<s>[INST] 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  
 [/INST] Generating Functions
$$
\begin{align}
\sum_{n=0}^\infty\sum_{k=0}^n(-1)^k\binom{n-k}{k}x^n
&=\sum_{k=0}^\infty\sum_{n=k}^\infty(-1)^k\binom{n-k}{k}x^n\\
&=\sum_{k=0}^\infty\sum_{n=0}^\infty(-1)^k\binom{n}{k}x^{n+k}\\
&=\sum_{n=0}^\infty x^n(1-x)^n\\
&=\frac1{1-x(1-x)}\\
&=\frac1{1-x+x^2}\\
&=\frac{1+x}{1+x^3}\\[6pt]
&=(1+x)(1-x^3+x^6-x^9+\dots)\\[11pt]
&=1+x-x^3-x^4+x^6+x^7-x^9-x^{10}+\dots
\end{align}
$$
Therefore,
$$
\sum_{k=0}^n(-1)^k\binom{n-k}{k}=\left\{\begin{array}{r}
1&\text{if }n\equiv0,1\pmod6\\
0&\text{if }n\equiv2,5\pmod6\\
-1&\text{if }n\equiv3,4\pmod6
\end{array}\right.
$$
</s>