a:Sửa đề: \(S=2+2^2+\cdots+2^{2024}\)
Ta có: \(S=2+2^2+\cdots+2^{2024}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+\cdots+\left(2^{2023}+2^{2024}\right)\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+\cdots+2^{2023}\left(1+2\right)\)
\(=3\left(2+2^3+\cdots+2^{2023}\right)\) ⋮3
b: Ta có: \(S=2+2^2+\cdots+2^{2024}\)
\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+\cdots+\left(2^{2021}+2^{2022}+2^{2023}+2^{2024}\right)\)
\(=\left(2+2^2+2^3+2^4\right)+2^4\left(2+2^2+2^3+2^4\right)+\cdots+2^{2020}\left(2+2^2+2^3+2^4\right)\)
\(=30\left(1+2^4+\cdots+2^{2020}\right)=3\cdot10\cdot\left(1+2^4+\cdots+2^{2020}\right)\) ⋮10
=>S có chữ số tận cùng là 0
