Sửa đề \(\frac{3}{2}+\frac{5}{2^2}+\frac{9}{2^3}+...+\frac{2^{100}+1}{2^{100}}=\frac{2+1}{2}+\frac{2^2+1}{2^2}+\frac{2^3+1}{2^3}+...+\frac{2^{100}+1}{2^{100}}\)
= \(\left(1+1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)(100 hạng tử 1)
= \(100+\left(1-\frac{1}{2^{100}}\right)=101-\frac{1}{2^{100}}< 101\)(1)
Vì \(-\frac{1}{2^{100}}>-1\Rightarrow101-\frac{1}{2^{100}}>101-1\Rightarrow B>100\)(2)
Từ (1) và (2) => 100 < B < 101