Question

If $a_0,a_1,a_2,.....$ be the coefficients in the expansion of $(1+x+x^2{)}^n$ in ascending powers of x, then prove that :
$(a_0+a_3+a_6+...)=(a_1+a_4+a_7+....)==(a_2+a_5+a_8+...)=3^{n-1}$

Answer 1

Put $x=\omega$ and note that $1+\omega +{\omega }^2=0$
where $\omega =\frac{-1}{2}+i\frac{\sqrt{3}}{2}$ and ${\omega }^2=\frac{-1}{2}-i\frac{\sqrt{3}}{2}$
$\Delta =(a_0+a_3+a_6+...)+\omega (a_1+a_4+...)+{\omega }^2(a_2+a_5+...)$
or $\Delta =A+B\omega +C{\omega }^2=0$
Equating real and imaginary parts on the both sides
$A-\frac{B}{2}-\frac{C}{2}=0,\frac{\sqrt{3}}{2}(B-C)=0$
$\therefore B=C$ and hence $A=B$
$\therefore A=B=C$
Again putting $x=1$ in the given relation, we get
$3^n=$ Sum of all coefficients
$=A+B+C$
or $3^n=3A\therefore A=3^{n-1}=B=C$