If $(3 + x^{2008} + x^{2009})^{2010} = a_{0} + a_{1}x + a_{2}x^2 + a_{3}x^3 + ... + a_{n}x^n $, then find the value of $$a_{0} - \frac{1}{2}a_{1} - \frac{1}{2}a_{2} +a_{3} - \frac{1}{2}a_{4} - \frac{1}{2}a_{5} + a_{6} - ...$$
My attempt: I tried to substitute $x=1 $ and $x=-1 $ separately in the given equation, but that was not very helpful for the question. The positive terms are three places apart from each other. Please guide how to solve this question.
First put $x =\omega$. We have: $$(3+\omega^{2008}+\omega^{2009})^{2010}=(3+\omega +\omega^2)^{2010} = (2+(1+\omega+\omega^2))^{2010}$$ $$=2^{2010} = a_0 + a_1\omega +a_2\omega^2+a_3 + a_4\omega+…(1)$$Similarly, on putting $x =\omega^2$, we have :$$(3+\omega^2 + \omega^4)^{2010}= (3+\omega+\omega^2)^{2010} =(2+(1+\omega+\omega^2))^{2010}$$ $$=2^{2010} = a_0 +a_1\omega^2 +a_2\omega +a_3 + a_4\omega^2+…(2)$$
Adding $(1)$ and $(2)$, we get, $$2^{2011} = 2a_0-a_1-a_2 + 2a_3-a_4-…$$ which gives:$$\boxed{2^{2010} = a_0-\frac{1}{2}a_2-\frac{1}{2}a_2+a_3-\frac{1}{2}a_4…}$$