Let the given expansion be represented by the polynomial P(x):

P(x) = (1 - x + x²)ⁿ = a₀ + a₁x + a₂x² + ... + a_2nx²ⁿ

Deriving the Sum of Even Coefficients

To find the sum of coefficients with even indices (i.e., a₀ + a₂ + a₄ + ... + a_2n), we can evaluate the polynomial P(x) at specific values of x.

Step 1: Evaluate P(x) at x = 1

Substituting x = 1 into the polynomial expansion:

P(1) = a₀ + a₁(1) + a₂(1)² + ... + a_2n(1)²ⁿ

P(1) = a₀ + a₁ + a₂ + ... + a_2n

Now, let's evaluate the original expression at x = 1:

P(1) = (1 - 1 + 1²)ⁿ = (1 - 1 + 1)ⁿ = (1)ⁿ = 1

Therefore, the sum of all coefficients is 1:

\(\sum_{i=0}^{2n} a_i = a_0 + a_1 + a_2 + ... + a_{2n} = 1\)

Step 2: Evaluate P(x) at x = -1

Substituting x = -1 into the polynomial expansion:

P(-1) = a₀ + a₁(-1) + a₂(-1)² + a₃(-1)³ + ... + a_2n(-1)²ⁿ

P(-1) = a₀ - a₁ + a₂ - a₃ + ... + a_2n

Now, let's evaluate the original expression at x = -1:

P(-1) = (1 - (-1) + (-1)²)ⁿ = (1 + 1 + 1)ⁿ = (3)ⁿ = 3ⁿ

Therefore, we have:

\(\sum_{i=0}^{2n} a_i (-1)^i = a_0 - a_1 + a_2 - ... + a_{2n} = 3^n\)

Step 3: Combine the results

Let \(\text{Sum}_{\text{even}} = a_0 + a_2 + a_4 + ... + a_{2n}\)

Let \(\text{Sum}_{\text{odd}} = a_1 + a_3 + a_5 + ... + a_{2n-1}\)

From Step 1, we have:

\(\text{Sum}_{\text{even}} + \text{Sum}_{\text{odd}} = 1\)

From Step 2, we have:

\(\text{Sum}_{\text{even}} - \text{Sum}_{\text{odd}} = 3^n\)

To find \(\text{Sum}_{\text{even}}\), we add the two equations:

(Sum_even + Sum_odd) + (Sum_even - Sum_odd) = 1 + 3ⁿ

\(2 \cdot \text{Sum}_{\text{even}} = 1 + 3^n\)

Dividing by 2, we get:

\(\text{Sum}_{\text{even}} = \frac{1 + 3^n}{2}\)

Thus, \(\text{a}_0 + \text{a}_2 + \text{a}_4 + ... + \text{a}_{2n} = \frac{3^n + 1}{2}\).