<s>[INST] $n-\frac{n(n^2-1)}{2!}+\frac{n(n^2-1)(n^2-4)}{2!3!}-\frac{n(n^2-1)(n^2-4)(n^2-9)}{3!4!}+.....$ If $n\in \mathbb{N}$, and
$n-\dfrac{n(n^2-1)}{2!}+\dfrac{n(n^2-1)(n^2-4)}{2!3!}-\dfrac{n(n^2-1)(n^2-4)(n^2-9)}{3!4!}+\dots$
$=s_1$ when $n$ is even and $s_2$ when $n$ is odd
then prove that $s_1+s_2=0$
I know that when I put any even value (say $n=2$), I get $s_1$ and when I put any odd value (say $n=5$), I get $s_2$. And these add up to zero. But this is just verification, not a proof. I want to know how should I prove this.
 [/INST] A neat question.
In fact, this is a particular case of a more general identity. The left-hand side may be rewritten as 
\begin{gather}
n-\dfrac{n(n^2-1)}{2!}+\dfrac{n(n^2-1)(n^2-4)}{2!3!}-\dfrac{n(n^2-1)(n^2-4)(n^2-9)}{3!4!}+\cdots\\ = \sum_{k=0}^{n-1} (-1)^k{n\choose k+1} {n+k \choose k}  = \sum_{k=0}^{n-1} {n \choose n-1-k} {-n-1 \choose k},
\end{gather}
where ${x\choose k} = x(x-1)\cdots(x-k+1)/k!$, $k\ge 1$; ${x\choose 0} = 1$.
Now this is a standard fact that 
$$\sum_{k=0}^m {a \choose k}{b \choose m-k} = {a+b \choose m}.$$
It is well known for non-negative integers $a$, $b$, $m$, but can be easily shown for any $a,b\in \mathbb R$, $m\in \mathbb Z_+$ by noting that $(1+z)^a = \sum_{k\ge 0} {a\choose k} z^k$.
Returning to your problem, we can take $a=n$, $b=-n-1$, $m=n-1$, so 
\begin{gather}
\sum_{k=0}^{n-1} {n \choose n-1-k} {-n-1 \choose k} =  {-1 \choose n-1} = (-1)^{n-1},
\end{gather}
as required.
</s>