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

$$(1 + x - 2x^{2})^{6} = A_{0} + \sum_{r = 1}^{12} A_{r}X^{r}$$

Substitute x = 1 

$$(1 + 1 - 2(1)^{2})^{6} = A_{0} + A_{1}(1^1)+A_{2}(1^2)+A_{3}(1^3)+A_{4}(1^4)+......+A_{12}(1^12)$$

$$(0)^{6} = A_{0} + A_{1}+A_{2}+A_{3}+A_{4}+......+A_{12}$$

$$0 = A_{0} + A_{1}+A_{2}+A_{3}+A_{4}+......+A_{12}\rightarrow1$$

Substitute x = -1

$$(1 + (-1) - 2(-1)^{2})^{6} = A_{0} + A_{1}(-1)^1+A_{2}(-1)^2+A_{3}(-1)^3+A_{4}(-1)^4+......+A_{12}(-1)^12$$

$$(-2)^{6} = A_{0} - A_{1}+A_{2}-A_{3}+A_{4}+......+A_{12}$$

$$64 = A_{0} - A_{1}+A_{2}-A_{3}+A_{4}+......+A_{12}\rightarrow2$$

Adding eq. 1 and eq. 2

$$64 =2( A_{0}+A_{2}+A_{4}+A_{6}+A_{8}+A_{10}+A_{12})$$

$$32 =A_{0}+A_{2}+A_{4}+A_{6}+A_{8}+A_{10}+A_{12}$$     

{To calculate the value of $$A_{0}$$, substitute the original eq. with x=0 => $$(1 + 0 - 2(0)^{2})^{6} = A_{0}$$ =>$$A_{0}=1$$}

$$32 =1+A_{2}+A_{4}+A_{6}+A_{8}+A_{10}+A_{12}$$

$$A_{2}+A_{4}+A_{6}+A_{8}+A_{10}+A_{12}=31$$