1. Cho tổng N=1/4+2/4^2+3/4^3+.....+2024/4^2024. Chứng minh rằng N<1/2 2. Cho S=1+3^2+3^4+3^6+....+3^2020. Chứng minh rằng S chia hết cho 8 Giúp mk với ạaaaa mk đang cần gấp!!!

 \begin{array}{l}
N=\dfrac{1}{4} +\dfrac{2}{4^{2}} +\dfrac{3}{4^{3}} +...+\dfrac{2024}{4^{2024}}\\
4N=1+\dfrac{1}{4} +\dfrac{2}{4^{2}} +\dfrac{3}{4^{3}} +...+\dfrac{2024}{4^{2023}}\\
4N-N\ =\ \left( 1+\dfrac{1}{4} +\dfrac{2}{4^{2}} +\dfrac{3}{4^{3}} +...+\dfrac{2024}{4^{2023}}\right) -\left(\dfrac{1}{4} +\dfrac{2}{4^{2}} +\dfrac{3}{4^{3}} +...+\dfrac{2024}{4^{2024}}\right)\\
3N\ =\ 1+\dfrac{1}{4} +\dfrac{1}{4^{2}} +...+\dfrac{1}{4^{2022}} -\dfrac{2024}{4^{2024}}\\
Đặt\ A\ =\ 1+\dfrac{1}{4} +\dfrac{1}{4^{2}} +...+\dfrac{1}{4^{2023}}\\
4A\ =\ 4+1+\dfrac{1}{4} +...+\dfrac{1}{4^{2021}}\\
4A\ -\ A\ =\left( \ 4+1+\dfrac{1}{4} +...+\dfrac{1}{4^{2021}}\right) -\left( 1+\dfrac{1}{4} +\dfrac{1}{4^{2}} +...+\dfrac{1}{4^{2023}}\right)\\
3A\ =\ 4-\dfrac{1}{4^{2023}}\\
A\ =\ \dfrac{4}{3} -\dfrac{1}{4^{2023} .3}\\
\Longrightarrow \ 3S\ =\ \ \dfrac{4}{3} -\dfrac{1}{4^{2023} .3} \ -\dfrac{2024}{4^{2024}}\\
\Longrightarrow \ S\ =\ \ \left(\dfrac{4}{3} -\dfrac{1}{4^{2023} .3} \ -\dfrac{2024}{4^{2024}}\right) :3\ =\ \dfrac{4}{9} -\dfrac{1}{4^{2023} .9} \ -\dfrac{2024}{3.4^{2024}} < \dfrac{4}{9} < \dfrac{1}{2}\\
\Longrightarrow \ ĐPCM\\
\\
S=3+3^{2} +3^{3} +3^{4} +...+3^{2020}\\
S=\left( 3+3^{2} +3^{3} +3^{4}\right) +...+3^{2020}\\
S=3\left( 1+3+3^{2} +3^{3}\right) +...+3^{2016}\left( 1+3+3^{2} +3^{3}\right)\\
S=\ 3.40+...+3^{2016} .40\\
S=40\left( 3+3^{5} +...+3^{2016}\right)\\
40\ \vdots 8\ \Longrightarrow \ B\vdots 4
\end{array}