Question

[mac mai trang]

thu gọn a=1/2-1/2^2+1/2^3-1/2^4+...+1/2^2023-1/2^2024

Answer 1

[Nguyễn Bảo An]

\(A=\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3-\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{2023}-\left(\dfrac{1}{2}\right)^{2024}\)

\(A=\dfrac{2}{2^2}-\dfrac{1}{2^2}+\dfrac{2}{2^4}-\dfrac{1}{2^4}+...+\dfrac{2}{2^{2024}}-\dfrac{1}{2^{2024}}\)

\(A=\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+...+\dfrac{1}{2^{2024}}\)

\(A=\dfrac{2^{2022}}{2^{2024}}+\dfrac{2^{2020}}{2^{2024}}+\dfrac{2^{2018}}{2^{2024}}+...+\dfrac{1}{2^{2024}}\)

\(2^2A=\dfrac{2^{2024}}{2^{2024}}+\dfrac{2^{2022}}{2^{2024}}+\dfrac{2^{2020}}{2^{2024}}+...+\dfrac{2^2}{2^{2024}}\)

\(\Rightarrow4A-A=3A=1-\dfrac{2}{2^{2024}}-\dfrac{1}{2^{2024}}\)

\(3A=1-\left(\dfrac{2}{2^{2024}}+\dfrac{1}{2^{2024}}\right)\)

\(3A=1-\dfrac{3}{2^{2024}}\)

\(A=\dfrac{1-\dfrac{3}{2^{2024}}}{3}\)

\(A=\dfrac{3\left(\dfrac{1}{3}-\dfrac{1}{2^{2024}}\right)}{3}\)

\(A=\dfrac{1}{3}-\dfrac{1}{2^{2024}}\)

Answer 2

[mac mai trang]

giúp mk vs các bn. chiều nay mk phải nộp r