Sửa đề: \(A=75\left(4^{2023}+4^{2022}+...+4^2+5\right)+25\)
Đặt \(B=4^{2023}+4^{2022}+...+4^2+5\)
=>\(B=4^{2023}+4^{2022}+...+4^2+4+1\)
=>\(4B=4^{2024}+4^{2023}+...+4^3+4^2+4\)
=>\(4B-B=4^{2024}+4^{2023}+...+4^3+4^2+4-4^{2023}-4^{2022}-...-4^2-4-1\)
=>\(3B=4^{2024}-1\)
=>\(B=\dfrac{4^{2024}-1}{3}\)
\(A=75\cdot B+25\)
\(=25\left(4^{2024}-1\right)+25=25\cdot4^{2024}⋮4^{2024}\)