\(2B=2\left(\frac{1}{2}+\frac{2}{2^2}+...+\frac{100}{2^{100}}\right)\)
\(2B=1+1+\frac{3}{2}+...+\frac{100}{2^{99}}\)
\(2B-B=\left(2+\frac{3}{2}+...+\frac{100}{2^{99}}\right)-\left(\frac{1}{2}+\frac{2}{2^2}+...+\frac{100}{2^{100}}\right)\)
\(B=2-\frac{1}{2}+\frac{2}{2^2}+\frac{100}{2^{100}}<2\)
=>B<2(đpcm)
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