Question

If $ (1+x+x^2)^n = a_o +a_1x +a_2x^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, $\left(1+x+x^{2}\right)^{n}$
$=a_{0}+a_{1}x +a_{2}x^{2}$
$+a_{3}x^{3}+\ldots+a_{2n} x^{2n}$
$\left(1+x+x^{2}\right)^{n} =\left(a_{0}+a_{3}x^{6}+a_{9}x^{9}+\ldots\right)$
$+x\left(a_{1}+a_{4}x^{3}+a_{7}x^{6}+\ldots\right)+x^{2}\left(a_{2}+a_{5}x^{3}+a_{8}x^{6}+\ldots\right)$
Let $E_{1}=a_{0}+a_{3}+a_{9}+\ldots$
$E_{2}=a_{1}+a_{4}+a_{7}+\ldots$
and $E_{3}=a_{2}+a_{5}+a_{8}+\ldots$
put $x=1, \omega, \omega^{2}$ respectively, we get
$\left(1+1+1\right)^{n} =E_{1}+E_{2}+E_{3} $
$\Rightarrow 3^{n}=E_{1}+E_{2}+E_{3} \ldots\left(i\right)$
$\left(1+\omega+\omega^{2}\right)^{n} =E_{1}+\omega\,E_{2}+\omega^{2}E_{3}$
$\Rightarrow 0=E_{1}+\omega\, E_{2}+\omega^{2}E_{3} \ldots\left(ii\right)$
and $\left(1+\omega^{2}+\omega^{4}\right)^{n} =E_{1}+\omega^{2}E_{2}+\omega^{4}E_{3}$
$\Rightarrow 0=E_{1}+\omega^{2}E_{2}+\omega\,E_{3} \ldots\left(iii\right)$
On adding Eqs. $\left(i\right)$, $\left(ii\right)$ and $\left(iii\right)$, we get
$3^{n}=3E_{1}+\left(1+\omega+\omega^{2}\right)E_{2}+\left(1+\omega^{2}+\omega\right)E_{3}$
$\Rightarrow 3^{n}=3E_{1}+0+0$
$\Rightarrow E_{1}=3^{n-1}$