Đặt \(A=\frac13+\frac{2}{3^2}+\frac{3}{3^3}+\cdots+\frac{100}{3^{100}}\)
=>\(3A=1+\frac23+\frac{3}{3^2}+\cdots+\frac{100}{3^{99}}\)
=>\(3A-A=1+\frac23+\frac{3}{3^2}+\cdots+\frac{100}{3^{99}}-\frac13-\frac{2}{3^2}-\cdots-\frac{100}{3^{100}}\)
=>\(2A=1+\frac13+\frac{1}{3^2}+\cdots+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
Đặt \(B=\frac13+\frac{1}{3^2}+\cdots+\frac{1}{3^{99}}\)
=>\(3B=1+\frac13+\cdots+\frac{1}{3^{98}}\)
=>3B-B\(=1+\frac13+\cdots+\frac{1}{3^{98}}-\frac13-\frac{1}{3^2}-\cdots-\frac{1}{3^{99}}\)
=>\(2B=1-\frac{1}{3^{99}}=\frac{3^{99}-1}{3^{99}}\)
=>\(B=\frac{3^{99}-1}{2\cdot3^{99}}\)
Ta có: \(2A=1+\frac13+\frac{1}{3^2}+\cdots+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(=1+\frac{3^{99}-1}{2\cdot3^{99}}-\frac{100}{3^{100}}=\frac{2\cdot3^{100}+3^{100}-3-100}{3^{100}\cdot2}=\frac{3\cdot3^{100}-103}{2\cdot3^{100}}=\frac32-\frac{103}{2\cdot3^{100}}<\frac32\)
=>\(A<\frac34\)
