1)}{3}\right)\tag{5}
\end{align}
$$
A: This is much easier if we instead compute the sequence $a_m := \sum_{i=0}^\infty y^i {{m-i} \choose i}$ (note that most terms are zero).  Then it is immediate that $a_{m+2}= a_{m+1}+y a_m$, from comparing terms and using $  [y^{i-1}]a_m + [y^i] a_{m+1} = {{m-(i-1)}\choose{i-1}}+{{m+1-i}\choose{i}} = {{m+2-i}\choose{i}} = [y^i]a_{m+2}$.
Substituting $y=-1$, it is not hard to see that the sequence $a_m$ has period $6$ (for example, the characteristic polynomial is cyclotomic).  So $a_{2n+1}$ has period $3$.
This approach was inspired by looking at generating functions, which are a very powerful tool when you don't know what's going on.  I've included that approach below.

Side note: the case $y=-1/4$ is particularly interesting, as we get the very clean answer of $a_m = (m+1)2^{-m}$.

Alternate solution, using generating functions:
$$f(x) := \sum_m a_m x^m$$
$$ = \sum_i y^i \sum_{m\geq 2i} {{m-i}\choose i} x^m = \sum_i (xy)^i \sum_{m'\geq i} {m'\choose i} x^{m'}$$
$$ = \sum_i (x y)^i \frac{x^i}{(1-x)^{i+1}} = \frac{1}{1-x} \sum_i \left(\frac{x^2y}{1-x} \right)^i$$
$$ = \frac{1}{1-x} \frac{1}{1-\frac{x^2 y}{1-x}} = \frac{1}{1-x-x^2 y}$$
Finally, we can read off the coefficients of $f(x)$ using partial fractions.
A: 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.

A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{I_{n}\equiv\sum_{k = 0}^{n}\pars{-1}^{k}{ 2n + 1 - k \choose k}:\ {\large ?}}$

Since $\quad\ds{{ 2n + 1 - k \choose k} = 0}\quad$ when
  $\quad\ds{k > 2n + 1 - k\ \imp\ k> n + \half\ \imp\ k \geq n + 1}\quad$ it's true that
  $$
\sum_{k = 0}^{n}\pars{-1}^{k}{ 2n + 1 - k \choose k}
=\sum_{k = 0}^{\color{#c00000}{\Large\infty}}\pars{-1}^{k}{ 2n + 1 - k \choose k}
$$

With $\ds{a > 1}$:
\begin{align}
I_{n}&=\sum_{k = 0}^{\infty}\pars{-1}^{k}
\oint_{\verts{z}\ =\ a}
{\pars{1 + z}^{2n + 1 - k} \over z^{k + 1}}\,{\dd z \over 2\pi\ic}
=\oint_{\verts{z}\ =\ a}{\pars{1 + z}^{2n + 1} \over z}\sum_{k = 0}^{\infty}
\bracks{-\,{1 \over z\pars{1 + z}}}^{k}\,{\dd z \over 2\pi\ic}
\\[5mm]&=\oint_{\verts{z}\ =\ a}{\pars{1 + z}^{2n + 1} \over z}
{1 \over 1 + 1/\bracks{z\pars{1 + z}}}\,{\dd z \over 2\pi\ic}
=\oint_{}{\pars{1 + z}^{2n + 2} \over z^{2} + z + 1}\,{\dd z \over 2\pi\ic}
\\[5mm]&=\oint_{\verts{z}\ =\ a}
{\pars{1 + z}^{2n + 2} \over \pars{z - r}\pars{z - r^{*}}}
\,{\dd z \over 2\pi\ic}\qquad\mbox{where}\quad
r \equiv \exp\pars{2\pi\ic \over 3} = -\,\half + {\root{3} \over 2}\,\ic
\end{align}

Then,
  \begin{align}
I_{n}&={\pars{1 + r}^{2n + 2} \over r - r^{*}}
+{\pars{1 + r^{*}}^{2n + 2} \over r^{*} - r}
=2\,\Re\bracks{\pars{1 + r}^{2n + 2} \over r - r^{*}}
=2\,\Re\bracks{\pars{1 + r}^{2n + 2} \over 2\ic\,\Im r}
\\[5mm]&={1 \over \Im r}\,\Im\pars{1 + r}^{2n + 2}
={2\root{3} \over 3}\,\Im\pars{\half + {\root{3} \over 2}\,\ic}^{2n + 2}
={2\root{3} \over 3}\,\Im\bracks{\exp\pars{\pi\ic \over 3}}^{2n + 2}
\\[5mm]&={2\root{3} \over 3}\,\Im\exp\pars{\pars{2n + 2}\pi\ic \over 3}
\end{align}

$$\color{#66f}{\large
I_{n}\equiv\color{#66f}{\large\sum_{k = 0}^{n}\pars{-1}^{k}{ 2n + 1 - k \choose k}}
=\color{#66f}{\large{2\root{3} \over 3}\,\sin\pars{\bracks{2n + 2}\pi \over 3}}}
$$
A: 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, $$\displaystyle \sum_{r=0}^{n+1} (-1)^{r} \dbinom{n+1}{r} \sin \left( \dfrac{r \pi}{3} \right) = \Im \left( \sum_{r=0}^{n+1} (-1)^{r} \dbinom{n+1}{r} e^{\frac{i r \pi}{3}} \right) = (-1)^{n+1} \sin \left( \dfrac{2 (n+1) \pi}{3} \right) $$  
$$ \implies \text{S} = \dfrac{2}{\sqrt{3}} \sin \left( \dfrac{2 (n+1) \pi}{3} \right) \quad \square $$     
A: Using Maple I am obtaining
$$ \left( 1/6\,i\sqrt {3}+1/2 \right)  \left( -1/2+1/2\,i\sqrt {3}
 \right) ^{n}+ \left( 1/2-1/6\,i\sqrt {3} \right)  \left( -1/2-1/2\,i
\sqrt {3} \right) ^{n}
$$
and it is rewritten as
$$1/6\, \left( i\sqrt {3}+3 \right)  \left( -1 \right) ^{n}{e}^{-1/3\,i
\pi \,n}-1/6\, \left( -3+i\sqrt {3} \right)  \left( -1 \right) ^{n}{e}
^{1/3\,i\pi \,n}
$$
Then your conjecture is correct. It is verified using Maple.
