Sum of the series $\binom{n}{0}-\binom{n-1}{1}+\binom{n-2}{2}-\binom{n-3}{3}+..........$ 
The sum of the series $$\binom{n}{0}-\binom{n-1}{1}+\binom{n-2}{2}-\binom{n-3}{3}+..........$$

$\bf{My\; Try::}$ We can write it as $\displaystyle \binom{n}{0} = $ Coefficient of $x^0$ in $(1+x)^n$ 
Similarly $\displaystyle \binom{n-1}{1} = $ Coefficient of $x^1$ in $(1+x)^{n-1}$
Similarly $\displaystyle \binom{n-2}{2} = $ Coefficient of $x^2$ in $(1+x)^{n-2}$
Now, how can I solve it after that, Help Required, Thanks  

For any polynomial $f(x) = \sum\limits_{k=0}^{\deg f} a_k x^k$, we will use $[x^k] f(x)$ to denote $a_k$, the coefficient of $x^k$ in $f(x)$. Notice
$$\binom{n-k}{k} = [x^k](1+x)^{n-k} = [x^n] (x+x^2)^{n-k}$$
We have
$$\require{cancel}
\begin{align}\sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k\binom{n-k}{k}
&= \sum_{k=0}^n (-1)^k \binom{n-k}{k}\\ 
&= \sum_{k=0}^n [x^n] (x+x^2)^{n-k}(-1)^k
= [x^n]\left(\sum_{k=0}^n (x+x^2)^{n-k}(-1)^k\right)\\
&= [x^n] \frac{(x+x^2)^{n+1}-(-1)^{n+1}}{1+x+x^2}
= [x^n] \frac{[\color{red}{\cancel{\color{gray}{(x+x^2)^{n+1}}}} + (-1)^n](1-x)}{1-x^3}\\
&= (-1)^n [x^n] \frac{1-x}{1-x^3}\\
&= (-1)^n [x^n]\left((1 - x) + x^3(1-x) + x^6(1-x) + \cdots\right)
\end{align}
$$
Based on last expression, it is clear the sum depends only on $n \pmod 6$. 
Its value is given by following formula:
$$\sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k\binom{n-k}{k}
= \begin{cases}
+1, & n \equiv 0, 1 \pmod 6\\
-1, & n \equiv 3, 4 \pmod 6\\
0, & n \equiv 2, 5 \pmod 6
\end{cases}$$
