Question

If ${\left(1+x+x^2\right)}^n=a_0+a_{1}x+a_2{x}^2.........+a_{2n}{x}^{2n}$ then the value of $a_0+a_3+a_6+.....$ is equal to

• A. $a_1+a_4+a_7$
• B. $a_2+a_5+a_8$
• C. $3^{n-1}$
• D. $\frac{2^n}{3}$

Answer 1

The correct option is C $3^{n-1}$

Let $x=1,$ then

$3^n=a_0+a_1+a_2+a_3+...+a_{2n}$

$=(a_0+a_3+a_6+...)+(a_1+a_4+...)+(a_2+a_5+...)$ ........$\left(1\right)$

Now,let $x=\omega$

${(1+\omega +{\omega }^2)}^n=a_0+a_1\omega +a_2{\omega }^2+a_3{\omega }^3+a_4{\omega }^4+...+a_{2n}{\omega }^{2n}$

$0=(a_0+a_3+a_6+...)+(a_1+a_4+...)\omega +(a_2+a_5+...){\omega }^2$ .........$\left(2\right)$

Put $x={\omega }^2$ we have

$(1+{\omega }^2+{\omega }^4)=a_0+a_1{\omega }^2+a_2{\omega }^4+a_3{\omega }^6+a_4{\omega }^8+...+a_{2n}{\left({\omega }^2\right)}^{2n}$

$0=(a_0+a_3+a_6+...)+(a_1+a_4+...){\omega }^2+(a_2+a_5+...)\omega$ ...........$\left(3\right)$

Now, adding $\left(1\right),\left(2\right)$ and $\left(3\right)$ we get

$3^n=3(a_0+a_3+a_6+...)+(1+\omega +{\omega }^2)(a_1+a_4+...)+(1+{\omega }^2+\omega )(a_2+a_5+...)$

$3^n=3(a_0+a_3+a_6+...)+0+0$

$\therefore a_0+a_3+a_6+...=\frac{3^n}{3}=3^{n-1}$