I am stuck in the following problem that says:
If $(1+x)^n=a_0+a_1x+a_2x^2+........,$ show that $a_0+a_3+a_6+........=\frac23(2^{n-1}+\cos \frac{n\pi}{3})$
My try: Let $(1+x)^n=a_0+a_1x+a_2x^2+........,\tag1$ From (1), we get by using the fact $$1+i=\sqrt 2(\cos \frac{\pi}{4}+\sin \frac{\pi}{4})$$ and putting $x=i$ in both sides of (1) and comparing both sides of (1) $$a_0-a_2+a_4-........=2^{\frac n2 } \cos \frac{n \pi}{4}$$ and $$a_1-a_3+a_5-........=2^{\frac n2 } \sin \frac{n \pi}{4}$$.
Now, putting $x=w$ where $w$ is a cube root of unity , I get from (1), $$(1+w)^n=a_0+a_3+..............\implies a_0+a_3+...........=(-w^2)^n$$. I used the fact $1+w+w^2=0,w^3=1$
Now , I am stuck. Can someone help me in the right direction?
Notice that $a_k = \binom{n}{k}$. You want to compute :
$$ \sum_{0\le 3k\le n} \binom{n}{3k}$$ You have a sum of binomial coefficients so you will want to use the 3rd roots of unity 1, j and $j^2$. Let :
$$ A =\sum_{0\le 3k\le n} \binom{n}{3k}$$ $$ B =\sum_{0\le 3k+1\le n} \binom{n}{3k+1}$$ $$ C =\sum_{0\le 3k+2\le n} \binom{n}{3k+2}$$
Notice that : $$ 2^n = (1+1)^n = \sum_{k=0}^n \binom{n}{k} = A+B+C$$ $$ (1+j)^n = \sum_{k=0}^{n} \binom{n}{k}j^k = A + jB + j^2C$$ $$ (1+j^2)^n = \sum_{k=0}^{n}\binom{n}{k} j^{2k} = A + j^2b + jC$$
Furthermore, $1+j+j^2 = 0$. So that $$2^n + (1+j)^n + (1+j^2)^n = 3A$$ $j = e^{2i\pi/3}= -1/2 + i\dfrac{\sqrt{3}}{2}$ and $j^2 = e^{-2i\pi/3} = -1/2 - i\dfrac{\sqrt{3}}{2}$
So that : $$ A = \dfrac{2^n + \left(1/2 + i\frac{\sqrt{3}}{2}\right)^n + \left(1/2 -i\frac{\sqrt{3}}{2}\right)^n}{3}$$ $$ = \dfrac{2^n+e^{ni\pi/3}+ e^{-ni\pi/3}}{3}= \dfrac{2^n + 2\cos(n\pi/3)}{3}$$
Which is the desired result.