If ${\left(1+\mathrm{x}+{\mathrm{x}}^2\right)}^{\mathrm{n}}={\mathrm{a}}_0+{\mathrm{a}}_1\mathrm{x}+{\mathrm{a}}_2{\mathrm{x}}^2+\dots ..+{\mathrm{a}}_{2\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}}\text{ then }{\mathrm{a}}_0^2-{\mathrm{a}}_1^2+{\mathrm{a}}_2^2-{\mathrm{a}}_3^2+\dots ..+{\mathrm{a}}_{2\mathrm{n}}^2$ is equals 

$$\begin{gathered} {\mathrm{a}}_0^2-{\mathrm{a}}_1^2+{\mathrm{a}}_2^2\dots .+{\mathrm{a}}_{\mathrm{n}}^2= \mathrm{coefficient} \mathrm{of} {\mathrm{x}}^{2\mathrm{n}}\text{ in }{\left(1+\mathrm{x}+{\mathrm{x}}^2\right)}^{\mathrm{n}}{\left({\mathrm{x}}^2-\mathrm{x}+1\right)}^{\mathrm{n}} \\ {\left(1+x^2+x^4\right)}^n\text{=}\underset{r=0}{\overset{2n}{\sum {\text{a}}_r{\text{x}}^{2r_{ }}}} \\ \therefore x^{2n}coefficient is a_n \end{gathered}$$