If (1 + x + x^2)^n = a_0 + a_1x + a_2x^2 + do | vedclass

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(B) Given the expansion: $(1 + x + x^2)^n = a_0 + a_1x + a_2x^2 + \dots + a_{2n}x^{2n}$.Let $f(x) = (1 + x + x^2)^n = \sum_{r=0}^{2n} a_r x^r$.To find

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$3^{n-1}$

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(B) Given the expansion: $(1 + x + x^2)^n = a_0 + a_1x + a_2x^2 + \dots + a_{2n}x^{2n}$.Let $f(x) = (1 + x + x^2)^n = \sum_{r=0}^{2n} a_r x^r$.To find the sum of coefficients $a_0 + a_3 + a_6 + \dots$,we use the roots of unity property.Let $\omega$ be a complex cube root of unity,such that $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.Evaluating $f(x)$ at $x = 1, \omega, \omega^2$:$f(1) = (1 + 1 + 1)^n = 3^n = a_0 + a_1 + a_2 + a_3 + \dots + a_{2n}$$f(\omega) = (1 + \omega + \omega^2)^n = 0^n = 0 = a_0 + a_1\omega + a_2\omega^2 + a_3 + a_4\omega + a_5\omega^2 + \dots$$f(\omega^2) = (1 + \omega^2 + \omega^4)^n = (1 + \omega^2 + \omega)^n = 0^n = 0 = a_0 + a_1\omega^2 + a_2\omega + a_3 + a_4\omega^2 + a_5\omega + \dots$Adding these three equations:$f(1) + f(\omega) + f(\omega^2) = 3^n + 0 + 0 = 3(a_0 + a_3 + a_6 + \dots)$Therefore,$a_0 + a_3 + a_6 + \dots = \frac{3^n}{3} = 3^{n-1}$.

If $(1 + x + x^2)^n = a_0 + a_1x + a_2x^2 + \dots + a_{2n}x^{2n}$,then $a_0 + a_3 + a_6 + \dots =$

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