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  

[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 $\frac{1}{2}=\cos\frac{\pi}{3}$,
$$ U_n\left(\frac{1}{2}\right) = \frac{\sin((n+1)\pi/3)}{\sin(\pi/3)}.\tag{2} $$
