Xét $(x+1)^{2n} = C_{2n}^0+x C_{2n}^1+x^2C_{2n}^2+x^3C_{2n}^3+ x^4C_{2n}^4-\dots + x^{2n} C_{2n}^{2n}$
Xét $(x-1)^{2n} = C_{2n}^0-x C_{2n}^1+x^2C_{2n}^2-x^3C_{2n}^3+x^4C_{2n}^4- \dots + x^{2n} C_{2n}^{2n}$
$\Rightarrow (x+1)^{2n}+(x-1)^{2n} = 2\bigg (C_{2n}^0+x^2C_{2n}^2+x^4C_{2n}^4+ \dots + x^{2n} C_{2n}^{2n} \bigg )$
Thay $x=3$ ta được
$2^{4n}+2^{2n}=2\bigg (C_{2n}^0+3^2C_{2n}^2+3^4C_{2n}^4+ \dots + 3^{2n} C_{2n}^{2n} \bigg )$
$\Rightarrow C_{2n}^0+3^2C_{2n}^2+3^4C_{2n}^4+ \dots + 3^{2n} C_{2n}^{2n} =2^{2n-1}.(2^{2n}+1)$