If \(\left(1-x+x^{2}\right) n=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{2 n} x^{2 n}\) then \(a_{0}+a_{2}\) \(+a_{4}+\cdots+a_{2 n}=\) (a) \(\left[\left(3^{\mathrm{n}}-1\right) / 2\right]\) (b) \(\left[\left(1-3^{\mathrm{n}}\right) / 2\right]\) (c) \(\left[\left(3^{\mathrm{n}}+1\right) / 2\right]\) (d) \(\left[\left(3^{\mathrm{n}+1}\right) / 2\right]\)

In this step, we need to find a value for x that extracts the even-indexed coefficients when the polynomial is evaluated using that value. One such value is a cube root of unity, ω, which satisfies \(ω^3 = 1\) and \(ω \neq 1\). Notice that \(ω^2\) is also a cube root of unity. So, we have: \(ω^3 = (ω^2)^3 = 1\)

Now, substitute x with ω and \(ω^2\) into the given polynomial expression: \(P(ω) = \left(1-ω+ω^{2}\right) n = a_0 + a_1ω + a_2ω^2 +...+ a_{2n}ω^{2n}\) \(P(ω^2) = \left(1-ω^2+ω^{4}\right) n = a_0 + a_1ω^2 + a_2ω^4 +...+ a_{2n}ω^{4n}\)

Now add the two polynomials together: \(P(ω) + P(ω^2) = \left[a_0 + a_1ω + a_2ω^2 +...+ a_{2n}ω^{2n}\right] + \left[a_0 + a_1ω^2 + a_2ω^4 +...+ a_{2n}ω^{4n}\right]\) This simplifies to: \(P(ω) + P(ω^2) = 2[a_0 + a_2(ω^2 + ω^4) +...+ a_{2n}(ω^{2n} + ω^{4n})]\)

Now evaluate n using ω and ω^2: \(n(1 - ω + ω^2) = n(1 - ω^2 + ω^4)\) From the properties of cube roots of unity, we know that \(1 + ω + ω^2 = 0\), so we have: \(n(1 - ω) = n(1 - ω^2)\) Notice that: \(1 - ω = ω^2(1 - ω^2)\) and \(1 - ω^2 = ω(1 - ω)\) Thus: \(nω^2 = nω\) Which means: \(n = (1 - ω)n\)

We evaluate the sum using the expression found in Step 3: \(P(ω) + P(ω^2) = 2[a_0 + a_2(ω^2 + ω^4) +...+ a_{2n}(ω^{2n} + ω^{4n})]\) Since \(n = (1 - ω)n\), \(P(1) + P(ω^2) = 2\left[\frac{(1 - ω)n}{1 - ω}\right]\) \(2\left[\frac{-ωn}{1 - ω}\right]\) Using the property \(ω^3 = 1\), we get: \(2\left[\frac{(-ω)(1 - ω^3)n}{1 - ω}\right] = 2\left[\frac{(1 + ω)n}{1 - ω}\right]\) Thus, the sum of even-indexed coefficients is: \(a_0 + a_2 + a_4 + ... + a_{2n} = \frac{(1 - ω^4)n}{1 - ω}\) The denominator can be written as: \(1 - ω = (1 - ω)(1 + ω + ω^2) = (1 - ω^3)\) So, \(a_0 + a_2 + a_4 + ... + a_{2n} = \frac{(1 - ω^4)n}{1 - ω^3} = \frac{(1 - ω)n}{1 - ω}\) Finally, by comparing this expression to the given options, we can see that it matches option (c): \(a_0 + a_2 + a_4 + ... + a_{2n} = \left[\left(3^{\mathrm{n}}+1\right) / 2\right]\)