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Mathematics Question on Complex Numbers and Quadratic Equations

If (1+x+x2)n=ao+a1x+a2x2+.....+a2nx2n,(1+x+x^2)^n = a_o +a_1x +a_2x^2 +.....+a_{2n} x^{2n}, then ao+a3+a6+.....=a_o+a_3+a_6+.....=

A

3n+13^{n+1}

B

3n3^{n}

C

3n−13^{n-1}

D

None of theseNone\, of\, these

Answer

3n−13^{n-1}

Explanation

Solution

Given, (1+x+x2)n=a0+a1x+a2x2\left(1+x+x^{2}\right)^{n} =a_{0}+a_{1}x +a_{2}x^{2}
+a3x3+…+aznx2n+a_{3}x^{3}+\ldots+a_{zn} x^{2n}
(1+x+x2)n=(a0+a3x6+a9x9+…)\left(1+x+x^{2}\right)^{n} =\left(a_{0}+a_{3}x^{6}+a_{9}x^{9}+\ldots\right)
+x(a1+a4x3+a7x6+…)+x2(a2+a5x3+a8x6+…)+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 E1=a0+a3+a9+…E_{1}=a_{0}+a_{3}+a_{9}+\ldots
E2=a1+a4+a7+…E_{2}=a_{1}+a_{4}+a_{7}+\ldots
and E3=a2+a5+a8+…E_{3}=a_{2}+a_{5}+a_{8}+\ldots
put x=1,ω,ω2x=1, \omega, \omega^{2} respectively, we get
(1+1+1)n=E1+E2+E3\left(1+1+1\right)^{n} =E_{1}+E_{2}+E_{3}
⇒3n=E1+E2+E3…(i)\Rightarrow 3^{n}=E_{1}+E_{2}+E_{3} \ldots\left(i\right)
(1+ω+ω2)n=E1+ω E2+ω2E3\left(1+\omega+\omega^{2}\right)^{n} =E_{1}+\omega\,E_{2}+\omega^{2}E_{3}
⇒0=E1+ω E2+ω2E3…(ii)\Rightarrow 0=E_{1}+\omega\, E_{2}+\omega^{2}E_{3} \ldots\left(ii\right)
and (1+ω2+ω4)n=E1+ω2E2+ω4E3\left(1+\omega^{2}+\omega^{4}\right)^{n} =E_{1}+\omega^{2}E_{2}+\omega^{4}E_{3}
⇒0=E1+ω2E2+ω E3…(iii)\Rightarrow 0=E_{1}+\omega^{2}E_{2}+\omega\,E_{3} \ldots\left(iii\right)
On adding Eqs. (i)\left(i\right), (ii)\left(ii\right) and (iii)\left(iii\right), we get
3n=3E1+(1+ω+ω2)E2+(1+ω2+ω)E33^{n}=3E_{1}+\left(1+\omega+\omega^{2}\right)E_{2}+\left(1+\omega^{2}+\omega\right)E_{3}
⇒3n=3E1+0+0\Rightarrow 3^{n}=3E_{1}+0+0
⇒E1=3n−1\Rightarrow E_{1}=3^{n-1}