<s>[INST] Summation closed form I have the following sum, 
$$\sum_{j=0}^{\lfloor\frac{i+n-1}{n+2}\rfloor}(-1)^{j}\binom{n}{j}\binom{i+n-j(n+2)-1}{n-1}+\sum_{j=0}^{\lfloor\frac{i+n-2}{n+2}\rfloor}2(-1)^{j}\binom{n}{j}\binom{i+n-j(n+2)-2}{n-1}+\sum_{j=0}^{\lfloor\frac{i+n-3}{n+2}\rfloor}(-1)^{j}\binom{n}{j}\binom{i+n-j(n+2)-3}{n-1}$$
Notice that the upper bound of each sum is slightly different in each term. I wonder if there is a closed form for this? If we don't have the last binomial in each summation then there would be a closed form but at this level I am not sure how to get the closed form. 
Edit: $n$ is non-negative integer, and the constraint on $i$ is as follow: $0\leqslant i\leqslant n(n+1)+2$.
 [/INST] Note that we can better write each term as
$$
\eqalign{
  & S(k,n)=\sum\limits_{j = 0}^{\left\lfloor {{{i + n - k} \over {n + 2}}} \right\rfloor } {\left( { - 1} \right)^{\,j} \left( \matrix{
  n \cr 
  j \cr}  \right)\left( \matrix{
  i + n - k - j\left( {n + 2} \right) \cr 
  n - 1 \cr}  \right)}  =   \cr 
  &  = \sum\limits_{0\, \le j} {\left[ {0 \le i + n - k - j\left( {n + 2} \right)} \right]\left( { - 1} \right)^{\,j} \left( \matrix{
  n \cr 
  j \cr}  \right)\left( \matrix{
  i + n - k - j\left( {n + 2} \right) \cr 
  n - 1 \cr}  \right)}  =  \quad \quad (1)  \cr 
  &  = \sum\limits_{0\, \le j} {\left[ {0 \le i + n - k - j\left( {n + 2} \right)} \right]\left( { - 1} \right)^{\,j} \left( \matrix{
  n \cr 
  j \cr}  \right)\left( \matrix{
  i + n - k - j\left( {n + 2} \right) \cr 
  i + 1 - k - j\left( {n + 2} \right) \cr}  \right)}  =  \quad \quad (2)  \cr 
  &  = \left[ {1 \le n} \right]\sum\limits_{0\, \le j} {\left( { - 1} \right)^{\,j} \left( \matrix{
  n \cr 
  j \cr}  \right)\left( \matrix{
  i + n - k - j\left( {n + 2} \right) \cr 
  i + 1 - k - j\left( {n + 2} \right) \cr}  \right)}  \quad \quad (3) \cr} 
$$
where $[P]$ denotes the Iverson bracket
and where:
 - (1) we replace the upper bound with Iverson bracket;
 - (2) we can apply reflection, since the upper term of the binomial is non-negative;
 - (3) we can omit the Iverson bracket, because for $1 \le n$ it is implicit in the binomial.
Note that $S(k,n)$ is equivalent to
$$
S(k,n) = N_{\,b} (i + 1 - k,\,n + 1,n)
$$
where
$$
N_b (s,r,m)\quad \left| {\;0 \leqslant \text{integers  }s,m,r} \right.\quad  =
\sum\limits_{\left( {0\, \leqslant } \right)\,\,k\,\,\left( { \leqslant \,\frac{s}{r+1}\, \leqslant \,m} \right)} 
{\left( { - 1} \right)^k \binom{m}{k}
 \binom
 { s + m - 1 - k\left( {r + 1} \right) } 
 { s - k\left( {r + 1} \right)}\ }
$$
is the number discussed in this related post, and is also called the "r-nomial coefficient" since
$$
F_b (x,r,m) = \sum\limits_{0\,\, \leqslant \,\,s\,\,\left( { \leqslant \,\,r\,m} \right)} {N_b (s,r,m)\;x^{\,s} }
  = \left( {1 + x +  \cdots  + x^{\,r} } \right)^m  = \left( {\frac{{1 - x^{\,r + 1} }}{{1 - x}}} \right)^m 
$$
Now, the cumulative sum of $N_b$ has a similar expression
$$
M_{\,b} (t,\,r,m) = \sum\limits_{0\, \le s\; \le \,\,t} {N_{\,b} (s,\,r,m)}
  = \sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,{t \over {r + 1}}\, \le \,m} \right)} {\left( { - 1} \right)^k \binom {m}{k}
\binom{  t + m - k\left( {r + 1} \right)} {t - k\left( {r + 1} \right) } } 
$$
but this does not help to simplify much your sum, since it will always leave three terms.
The attempt to directly perform the sum will give:
$$
\eqalign{
  & S(n) = \left[ {1 \le n} \right]\sum\limits_{\left( {0\, \le } \right)k\,\left( { \le \,2} \right)} {\left( \matrix{
  2 \cr 
  k \cr}  \right)\sum\limits_{0\, \le j} {\left( { - 1} \right)^{\,j} \left( \matrix{
  n \cr 
  j \cr}  \right)\left( \matrix{
  i + n - k - j\left( {n + 2} \right) \cr 
  i + 1 - k - j\left( {n + 2} \right) \cr}  \right)} }  =   \cr 
  &  = \sum\limits_{\left( {0\, \le } \right)k\,\left( { \le \,2} \right)} {\sum\limits_{0\, \le j} {\left( { - 1} \right)^{\,j} \left( \matrix{
  2 \cr 
  k \cr}  \right)\left( \matrix{
  n \cr 
  j \cr}  \right)\left( \matrix{
  i + n - k - j\left( {n + 2} \right) \cr 
  i + 1 - k - j\left( {n + 2} \right) \cr}  \right)} }  =   \cr 
  &  = \sum\limits_{\left( {0\, \le } \right)k\,\left( { \le \,2} \right)} {\sum\limits_{0\, \le j} {\left( { - 1} \right)^{\,k + j} \left( \matrix{
  k - 3 \cr 
  k \cr}  \right)\left( \matrix{
  n \cr 
  j \cr}  \right)\left( \matrix{
  i + n - \left( {k + j} \right) - j\left( {n + 1} \right) \cr 
  i + 1 - \left( {k + j} \right) - j\left( {n + 1} \right) \cr}  \right)} }  =   \cr 
  &  = \sum\limits_{0\, \le l} {\sum\limits_{0\, \le j} {\left( { - 1} \right)^{\,l} \left( \matrix{
  l - j - 3 \cr 
  l - j \cr}  \right)\left( \matrix{
  n \cr 
  j \cr}  \right)\left( \matrix{
  i + n - l - j\left( {n + 1} \right) \cr 
  i + 1 - l - j\left( {n + 1} \right) \cr}  \right)} }  \cr} 
$$
and we cannot proceed to simplify by applying double correlation, due to the presence of $\left( { - 1} \right)^{\,l}$
We could instead simplify the related sum
$$
\eqalign{
  & S_{\, - } (n) = \sum\limits_{\left( {0\, \le } \right)k\,\left( { \le \,2} \right)} {\left( { - 1} \right)^{\,k} \left( \matrix{
  2 \cr 
  k \cr}  \right)S(k,n)}  =   \cr 
  &  = \sum\limits_{\left( {0\, \le } \right)k\,\left( { \le \,2} \right)} {\left( { - 1} \right)^{\,k} \left( \matrix{
  2 \cr 
  k \cr}  \right)\sum\limits_{} {\left( { - 1} \right)^{\,j} \left( \matrix{
  n \cr 
  j \cr}  \right)\left( \matrix{
  i + n - k - j\left( {n + 2} \right) \cr 
  i + 1 - k - j\left( {n + 2} \right) \cr}  \right)} }  =   \cr 
  &  = \sum\limits_{0\, \le j} {\left( { - 1} \right)^{\,j} \left( \matrix{
  n \cr 
  j \cr}  \right)\sum\limits_{\left( {0\, \le } \right)k\,\left( { \le \,2} \right)} {\left( \matrix{
  k - 3 \cr 
  k \cr}  \right)\left( \matrix{
  i + n - k - j\left( {n + 2} \right) \cr 
  i + 1 - k - j\left( {n + 2} \right) \cr}  \right)} }  =   \cr 
  &  = \sum\limits_{0\, \le j} {\left( { - 1} \right)^{\,j} \left( \matrix{
  n \cr 
  j \cr}  \right)\left( \matrix{
  i + n - 2 - j\left( {n + 2} \right) \cr 
  i + 1 - j\left( {n + 2} \right) \cr}  \right)}  \cr} 
$$
so that the actual sum may be reduced to two terms.   
Besides these, and other related manipulations, I do not see better ways.
</s>