If 1+x−2x26=∑r=012 Arxr , then value of A2 + A4 +A6+…+A12 is

Clearly $A_{0} = 1$

Putting $x = 1$ and $x = - 1,$ we get

$0 = A_{0} + A_{1} + A_{2} + \dots + A_{12}$

and $(- 2)^{6} = A_{0} - A_{1} + A_{2} + \dots + A_{12}$

Adding we get

$\begin{aligned} 64 = 2 ⌊A_{0} + A_{2} + \dots + A_{12}⌋ \\ \Rightarrow A_{2} + A_{4} + \dots + A_{12} = 31 \end{aligned}$

If ${(1 + x - 2 x^{2})}^{6} = \sum_{r = 0}^{12} A_{r} x^{r}$ , then value of $A_{2} + A_{4}$ $+ A_{6} + \dots + A_{12}$ is