Question

[Nhi Nguyễn Thị Tuệ]

1/2 + 1/2^2 + 1/2^3 + ... +1/2^2022

Tính nhanh

Gấp ạaa

Answer 1

[Vũ Đức Thịnh]

Đặt A = \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2022}}\)

\(\Rightarrow2.A=\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2022}}+\dfrac{1}{2^{2023}}\)

\(\Rightarrow2.A-A=\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2022}}+\dfrac{1}{2^{2023}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2022}}\right)\)

\(\Rightarrow A=\dfrac{1}{2^{2023}}-\dfrac{1}{2}\)

Answer 2

[Nguyễn Bảo Ngọc]

S = 1 + 1  + 1  + ...... + 1      < 1 
      2   2^2  2^3           2^20 

2S = 1 + 1 + 1  + ...... + 1 
              2   2^2           2^19 

2S - S = 1 - 1
                2^20
S = 1 - 1      < 1
          2^20

=> S < 1