Ta có: \(S=\frac{1}{5^2}+\frac{2}{5^3}+\cdots+\frac{99}{5^{100}}\)
=>\(5S=\frac15+\frac{2}{5^2}+\ldots+\frac{99}{5^{99}}\)
=>\(5S-S=\frac15+\frac{2}{5^2}+\cdots+\frac{99}{5^{99}}-\frac{1}{5^2}-\frac{2}{5^3}-\cdots-\frac{99}{5^{100}}=\frac15+\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)
=>\(4S=\frac15+\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)
Đặt \(A=\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}\)
=>\(5A=\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{98}}\)
=>\(5A-A=\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{98}}-\frac{1}{5^2}-\frac{1}{5^3}-\cdots-\frac{1}{5^{99}}=\frac15-\frac{1}{5^{99}}=\frac{5^{98}-1}{5^{99}}\)
=>\(4A=\frac{5^{98}-1}{5^{99}}\)
=>\(A=\frac{5^{98}-1}{4\cdot5^{99}}\)
Ta có: \(4S=\frac15+\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)
\(=\frac15+\frac{5^{98}-1}{4\cdot5^{99}}-\frac{99}{5^{100}}=\frac{5^{99}-5-396}{4\cdot5^{100}}+\frac15=\frac15+\frac{5^{99}-401}{4\cdot5^{100}}\)
=>\(4S=\frac15+\frac{1}{4\cdot5}-\frac{401}{4\cdot5^{100}}=\frac14-\frac{401}{4\cdot5^{100}}<\frac14\)
=>S<1/16
