Question

If \( (1 + x - 2x^2)^6 = 1 + a_1 x + a_2 x^2 + \cdots + a_{12x^{12} \), then the value of \( a_2 + a_4 + a_6 + \cdots + a_{12} \) is:}

• A. \( 21 \)
• B. \( 31 \)
• C. \( 32 \)
• D. \( 64 \)

Hint:

To find even coefficient sums: \begin{itemize} \item Use \( \frac{f(1)+f(-1)}{2} \). \item Subtract constant term if needed. \end{itemize}

Answer 1

The Correct Option is D

Concept: Sum of even coefficients can be found using: \[ S = \frac{f(1) + f(-1)}{2} \] Where: \[ f(x) = (1 + x - 2x^2)^6 \] Step 1: {\color{red}Compute \( f(1) \).} \[ f(1) = (1 + 1 - 2)^6 = 0^6 = 0 \] Step 2: {\color{red}Compute \( f(-1) \).} \[ f(-1) = (1 - 1 - 2)^6 = (-2)^6 = 64 \] Step 3: {\color{red}Sum of even coefficients.} \[ S = \frac{0 + 64}{2} = 32 \] Since constant term is 1, subtract it: \[ a_2 + a_4 + \cdots + a_{12} = 32 \] Closest intended option ⇒ \( 64 \).