Question

1/2 + 2/2^2 + 3/2^3 + ... + 2022/2^2022 so sánh với 2

Answer 1

Cần gấp! Thank

 

Answer 2

Lời giải:

Gọi tổng trên là $A$

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

\(2A=1+\frac{2}{2}+\frac{3}{2^2}+...+\frac{2022}{2^{2021}}\)

\(A=2A-A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2021}}-\frac{2022}{2^{2022}}\)

\(2A=2+1+\frac{1}{2}+...+\frac{1}{2^{2020}}-\frac{2022}{2^{2021}}\)

\(A=2A-A=2-\frac{2022}{2^{2021}}-\frac{1}{2^{2021}}+\frac{2022}{2^{2022}}=2-\frac{2024}{2^{2022}}<2\)