Ta có: \(S=\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots-\frac{100}{3^{100}}\)
=>\(3A=1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots-\frac{100}{3^{99}}\)
=>\(3A+A=1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots-\frac{100}{3^{99}}+\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots-\frac{100}{3^{100}}\)
=>\(4A=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
Đặt \(B=-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}\)
=>\(3B=-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{98}}\)
=>\(3B+B=-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{98}}-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}=-1-\frac{1}{3^{99}}=\frac{-3^{99}-1}{3^{99}}\)
=>\(4B=\frac{-3^{99}-1}{3^{99}}\)
=>\(B=\frac{-3^{99}-1}{4\cdot3^{99}}\)
Ta có: \(4A=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(=1+\frac{-3^{99}-1}{4\cdot3^{99}}-\frac{100}{3^{100}}=1+\frac{-3^{100}-3-400}{4\cdot3^{100}}=1-\frac14-\frac{403}{4\cdot3^{100}}<\frac34\)
=>\(A<\frac{3}{16}\)
mà \(\frac{3}{16}<\frac{3.2}{16}=\frac15\)
nên \(A<\frac15\)
