Question

Cho S=\(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+\dfrac{4}{4^4}+...+\dfrac{2023}{4^{2023}}\). Chứng minh S < \(\dfrac{1}{2}\)

Answer 1

=> 4S = 1 + 2/4 + 3/4^2 +...+ 2023/4^2022

=> 4S-S = 1 + (2/4-1/4) + (3/4^2 - 2/4^2) +...+ (2023/4^2022 - 2022/4^2022) - 2023/4^2023

=> 3S = 1 + 1/4 + 1/4^2 +...+ 1/4^2022 - 2023/4^2023

=> 12S = 4 + 1 + 1/4 +... + 1/4^2021 - 2023/4^2022

=> 12S - 3S = 4 + (1-1) + (1/4-1/4) +... + (1/4^2021 - 1/4^2021)  - 1/4^2022 - 2023/4^2022 + 2023/4^2023

=> 9S = 4 -  1/4^2022 - 2023/4^2022 + 2023/4^2023

= 4- 2024/4^2022 + 2023/4^2023

Do 2024/4^2022 > 2024/4^2023 > 2023/4^2023 nên - 2024/4^2022 + 2023/4^2023 < 0

=> 9S < 4 < 9/2

=> S < 1/2 (đpcm)

Answer 2

Cho S=1+3+3^2+....+3^2023

Chứng tỏ S chia hết cho 4