Cho 1/2 +1/2^2 + 1/2^3 +...+ 1/2^2022 so sánh S với 1 theo đề ta có

Theo đề, ta có:

$S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\mathrm{\dots }+\frac{1}{2^{2022}}$

Suy ra $2S=2.(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\mathrm{\dots }+\frac{1}{2^{2022}})$

$2S=\frac{2}{2}+\frac{2}{2^2}+\frac{2}{2^3}+\mathrm{\dots }+\frac{2}{2^{2022}}$

$2S=1+\frac{1}{2}+\frac{1}{2^2}+\mathrm{\dots }+\frac{1}{2^{2021}}$

Ta có $S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\mathrm{\dots }+\frac{1}{2^{2022}}$ và $2S=1+\frac{1}{2}+\frac{1}{2^2}+\mathrm{\dots }+\frac{1}{2^{2021}}$

$S-S=(1+\frac{1}{2}+\frac{1}{2^2}+\mathrm{\dots }+\frac{1}{2^{2021}})$ $-(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\mathrm{\dots }+\frac{1}{2^{2022}})$

Hay

S$=1+\frac{1}{2}+\frac{1}{2^2}+\mathrm{\dots }+\frac{1}{2^{2021}}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-\mathrm{\dots }-\frac{1}{2^{2022}}$

S= $=1+(\frac{1}{2}-\frac{1}{2})+(\frac{1}{2^2}-\frac{1}{2^2})+(\frac{1}{2^3}-\frac{1}{2^3})+\mathrm{\dots }$$+(\frac{1}{2^{2021}}-\frac{1}{2^{2021}})-\frac{1}{2^{2022}}$

S$=1-\frac{1}{2^{2022}}=\frac{2^{2022}-1}{2^{2022}}$