\(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+....+\dfrac{100}{3^{100}}< \dfrac{3}{4}\)
\(\Rightarrow3A=1+\dfrac{2}{3}+\dfrac{3}{3^2}+\dfrac{4}{3^3}+....+\dfrac{100}{3^{99}}\)
\(\Rightarrow3A-A=1+\left(\dfrac{2}{3}-\dfrac{1}{3}\right)+\left(\dfrac{3}{3^2}-\dfrac{2}{3^3}\right)+....+\left(\dfrac{100}{3^{99}}-\dfrac{99}{3^{99}}\right)-\dfrac{100}{3^{100}}\)
\(\Rightarrow2A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+....+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)
Đặt \(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+....+\dfrac{1}{3^{99}}\)
\(\Rightarrow3B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+....+\dfrac{1}{3^{98}}\)
\(\Rightarrow2B=1-\dfrac{1}{3^{99}}\Rightarrow B=\dfrac{\left(1-\dfrac{1}{3^{99}}\right)}{2}\)
Thay vào \(2A\) \(\Rightarrow2A=1+\dfrac{1}{2}-\dfrac{1}{\left(2.3^{99}\right)}-\dfrac{100}{3^{100}}< 1+\dfrac{1}{2}=\dfrac{3}{2}\)
\(\Rightarrow A< \dfrac{3}{4}\)
Chúc bạn học tốt!
