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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

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Correct option is B

Clearly A0=1Putting x = 1 and x =– 1, we get0=A0+A1+A2+…+A12and (−2)6=A0−A1+A2+…+A12Adding we get 64=2A0+A2+…+A12⇒A2+A4+…+A12=31

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

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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