Bài này phải sửa thành chứng minh:
$$C^{0}_{2004}+2^2C^{2}_{2004}+....+2^{2004}C^{2004}_{2004}=\frac{3^{2004}+1}{2}.$$
Ta có
$3^{2004}=(2+1)^{2004}=2^0C^{0}_{2004}+2^1C^{1}_{2004}+....+2^{2003}C^{2003}_{2004}+2^{2004}C^{2004}_{2004}$
$1=(2-1)^{2004}=2^0C^{0}_{2004}-2^1C^{1}_{2004}+....-2^{2003}C^{2003}_{2004}+2^{2004}C^{2004}_{2004}$
Cộng theo từng vế suy ra
$3^{2004}+1 = 2\left ( 2^0C^{0}_{2004}+2^2C^{2}_{2004}+....+2^{2004}C^{2004}_{2004} \right )$
Ta có
$$C^{0}_{2004}+2^2C^{2}_{2004}+....+2^{2004}C^{2004}_{2004}=\frac{3^{2004}+1}{2}.$$