B = \(\frac{4}{3^1}+\frac{10}{3^2}+\frac{28}{3^3}+...+\frac{3^{98}+1}{3^{98}}\)
B = \(\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{3^2}\right)+\left(1-\frac{1}{3^3}\right)+...+\left(1-\frac{1}{3^{98}}\right)\)
B = \(\left(1+1+1+...+1\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)\)
B = \(98-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)\)
=> B < 98 < 100
vậy B < 100
\(B=\frac{4}{3}+\frac{10}{3^2}+...+\frac{3^{98}+1}{3^{98}}\)
\(3B=3\left(\frac{4}{3}+\frac{10}{3^2}+...+\frac{3^{98}+1}{3^{98}}\right)\)
\(3B=4+\frac{10}{3}+...+\frac{3^{98}+1}{3^{97}}\)
\(3B=\left(4+\frac{10}{3}+...+\frac{3^{98}+1}{3^{97}}\right)-\left(\frac{4}{3}+\frac{10}{3^2}+...+\frac{3^{98}+1}{3^{98}}\right)\)
\(2B=4-\frac{3^{98}+1}{3^{98}}\)
\(B=2-\frac{3^{98}+1}{2.3^{98}}<2\)
mà 2<100
=>B<100
