A Binomial Expansion (Sum of Coeffients)

Question

If $(1+x+x^2)^n = a_{0}+a_{1}x + a_{2}x^{2} +\cdots +a_{2n}x^{2n}$, then find the value of $a_{0}+a_{3}+a_{6}+\cdots $.

Answer 1

HINT:

Set $x=1,\omega,\omega^2$ in the given identity where $\omega$ is a complex cube root of unity

Then add the three results

Use the fact $1+\omega+\omega^2=0$

To find $a_0+a_2+a_4+\cdots,$ set $x=1,-1$