Question

If \( (1 + x - 2x^2)^6 = 1 + a_1x + a_2x^2 + \ldots + a_{12}x^{12} \), then the value of \( a_2 + a_4 + a_6 + \ldots + a_{12} \) is:

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

Hint:

Always check if the constant term is included when calculating the sum of even degree coefficients. Subtract it if the question asks for only higher-degree terms (like \( x^2, x^4, \) etc.).

Answer 1

The Correct Option is B

Step 1: Expand the given expression.
We are given: \[ (1 + x - 2x^2)^6 \] Apply the multinomial expansion formula: \[ (1 + x + (-2x^2))^6 = \sum \frac{6!}{r!s!t!} (1)^r (x)^s (-2x^2)^t \] where \( r+s+t = 6 \). Simplifying: \[ = \sum \frac{6!}{r!s!t!} (-2)^t x^{s+2t} \]
Step 2: Find sum of even degree coefficients.
Substitute \( x=1 \) and \( x=-1 \) into the polynomial:
\( P(1) \) = sum of all coefficients.
\( P(-1) \) = sum of coefficients of even degree terms minus the sum of coefficients of odd degree terms.
Let:
Sum of even degree coefficients = \( E \),
Sum of odd degree coefficients = \( O \).
Then: \[ P(1) = E + O, \quad P(-1) = E - O \] Adding these two: \[ P(1) + P(-1) = 2E \quad \Rightarrow \quad E = \frac{P(1) + P(-1)}{2} \] Now, calculate: \[ P(1) = (1 + 1 - 2(1)^2)^6 = (0)^6 = 0 \] \[ P(-1) = (1 - 1 - 2(1)^2)^6 = (-2)^6 = 64 \] Thus, \[ E = \frac{0 + 64}{2} = 32 \]
Step 3: Adjust for constant term.
The even degree terms include the constant term (coefficient of \( x^0 \)), which is \( 1 \). Thus, \[ a_2 + a_4 + a_6 + \ldots + a_{12} = 32 - 1 = 31 \]
Final Answer: \( 31 \).