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If \((1 + x - 2x^2)^6 = A_0 + \sum_{r=1}^{12} A_r x^r\), then the value of \(A_2 + A_4 + A_6 + \cdots + A_{12}\) is

Question and options

If (1+x−2x2)6=A0+∑r=112Arxr(1 + x - 2x^2)^6 = A_0 + \sum_{r=1}^{12} A_r x^r, then the value of A2+A4+A6+⋯+A12A_2 + A_4 + A_6 + \cdots + A_{12} is
  1. 31
  2. 32
  3. 30
  4. 29

Answer and explanation

Correct answer: A - 31
f(1) = 0 and f(-1) = (-2)^6 = 64, so sum of even-index coefficients = (0+64)/2 = 32. Subtract A0 = 1: 31.

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