Question

If the expansion of $(1+x+x^2{)}^n$ be $a_0+a_{1}x+a_2{x}^2+.....+a_r{x}^r+.....+a_{2n}{x}^{2n}$,
show that
$a_0+a_3+a_6+.....=a_1+a_4+a_7+.....=a_2+a_5+a_8+....=3^{n-1}$.

Answer 1

Given $(1+x+x^2{)}^n=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4\dots \dots \dots (1)$

Let $\omega ,{\omega }^2,{\omega }^3$ be the cube root unity then $1+\omega +{\omega }^2=0$, and ${\omega }^3=1$, ${\omega }^4=\omega$, ${\omega }^5={\omega }^2$.....

For $x=x\omega ,(1+\omega x+{\omega }^2{x}^2{)}^n=a_0+a_1\omega x+a_2{\omega }^2{x}^2+a_3{x}^3+a_4\omega x^4+\dots \dots \dots (2)$

For $x=x{\omega }^2,(1+{\omega }^{2}x+\omega x^2{)}^n=a_0+a_1{\omega }^{2}x+a_2\omega x^2+a_3{x}^3+a_4\omega x^4+\dots \dots \dots (3)$

Put $x=1$ in $\left(1\right),\left(2\right),\left(3\right)$ and add to get

$3^n=3(a_0+a_3+a_6+\dots )$

Multiply $\left(1\right),\left(2\right),\left(3\right)$ by $1,{\omega }^2,\omega$ respectively, put $x=1$ and add to get

$3^n=3(a_1+a_4+a_7+\dots )$

Multiply $\left(1\right),\left(2\right),\left(3\right)$ by $1,\omega ,{\omega }^2$ respectively, put $x=1$ and add to get

$3^n=3(a_2+a_5+a_8+\dots )$

$\therefore a_0+a_3+a_6+\dots =a_1+a_4+a_7+\dots =a_2+a_5+a_8+\dots =3^{n-1}$