if ${\left(1-\mathrm{x}+{\mathrm{x}}^2\right)}^{\mathrm{n}}={\mathrm{a}}_0+{\mathrm{a}}_1\mathrm{x}+{\mathrm{a}}_2{\mathrm{x}}^2+\dots +{\mathrm{a}}_{2\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}}$ then ${\mathrm{a}}_0+{\mathrm{a}}_2+{\mathrm{a}}_4+\dots +{\mathrm{a}}_{2\mathrm{n}}$ is equal to

${\left(1-\mathrm{x}+{\mathrm{x}}^2\right)}^{\mathrm{n}}={\mathrm{a}}_0+{\mathrm{a}}_1\mathrm{x}+{\mathrm{a}}_2{\mathrm{x}}^2+\dots +{\mathrm{a}}_{2\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}}$ on putting X=1, we get

$(1-1+1{)}^{\mathrm{n}}={\mathrm{a}}_0+{\mathrm{a}}_1+{\mathrm{a}}_2+\dots +{\mathrm{a}}_{2\mathrm{n}}$ 

$\Rightarrow 1={\mathrm{a}}_0+{\mathrm{a}}_1+{\mathrm{a}}_2+\dots +{\mathrm{a}}_{2\mathrm{n}} .....\left(\mathrm{i}\right)$

Again, putting  x= - 1, we get

$3^{\mathrm{n}}={\mathrm{a}}_0-{\mathrm{a}}_1+{\mathrm{a}}_2-\dots +{\mathrm{a}}_{2\mathrm{n}}$

$\frac{3^{\mathrm{n}}+1}{2}={\mathrm{a}}_0+{\mathrm{a}}_2+{\mathrm{a}}_4+\dots +{\mathrm{a}}_{2\mathrm{n}}$