Giải thích các bước giải:
Ta có:
$B=\dfrac{4}{3}+\dfrac{10}{9}+\dfrac{28}{27}+...+\dfrac{3^{98}+1}{3^{98}}$
$\to B=\dfrac{3+1}{3}+\dfrac{3^2+1}{3^2}+\dfrac{3^3+1}{3^3}+...+\dfrac{3^{98}+1}{3^{98}}$
$\to B=(1+\dfrac13)+(1+\dfrac1{3^2})+(1+\dfrac1{3^3})+...+(1+\dfrac1{3^{98}})$
$\to B=98+\dfrac13+\dfrac1{3^2}+...+\dfrac{1}{3^{98}}$
$\to 3B=3\cdot 98+\dfrac11+\dfrac1{3}+...+\dfrac{1}{3^{97}}$
$\to 3B-B=3\cdot 98-98+\dfrac11-\dfrac1{3^{98}}$
$\to 2B=2\cdot 98+1-\dfrac1{3^{98}}$
$\to 2B<2\cdot 98+2$
$\to B<98+1$
$\to B<100$
