Đáp án:
$\begin{array}{l}
B = \dfrac{1}{5} - \dfrac{1}{{{5^2}}} + \dfrac{1}{{{5^3}}} - \dfrac{1}{{{5^4}}} + ... + \dfrac{1}{{{5^{99}}}} - \dfrac{1}{{{5^{100}}}}\\
a)5.B = 1 - \dfrac{1}{5} + \dfrac{1}{{{5^2}}} - \dfrac{1}{{{5^3}}} + ... + \dfrac{1}{{{5^{98}}}} - \dfrac{1}{{{5^{99}}}}\\
\Rightarrow 5B + B = 6B = 1 - \dfrac{1}{{{5^{100}}}}\\
\Rightarrow B = \dfrac{1}{6} - \dfrac{1}{{{{6.5}^{100}}}}\\
b)1 - 6B = {\left( {\dfrac{1}{5}} \right)^{n + 1}}\\
\Rightarrow 1 - \left( {1 - \dfrac{1}{{{5^{100}}}}} \right) = {\left( {\dfrac{1}{5}} \right)^{n + 1}}\\
\Rightarrow 1 - 1 + \dfrac{1}{{{5^{100}}}} = \dfrac{1}{{{5^{n + 1}}}}\\
\Rightarrow \dfrac{1}{{{5^{100}}}} = \dfrac{1}{{{5^{n + 1}}}}\\
\Rightarrow 100 = n + 1\\
\Rightarrow n = 99\\
\text{Vậy}\,n = 99\\
c)B = \dfrac{1}{6}.\left( {1 - \dfrac{1}{{{5^{100}}}}} \right)\\
Do:\dfrac{7}{{30}} = \dfrac{1}{6}.\dfrac{7}{5}\\
\dfrac{7}{5} > 1 > 1 - \dfrac{1}{{{5^{100}}}}\\
\Rightarrow \dfrac{1}{6}.\dfrac{7}{5} > \dfrac{1}{6}.\left( {1 - \dfrac{1}{{{5^{100}}}}} \right)\\
\Rightarrow \dfrac{7}{{30}} > B\\
\text{Vậy}\,B < \dfrac{7}{{30}}
\end{array}$
