Question

lf $(1+x+x^2{)}^n=a_0+a_{1}x+a_2{x}^2+\dots +a_{2n}{x}^{2n}$, then $a_0+a_3+a_6+...=$

• A. $3^n$
• B. $3^{n-1}$
• C. $3^{n-2}$
• D. $1$

Answer 1

Answer: (B)

$3^{n-1}$

Substituting $x=1$ we get,
$3^n=a_0+a_1+a_2+a_3...+a_{2n}...\left(i\right)$
Substituting $x=\omega$ we get, (where $w$ is the non-real cube root of unity)
$(1+\omega +{\omega }^2)=0$
$\Rightarrow a_0+a_1\left(\omega \right)+a_2\left({\omega }^2\right)+a_3\left({\omega }^3\right)+...+a_{2n}\left({\omega }^{2n}\right)=0...\left(ii\right)$
Substituting $x={\omega }^2$ we get,
$(1+{\omega }^2+{\omega }^4)$
$=(1+{\omega }^2+\omega {\omega }^3)$
$=(1+{\omega }^2+\omega )$
$=0$
$\Rightarrow a_0+a_1{\omega }^2+a_2{\omega }^4+a_3{\omega }^6...a_{2n}{\omega }^{4n}=0...\left(iii\right)$
Adding (i), (ii) and (iii) we get
$3^n=3a_0+a_1(1+\omega +{\omega }^2)+a_2(1+{\omega }^2+{\omega }^4)+a_3(1+{\omega }^3+{\omega }^6)...$
$3^n=3a_0+a_2(1+{\omega }^2+{\omega }^3\omega )+a_3(1+1+{\omega }^3{\omega }^3)...$
$3^n=3(a_0+a_3+...)$
$a_0+a_3+a_6...=3^{n-1}$