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  
 A: Denote this sum by $S_n$ then from $${n+1 \choose m} = {n \choose m} + {n \choose m-1}$$ it follows that $$S_{n+1}=S_{n}-S_{n-1}.$$ Now start from $S_0= S_1= 1$ and you will notice the pattern.
A: 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:
$$\sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k\binom{n-k}{k}
= \begin{cases}
+1, & n \equiv 0, 1 \pmod 6\\
-1, & n \equiv 3, 4 \pmod 6\\
0, & n \equiv 2, 5 \pmod 6
\end{cases}$$
A: Hint:
$$\binom{n}{0}= \mbox{ coefficient of } x^n \mbox{ in } (1+x)^n \\
\binom{n-1}{1}= \mbox{ coefficient of } x^n \mbox{ in } x^2(1+x)^{n-1} \\
\binom{n-2}{2}= \mbox{ coefficient of } x^n \mbox{ in } x^4(1+x)^{n-2} \\
...$$
Hint 2:
$$(1+x)^n-x^2(1+x)^{n-1}+x^4(1+x)^{n-2}-..=(1+x)^n \left(1-(\frac{x^2}{1+x})+ (\frac{x^2}{1+x})^2-(\frac{x^2}{1+x})^3 ...\right)$$
A: 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,
$$
\sum_{k=0}^n(-1)^k\binom{n-k}{k}=\left\{\begin{array}{r}
1&\text{if }n\equiv0,1\pmod6\\
0&\text{if }n\equiv2,5\pmod6\\
-1&\text{if }n\equiv3,4\pmod6
\end{array}\right.
$$
A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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

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


A: [Imported from a duplicate question]
Chebyshev polynomials of the second kind have the following representation:
$$ U_n(x)=\sum_{r\geq 0}\binom{n-r}{r}(-1)^r (2x)^{n-2r} \tag{1}$$
hence the wanted sum is just $U_n\left(\frac{1}{2}\right)$, and since $\frac{1}{2}=\cos\frac{\pi}{3}$,
$$ U_n\left(\frac{1}{2}\right) = \frac{\sin((n+1)\pi/3)}{\sin(\pi/3)}.\tag{2} $$
A: Here is an answer based upon a transformation of generating series.

We show 
\begin{align*}
\sum_{j=0}^k\binom{k-j}{j}(-1)^j
=\frac{(-1)^{\lfloor k/3\rfloor}+(-1)^{\lfloor (k+1)/3\rfloor}}{2}
\qquad\qquad k\geq 0\tag{1}
\end{align*}

where  $\lfloor x \rfloor$  denotes the floor function. We set as upper limit of the sum $j=k$ and use $\binom{p}{q}=0$ if $q>p$. We will also use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ in a series.

Note, the sum at the LHS of (1) is of the form
  \begin{align*}
  \sum_{j=0}^k\binom{k-j}{j}a_j
  \end{align*}
We can find in Riordan Array Proofs of Identities in Gould's Book by R. Sprugnoli in section 1.4 (A) a useful transformation formula:
Let $A(z)=\sum_{j=0}^\infty a_jz^j$ be a series, then the following holds
  \begin{align*}
 \frac{1}{1-z}A\left(\frac{z^2}{1-z}\right)
  =\sum_{k=0}^\infty\left(\sum_{j=0}^{k}\binom{k-j}{j}a_j\right)z^k
  \end{align*}
So, we have the following relationship
  \begin{align*}
[z^k]A(z)=a_k\qquad\longleftrightarrow\qquad
[z^k]\frac{1}{1-z}A\left(\frac{z^2}{1-z}\right)=\sum_{j=0}^{k}\binom{k-j}{j}a_j
\tag{2}\end{align*}

We obtain from (1) with $a_j=(-1)^j$ the generating function $A(z)$
\begin{align*}
A(z)=\sum_{j=0}^\infty(-z)^j=\frac{1}{1+z}
\end{align*}

and conclude according to (2)
  \begin{align*}
  \sum_{j=0}^k\binom{k-j}{j}(-1)^j&=[z^k]\frac{1}{1-z}\cdot\frac{1}{1+\frac{z^2}{1-z}}\tag{3}\\
  &=[z^k]\frac{1}{1-z+z^2}\tag{4}\\
  &=[z^k]\left(1+z-z^3-z^4+z^6+z^7-z^9-z^{10}+\cdots\right)\tag{5}\\
  &=\frac{(-1)^{\lfloor k/3\rfloor}+(-1)^{\lfloor (k+1)/3\rfloor}}{2}\tag{6}
  \end{align*}
  and the claim follows.

Comment:


*

*In (3) we apply the transformation formula (2).

*In (4) we do some simplifications.

*In (5) we expand the series with the help of Wolfram Alpha.

*In (6) we select the coefficient of $z^k$. 
Note: This binomial identity can also be found as (1.75) in R. Sprugnoli's paper.
