Question

Cho S= \(\dfrac{1}{2^0}+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\).Chứng tỏ rằng S<2

Answer 1

\(S=2S-S\)

\(=2\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^{2022}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^{2022}}\right)\)

\(=\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2021}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^{2022}}\right)\)

\(=2-\dfrac{1}{2^{2022}}< 2\left(đpcm\right)\)