Ta có :
\(B=\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+...+\frac{3^{98}+1}{3^{98}}\)
\(B=\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{9}\right)+\left(1+\frac{1}{27}\right)+...+\left(1+\frac{1}{3^{98}}\right)\)
\(B=\left(1+1+1+...+1\right)+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{3^{98}}\right)\)
\(B=97+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{3^{98}}\right)\)
gọi A là biểu thức trong ngoặc
Lại có :
\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{3^{98}}\)
\(\Leftrightarrow A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)\)
\(2A=1-\frac{1}{3^{98}}< 1\)
\(\Rightarrow A=\frac{1-\frac{1}{3^{98}}}{2}< \frac{1}{2}< 1\)
\(\Rightarrow A< 1\)
\(\Rightarrow B< 97+1=98< 100\)
vậy \(B< 100\)
