Question

If $(1+x+x^2{)}^n=a_0+a_{1}x+a_{2}x+....+a_{2n}{x}^{2n}$, then the value of $a_0+a_3+a_6+.=$

• A. $3^n$
• B. $3^{n-1}$
• C. $0$
• D. $noneofthese$

Answer 1

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

We have,

${(1+x+x^2)}^n=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+...+a_{2n}{x}^{2n}$

Now, put $x=1$ and we get,

${(1+1+1)}^n=a_0+a_1+a_2+a_3+...+a_{2n}$

$[(a_0+a_3+a_6+.....)+(a_1+a_4+a_5+.....)+(a_2+a_5+a_8+....)]=3^n.......\left(1\right)$

Put $x=\omega$

And we get,

${(1+\omega +{\omega }^2)}^n=a_0+a_1\omega +a_2{\omega }^2+....+a_n{\omega }^{2n}(\therefore 1+\omega +{\omega }^2=0)$

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

Finally putting $x={\omega }^2$ and we get,

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

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

Now,

Adding equation (1)+(2) and (3) to and we get,

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

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

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

Hence, this is the answer.