Question:

Let $P(x) = (1 + x - 2x^2)^6$. The given expansion is $P(x) = 1 + a_1 x + a_2 x^2 + \cdots + a_{12} x^{12}$. From this expansion, the constant term is $a_0 = 1$. Step 1: Evaluate $P(x)$ at $x=1$ and $x=-1$. $$P(1) = (1 + 1 - 2(1)^2)^6 = (1 + 1 - 2)^6 = 0^6 = 0$$ $$P(-1) = (1 + (-1) - 2(-1)^2)^6 = (1 - 1 - 2)^6 = (-2)^6 = 64$$ Step 2: Determine the sum of coefficients of even powers of $x$. The general form of the polynomial is $P(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_{12} x^{12}$. Evaluating $P(1)$ and $P(-1)$ in terms of coefficients: $$P(1) = a_0 + a_1 + a_2 + a_3 + \cdots + a_{12}$$ $$P(-1) = a_0 - a_1 + a_2 - a_3 + \cdots + a_{12}$$ Adding these two equations yields: $$P(1) + P(-1) = 2(a_0 + a_2 + a_4 + \cdots + a_{12})$$ Substitute the calculated values of $P(1)$ and $P(-1)$: $$0 + 64 = 2(a_0 + a_2 + a_4 + \cdots + a_{12})$$ $$64 = 2(a_0 + a_2 + a_4 + \cdots + a_{12})$$ $$32 = a_0 + a_2 + a_4 + \cdots + a_{12}$$ Step 3: Calculate the value of $a_2 + a_4 + a_6 + \cdots + a_{12}$. Since $a_0 = 1$, substitute this value into the equation from Step 2: $$32 = 1 + a_2 + a_4 + \cdots + a_{12}$$ Therefore, the expression $a_2 + a_4 + a_6 + \cdots + a_{12}$ has the value: $$a_2 + a_4 + a_6 + \cdots + a_{12} = 32 - 1 = 31$$