\(A=2+2^2+2^3+...+2^{2024}\)
\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{2021}+2^{2022}+2^{2023}+2^{2024}\right)\)
\(=\left(2+2^2+2^3+2^4\right)+2^4\left(2+2^2+2^3+2^4\right)+...+2^{2020}\left(2+2^2+2^3+2^4\right)\)
\(=30\left(1+2^4+...+2^{2020}\right)⋮10\)
=>Chữ số hàng đơn vị của A là 0
Lời giải:
$A=(2+2^2+2^3+2^4)+(2^5+2^6+2^7+2^8)+....+(2^{2021}+2^{2022}+2^{2023}+2^{2024})$
$=2(1+2+2^2+2^3)+2^5(1+2+2^2+2^3)+....+2^{2021}(1+2+2^2+2^3)$
$=(1+2+2^2+2^3)(2+2^5+...+2^{2021})$
$=15(2+2^5+...+2^{2021})\vdots 15\vdots 5$
Hiển nhiên $A$ cũng chia hết cho 2
$\Rightarrow A\vdots 2; 5\Rightarrow A\vdots 10$
$\Rightarrow A$ tận cùng là $0$
