Question

B=\(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\). Chứng minh rằng B<1

Answer 1

\(B=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)

\(2B=1+\dfrac{1}{2}+\dfrac{1}{4}+....+\dfrac{1}{2^{99}}\)

\(=>B=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}...+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\right)\)

\(B=1-\dfrac{1}{2^{99}}< 1=>B< 1\)

Answer 2

\(B=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\) 

\(2B=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\) 

\(2B-B=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)\) 

\(B=1-\dfrac{1}{2^{100}}< 1\left(đpcm\right)\) 

Vậy \(B< 1\)