<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] $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
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 \newcommand{\dd}{\mathrm{d}}
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{{n \choose 0} - {n - 1 \choose 1} + {n - 2 \choose 2} -
{n-3 \choose 3} + \cdots:\ ?}$.

\begin{align}
\sum_{k = 0}^{n}\pars{-1}^{k}{n - k \choose k} & =
\sum_{k = 0}^{\infty}\pars{-1}^{k}{n - k \choose n - 2k} =
\sum_{k = 0}^{\infty}\pars{-1}^{k}
\oint_{\verts{z}\ =\ 1^{-}}{\pars{1 + z}^{n - k} \over z^{n - 2k + 1}}
\,{\dd z \over 2\pi\ic}
\\[5mm] & =
\oint_{\verts{z}\ =\ 1^{-}}{\pars{1 + z}^{n} \over z^{n + 1}}
\sum_{k = 0}^{\infty}\pars{-\,{z^{2} \over 1 + z}}^{k}\,{\dd z \over 2\pi\ic}
\\[5mm] & =
\oint_{\verts{z}\ =\ 1^{-}}{\pars{1 + z}^{n} \over z^{n + 1}}
{1 \over 1 + z^{2}/\pars{1 + z}}\,{\dd z \over 2\pi\ic}
\\[5mm] & =
\oint_{\verts{z}\ =\ 1^{-}}
{\pars{1 + z}^{n + 1} \over z^{n + 1}\pars{z^{2} + z + 1}}\,{\dd z \over 2\pi\ic}
\,\,\,\stackrel{z\ \mapsto\ 1/z}{=}\,\,\,
\oint_{\verts{z}\ =\ 1^{\color{#f00}{+}}}
{\pars{1 + z}^{n + 1} \over z^{2} + z + 1}\,{\dd z \over 2\pi\ic}
\\[5mm] & =
{\pars{1 + r}^{n + 1} \over 2r + 1} +
{\pars{1 + \bar{r}}^{\, n + 1} \over 2\bar{r} + 1}\quad
\mbox{where}\quad
r \equiv -\,{1 \over 2} + {\root{3} \over 2}\,\ic =
\exp\pars{{2\pi \over 3}\,\ic}
\end{align}

Then $\ds{\pars{~\mbox{note that}\ 1 + r = \exp\pars{{\pi \over 3}\,\ic}~}}$,
\begin{align}
\sum_{k = 0}^{n}\pars{-1}^{k}{n - k \choose k} & =
2\,\Re\pars{\bracks{1 + r}^{n + 1} \over 2r + 1} =
2\,\Re\pars{\expo{\bracks{n + 1}\pi\ic/3} \over \root{3}\ic}
\\[5mm] & =
\bbox[10px,border:1px groove navy]{{2\root{3} \over 3}
\,\sin\pars{\bracks{n + 1}\pi \over 3}} =
\bbox[10px,border:1px groove navy]{%
{\root{3} \over 3}\,\sin\pars{n\pi \over 3} + \cos\pars{n\pi \over 3}} 
\end{align}


This result generates the sequence $\ds{\pars{~\mbox{starting with}\ n = 0~}}$:

$$
\underbrace{1, 1, 0, -1, -1, 0}_{},\
\underbrace{1, 1, 0, -1, -1, 0}_{},\ \underbrace{1, 1, 0, -1, -1, 0}\ldots
$$

because
  $\ds{\sin\pars{\bracks{n + 1}\pi \over 3} =
\sin\pars{\bracks{n + \color{#f00}{6} + 1}\pi \over 3} =
\sin\pars{\bracks{n + 7}\pi \over 3}}$.


</s>