Tính 

- Hướng dẫn giải

Đáp án cần chọn là: D

$\begin{array}{l}M=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\mathrm{...}+\frac{1}{2^{99}}+\frac{1}{2^{100}}\\ 2M=2.\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\mathrm{...}+\frac{1}{2^{99}}+\frac{1}{2^{100}}\right)\\ =2+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\mathrm{...}+\frac{1}{2^{99}}\end{array}$

Ta có : 

$\begin{array}{l}\\ \begin{array}{l}M=2M-M\\ =\left(2+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\mathrm{...}+\frac{1}{2^{98}}+\frac{1}{2^{99}}\right)\\ -\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\mathrm{...}+\frac{1}{2^{99}}+\frac{1}{2^{100}}\right)\\ =2-\frac{1}{2^{100}}\\ =\frac{2^{101}-1}{2^{100}}\end{array}\end{array}$