<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] 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.
</s>