If $(1+x+x^2)^{25} =\sum^{50}_{r=0} a_r x^r$ then .... If $$(1+x+x^2)^{25} =\sum^{50}_{r=0} a_r x^r$$ 
then find :
$\sum^{16}_{r=0} a_{3r} =$
My approach : 
let (1+x) =t  therefore, 
$(1+x+x^2)^{25} =\sum^{50}_{r=0} a_r x^r$
=$(1+x+x^2)^{25} = (t+x^2)^{25}$
$= ^{25}C_0t^{25} +^{25}C_1 t^{24}x^2 +^{25}C_2 t^{23}x^4 +\cdots + x^{50}$
Also expanding the right hand side which is $\sum^{50}_{r=0} a_r x^r = a_0x^0 +a_1x+a_2x^2 +\cdots a_{50}x^{50}$
But unable to correlate this with the  given problem i.e how to find $\sum^{16}_{r=0} a_{3r}$ please guide further on this. thanks.

Hint: Evaluate at the two roots of the equation $1+x+x^2=0$ and at $x=1$, that is, at the three cube roots of unity, and add. The key fact is that if $\omega$ is one of the roots of $1+x+x^2=0$, then $\omega^2$ is the other, and $1^k+\omega^k+\omega^{2k}=0$ unless $k$ is divisible by $3$.
