Đáp án:
$\begin{array}{l}
a)A = \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{49}}}} + \frac{1}{{{2^{50}}}}\\
\Rightarrow 2A = 1 + \frac{1}{2} + \frac{1}{{{2^2}}} + ... + \frac{1}{{{2^{49}}}}\\
\Rightarrow 2A - A = A = 1 - \frac{1}{{{2^{50}}}} < 1\\
Vậy\,A < 1\\
b)B = \frac{1}{3} + \frac{1}{{{3^2}}} + ... + \frac{1}{{{3^{100}}}}\\
\Rightarrow 3B = 1 + \frac{1}{3} + ... + \frac{1}{{{3^{99}}}}\\
\Rightarrow 2B = 1 - \frac{1}{{{3^{100}}}}\\
\Rightarrow B = \frac{1}{2} - \frac{1}{{{{2.3}^{100}}}} < \frac{1}{2}\\
c)TT:C < \frac{1}{3}
\end{array}$
