Finding sum multinomial

Question

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.

Answer 1

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.

Answer 2

Similar to Markus:

Letting

$$ \begin{align*} f(x)&=(x^{2016}+x^{2008}+2)^{2010}=\sum_{j=0}^na_jx^j\\ h&=a_0 -\frac12 a_1 -\frac12 a_2 +a_3 -\frac12 a_4 -\frac12 a_5 \cdots \\ \theta &= e^{i2 \pi/3}=-\frac12 + i \frac{\sqrt{3}}{2} \end{align*} $$

Then, because $\operatorname{Re} [\theta^{3k}]=1 $ and $\operatorname{Re} [\theta^{3k+1}]=\operatorname{Re} [\theta^{3k+2}]=-\frac12 $:

$$ \begin{align*} h&= \operatorname{Re}[f(\theta)]\\ &= \operatorname{Re}[(1 + \theta +2)^{2010}]\\ &= \operatorname{Re}[(\frac52 +i \frac{\sqrt{3}}{2})^{2010}]\\ \end{align*} $$

Putting the number in polar form we can find that the resulting angle (multiplied by $2010$) falls in the third quadrant, hence $h<0$. But still we need a calculator to evaluate that, it seems.