If ${(1+\mathrm{x}-2{\mathrm{x}}^2)}^6=1+{\mathrm{a}}_1\mathrm{x}+{\mathrm{a}}_2{\mathrm{x}}^2+\dots +{\mathrm{a}}_{12}{\mathrm{x}}^{12}$  then ${\mathrm{a}}_2+{\mathrm{a}}_4+{\mathrm{a}}_6+\dots +{\mathrm{a}}_{12}=$

Given,

${(1+\mathrm{x}-2{\mathrm{x}}^2)}^6=1+{\mathrm{a}}_1\mathrm{x}+{\mathrm{a}}_2{\mathrm{x}}^2+\dots +{\mathrm{a}}_{12}{\mathrm{x}}^{12}$

Putting x = 1, we get

$0=1+{\mathrm{a}}_1+{\mathrm{a}}_2+\dots +{\mathrm{a}}_{12}---\left(1\right)$

Putting x = – 1, we get

$64=1-{\mathrm{a}}_1+{\mathrm{a}}_2-\dots +{\mathrm{a}}_{12}----\left(2\right)$

Adding Eq. (1) and (2), we get

$\begin{array}{rl} & 64=2(1+{\mathrm{a}}_2+{\mathrm{a}}_4+\dots )\\ & \therefore {\mathrm{a}}_2+{\mathrm{a}}_4+{\mathrm{a}}_6+\dots +{\mathrm{a}}_{12}=31\end{array}$