<s>[INST] Finding sum multinomial 
I did put x=$w, w^2 ,i ,-i$ but nothing of type is fetting 
formed. How come 1/2 is remaining constant.
That means because of some substitution,
$2a_o= a_1+ a_2$ is happening.
Also tried putting x=ix.
 [/INST] We consider
\begin{align*}
f(x)&:=(x^{2016}+x^{2008}+2)^{2010}=\sum_{j=0}^na_jx^j\\
g(x)&:=\frac{1}{3}\left(f(x)+f(xe^{2\pi i/3})+f(xe^{4\pi i/3}\right)=\sum_{{j=0\  }\atop{\ \ j\equiv 0(3)}}^n a_jx^j
\end{align*}

We obtain
  \begin{align*}
&\color{blue}{a_0-\frac{1}{2}a_1-\frac{1}{2}a_2+a_3-\frac{1}{2}a_4-\frac{1}{2}a_5+\cdots}\\
&\qquad=\sum_{{j=0\  }\atop{\ \ j\equiv 0(3)}}^n a_j-\frac{1}{2}\sum_{{j=0\  }\atop{\ \ j\not\equiv 0(3)}}^n a_j\\
&\qquad=g(1)-\frac{1}{2}\left(f(1)-g(1)\right)\\
&\qquad=\frac{3}{2}g(1)-\frac{1}{2}f(1)\\
&\qquad=\frac{1}{2}\left(f(e^{2\pi i/3})+f(e^{4\pi i/3})\right)\\
&\qquad=\frac{1}{2}\left(e^{2\pi i/3\cdot2016}+e^{2\pi i/3\cdot2008}+2\right)^{2010}\\
&\qquad\qquad+\frac{1}{2}\left(e^{4\pi i/3\cdot2016}+e^{4\pi i/3\cdot2008}+2\right)^{2010}\\
&\qquad=\frac{1}{2}\left(1+\left(-\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)+2\right)^{2010}\\
&\qquad\qquad+\frac{1}{2}\left(1+\left(-\frac{1}{2}-\frac{i\sqrt{3}}{2}\right)+2\right)^{2010}\\
&\qquad=\frac{1}{2^{2011}}\left(5+i\sqrt{3}\right)^{2010}+\frac{1}{2^{2011}}\left(5-i\sqrt{3}\right)^{2010}\\
&\qquad\,\,\color{blue}{<0}
\end{align*}
where the last line was calculated with the help of Wolfram Alpha. We conclude option (2) is valid.

</s>