\(B=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+\dfrac{4}{3^4}+...+\dfrac{100}{3^{100}}< \dfrac{3}{4}\)
\(3B=1+\dfrac{2}{3}+\dfrac{3}{3^2}+\dfrac{4}{3^3}+...+\dfrac{100}{3^{99}}\)
\(3B-B=\left(1+\dfrac{2}{3}+\dfrac{3}{3^2}+\dfrac{4}{3^3}+...+\dfrac{100}{3^{99}}\right)-\left(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+\dfrac{4}{3^4}+...+\dfrac{100}{3^{100}}\right)\)
\(2B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)
\(6B=3+1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\)
\(6B-2B=\left(3+1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\right)\)
\(4B=3-\dfrac{100}{3^{99}}-\dfrac{1}{3^{99}}+\dfrac{100}{3^{100}}\)
\(4B=3-\dfrac{300}{3^{100}}-\dfrac{3}{3^{100}}+\dfrac{100}{3^{100}}\)
\(4B=3-\dfrac{203}{3^{100}}< 3\)
\(\Rightarrow C< \dfrac{3}{4}\left(đpcm\right)\)
nhân 3 trừ đi sau đó xét cái sau sẽ thấy B<3/4
Lười lắm
\(B=\dfrac{1}{3}+\dfrac{2}{3^2}+...+\dfrac{100}{3^{100}}\\ \Rightarrow3B=1+\dfrac{2}{3}+....+\dfrac{100}{3^{99}}\\ \Rightarrow2B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+....+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\\ A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+..+\dfrac{1}{3^{99}}\\ \Rightarrow3A=3+1+..+\dfrac{1}{3^{98}}\\ \Rightarrow2.A=3-\dfrac{1}{3^{99}}< 3\\ \Rightarrow A< \dfrac{3}{2}\\ \Rightarrow2B=A-\dfrac{100}{3^{100}}< \dfrac{3}{2}\\ \Rightarrow B< \dfrac{3}{4}\)
