First slide

Algebra of complex numbers

Question

If ${(1 + x + x^{2})}^{n} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{2 n} x^{2 n},$ then value of $a_{0} + a_{3} + a_{6} + \dots$ is

Moderate

A

1
B

$2^{n}$
C

$2^{n - 1}$
D

$3^{n - 1}$

Solution

Putting $x = 1, ω, ω^{2}$ in
${(1 + x + x^{2})}^{n} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{2 n} x^{2 n},$ we get
$\begin{array}{l} (1 + 1 + 1)^{n} = a_{0} + a_{1} + a_{2} + a_{3} + \dots + a_{2 n}, \\ {(1 + ω + ω^{2})}^{n} = a_{0} + a_{1} ω + a_{2} ω^{2} + a_{3} ω^{3} + \dots + a_{2 n} ω^{2 n}, \end{array}$
and ${(1 + ω^{2} + ω^{4})}^{n} = a_{0} + a_{1} ω^{2} + a_{2} ω^{4} + a_{3} ω^{6} + \dots + a_{2 n} ω^{4 n}$
Adding the above three equations and using $1 + ω + ω^{2} = 0, ω^{3} = 1$ we get
$3^{n} = 3 (a_{0} + a_{3} + a_{6} + . . . .)$
$\Rightarrow a_{0} + a_{3} + a_{6} + . . . = 3^{n - 1}$

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