If ${\left(1+\mathrm{x}-2{\mathrm{x}}^2\right)}^6=1+{\mathrm{a}}_1\mathrm{x}+{\mathrm{a}}_2{\mathrm{x}}^2+{\mathrm{a}}_3{\mathrm{x}}^3+\cdots ,$ then the value of ${\mathrm{a}}_2+{\mathrm{a}}_4+{\mathrm{a}}_6+\cdots +{\mathrm{a}}_{12}$ will be

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

Putting x = 1, we get

$0=1+{\mathrm{a}}_1+{\mathrm{a}}_2+{\mathrm{a}}_3+\dots +{\mathrm{a}}_{12}$   …….(1)

Putting x = -1, we get

$64=1-{\mathrm{a}}_1+{\mathrm{a}}_2-{\mathrm{a}}_3+\dots +{\mathrm{a}}_{12}$  ……..(2)

(1) + (2) gives

$\begin{array}{l}64=2\left[1+{\mathrm{a}}_2+{\mathrm{a}}_4+\dots {\mathrm{a}}_{12}\right]\\ 1+{\mathrm{a}}_2+{\mathrm{a}}_4+\dots +{\mathrm{a}}_{12}=32\\ {\mathrm{a}}_2+{\mathrm{a}}_4+\dots +{\mathrm{a}}_{32}=31\end{array}$