The sum of the coefficients of even power of x in the expansion of ${\left(1+\mathrm{x}+{\mathrm{x}}^2+{\mathrm{x}}^3\right)}^5$ is

${\left(1+\mathrm{x}+{\mathrm{x}}^2+{\mathrm{x}}^3\right)}^5={\mathrm{a}}_0+{\mathrm{a}}_1\mathrm{x}+{\mathrm{a}}_2{\mathrm{x}}^2+{\mathrm{a}}_3{\mathrm{x}}^3+\cdots +{\mathrm{a}}_{15}{\mathrm{x}}^{15}$

Putting x = 1 and x = -1 alternatively, we have

$\begin{array}{l}{\mathrm{a}}_0+{\mathrm{a}}_1+{\mathrm{a}}_2+{\mathrm{a}}_3+\cdots +{\mathrm{a}}_{15}=4^5 \left(1\right)\\ {\mathrm{a}}_0-{\mathrm{a}}_1+{\mathrm{a}}_2-{\mathrm{a}}_3+\cdots -{\mathrm{a}}_{15}=0 \left(2\right)\end{array}$

Adding (1) and (2), we have

$\begin{array}{c}2\left({\mathrm{a}}_0+{\mathrm{a}}_2+{\mathrm{a}}_4+\cdots +{\mathrm{a}}_{14}\right)=4^5\\ \mathrm{or} {\mathrm{a}}_0+{\mathrm{a}}_2+{\mathrm{a}}_4+\cdots +{\mathrm{a}}_{14}=2^9=512\end{array}$