Q74 / 155 · WBJEE 2025 · Mathematics

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

  • A
    $21$
  • Correct answer
    B
    $31$
  • C
    $32$
  • D
    $64$

Solution

$\begin{aligned} & \left(1+x-2 x^2\right)^6=1+a_1 x+a_2 x^2+\ldots+a_{12} x^{12} \\ & x=1 \Rightarrow 0=1+a_1+a_2+\ldots+a_{12} \\ & x=-1 \Rightarrow 64=1-a_1+a_2+\ldots+a_{12} \\ & \text { Adding } 64=2\left(1+a_2+a_4+\ldots+a_{12}\right) \\ & a_2+a_4+\ldots+a_{12}=31\end{aligned}$

Hence, the correct answer is option: 'B'.

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