Question

If $(1+x+x^2{)}^n=a_o+a_{1}x+a_2{x}^2+.....+a_{2n}{x}^{2n},$ then $a_o+a_3+a_6+.....=$

• A. $3^{n+1}$
• B. $3^n$
• C. $3^{n-1}$
• D. None of these

Answer 1

Answer: (C)

Given, ${\left(1+x+x^2\right)}^n$
$=a_0+a_{1}x+a_2{x}^2$
$+a_3{x}^3+\dots +a_{2n}{x}^{2n}$
${\left(1+x+x^2\right)}^n=\left(a_0+a_3{x}^6+a_9{x}^9+\dots \right)$
$+x\left(a_1+a_4{x}^3+a_7{x}^6+\dots \right)+x^2\left(a_2+a_5{x}^3+a_8{x}^6+\dots \right)$
Let $E_1=a_0+a_3+a_9+\dots$
$E_2=a_1+a_4+a_7+\dots$
and $E_3=a_2+a_5+a_8+\dots$
put $x=1,\omega ,{\omega }^2$ respectively, we get
${\left(1+1+1\right)}^n=E_1+E_2+E_3$
$\Rightarrow 3^n=E_1+E_2+E_3\dots \left(i\right)$
${\left(1+\omega +{\omega }^2\right)}^n=E_1+\omega E_2+{\omega }^2{E}_3$
$\Rightarrow 0=E_1+\omega E_2+{\omega }^2{E}_3\dots \left(ii\right)$
and ${\left(1+{\omega }^2+{\omega }^4\right)}^n=E_1+{\omega }^2{E}_2+{\omega }^4{E}_3$
$\Rightarrow 0=E_1+{\omega }^2{E}_2+\omega E_3\dots \left(iii\right)$
On adding Eqs. $\left(i\right)$, $\left(ii\right)$ and $\left(iii\right)$, we get
$3^n=3E_1+\left(1+\omega +{\omega }^2\right)E_2+\left(1+{\omega }^2+\omega \right)E_3$
$\Rightarrow 3^n=3E_1+0+0$
$\Rightarrow E_1=3^{n-1}$