help vs mn
Ta có: \(M=\frac13+\frac{2}{3^2}+\frac{3}{3^3}+\cdots+\frac{2022}{3^{2022}}+\frac{2023}{3^{2023}}\)
=>\(3M=1+\frac23+\frac{3}{3^2}+\cdots+\frac{2022}{3^{2021}}+\frac{2023}{3^{2022}}\)
=>\(3M-M=1+\frac23+\frac{3}{3^2}+\cdots+\frac{2022}{3^{2021}}+\frac{2023}{3^{2022}}-\frac13-\frac{2}{3^2}-\cdots-\frac{2022}{3^{2022}}-\frac{2023}{3^{2023}}\)
=>\(2M=1+\frac13+\frac{1}{3^2}+\cdots+\frac{1}{3^{2021}}+\frac{1}{3^{2022}}-\frac{2023}{3^{2023}}\)
Đặt \(A=1+\frac13+\frac{1}{3^2}+\cdots+\frac{1}{3^{2022}}\)
=>\(\frac{A}{3}=\frac13+\frac{1}{3^2}+\frac{1}{3^3}+\cdots+\frac{1}{3^{2023}}\)
=>\(A-\frac{A}{3}=1+\frac13+\frac{1}{3^2}+\cdots+\frac{1}{3^{2022}}-\frac13-\frac{1}{3^2}-\frac{1}{3^3}-\cdots-\frac{1}{3^{2023}}\)
=>\(A\cdot\frac23=1-\frac{1}{3^{2023}}\)
=>\(A=\frac32-\frac{1}{3^{2023}}:\frac23=\frac32-\frac{1}{2\cdot3^{2022}}\)
\(2M=A-\frac{2023}{3^{2023}}\)
=>\(2M=\frac32-\frac{1}{2\cdot3^{2022}}-\frac{2023}{3^{2023}}=\frac32-\frac{3}{2\cdot3^{2023}}-\frac{4046}{2\cdot3^{2023}}=\frac32-\frac{4049}{2\cdot3^{2023}}\)
=>\(2M=\frac{3^{2023}\cdot3-4049}{2\cdot3^{2023}}\)
=>\(4M=\frac{6\cdot3^{2023}-8098}{2\cdot3^{2023}}=3-\frac{8098}{2\cdot3^{2023}}\)
=>4M<3
=>\(M<\frac34\)
