Polynomial Expansion Coefficients Explained

This problem involves finding the sum of specific coefficients in the expansion of a polynomial. We are given the polynomial $P(x) = (1 + x - 2x^2)^6$. The expansion is provided in the form $P(x) = 1 + a_1x + a_2x^2 + \dots + a_{12}x^{12}$. Our goal is to calculate the value of $S = a_2 + a_4 + \dots + a_{12}$.

Strategy: Evaluating the Polynomial

A useful method to find sums of coefficients involves evaluating the polynomial at specific points, namely $x=1$ and $x=-1$. Let's apply this strategy.

Calculation Step 1: Evaluate P(1)

Substitute $x=1$ into the polynomial expression:

Simplify the expression inside the parenthesis:

From the expansion $P(x) = 1 + a_1x + a_2x^2 + \dots + a_{12}x^{12}$, substituting $x=1$ gives:

Equating the two results for $P(1)$:

Calculation Step 2: Evaluate P(-1)

Substitute $x=-1$ into the polynomial expression:

Simplify the expression inside the parenthesis:

From the expansion $P(x) = 1 + a_1x + a_2x^2 + \dots + a_{12}x^{12}$, substituting $x=-1$ gives:

Equating the two results for $P(-1)$:

Calculation Step 3: Combining Equations to Find the Sum

Let $E = a_2 + a_4 + \dots + a_{12}$ (the sum we want to find). Let $O = a_1 + a_3 + \dots + a_{11}$ (the sum of coefficients of odd powers).

Rewrite Equation 1 using $E$ and $O$: $1 + (a_1 + a_3 + \dots + a_{11}) + (a_2 + a_4 + \dots + a_{12}) = 0$ $1 + O + E = 0$ Rearranging gives: $O + E = -1$ (Equation 3)

Rewrite Equation 2 using $E$ and $O$: $1 - (a_1 + a_3 + \dots + a_{11}) + (a_2 + a_4 + \dots + a_{12}) = 64$ $1 - O + E = 64$ Rearranging gives: $-O + E = 63$ (Equation 4)

Now, we solve the system of linear equations formed by Equation 3 and Equation 4:

Add Equation 3 and Equation 4:

Solve for $E$ by dividing by 2:

So, the sum $a_2 + a_4 + \dots + a_{12}$ is 31.