If (1+x)^n = p_0 + p_1 x + p_2 x^2 + ldots + | vedclass

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Question

(A) Let $\omega$ be a complex cube root of unity,such that $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.Consider the expansion $(1+x)^n = \sum_{k=0}

Vedclass · Accepted Answer

$\frac{1}{3} \left[ 2^n + 2 \cos \frac{n \pi}{3} \right]$

Vedclass · Answer

(A) Let $\omega$ be a complex cube root of unity,such that $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.Consider the expansion $(1+x)^n = \sum_{k=0}^n p_k x^k$.Substituting $x = 1, \omega, \omega^2$:$S_0 = (1+1)^n = 2^n = p_0 + p_1 + p_2 + p_3 + \ldots$$S_1 = (1+\omega)^n = p_0 + p_1 \omega + p_2 \omega^2 + p_3 \omega^3 + \ldots = p_0 + p_1 \omega + p_2 \omega^2 + p_3 + \ldots$$S_2 = (1+\omega^2)^n = p_0 + p_1 \omega^2 + p_2 \omega^4 + p_3 \omega^6 + \ldots = p_0 + p_1 \omega^2 + p_2 \omega + p_3 + \ldots$Using the property $p_0 + p_3 + p_6 + \ldots = \frac{1}{3} [S_0 + S_1 + S_2]$:$S_1 + S_2 = (1+\omega)^n + (1+\omega^2)^n = (-\omega^2)^n + (-\omega)^n = (-1)^n (\omega^{2n} + \omega^n)$.Using $1+\omega = -\omega^2$ and $1+\omega^2 = -\omega$:$S_1 + S_2 = 2 \cos \frac{n \pi}{3} \cdot (-1)^n$ is not quite right; rather,$(1+\omega)^n + (1+\omega^2)^n = 2 \cos \frac{n \pi}{3}$.Thus,$p_0 + p_3 + p_6 + \ldots = \frac{1}{3} [2^n + 2 \cos \frac{n \pi}{3}]$.

If $(1+x)^n = p_0 + p_1 x + p_2 x^2 + \ldots + p_n x^n$,then the value of $p_0 + p_3 + p_6 + \ldots$ is equal to:

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