2A= \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
=> 2A - A= \(\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{^{2^{100}}}\right)\)
=> A = \(1-\frac{1}{2^{100}}\)< 1
=> A< 1
đúng nhé
A=\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{100}}\)
2A=\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}\)
2A-A=\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}\)\(-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{100}}\right)\)
A=\(1-\frac{1}{2^{100}}\)
Vì \(1-\frac{1}{2^{100}}\)< \(1\)
Nên \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{100}}\)< \(1\)
Vậy A <\(1\)