Đề bài
So sánh:
a] \[A = \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{2020}}}} + \frac{1}{{{2^{2021}}}}\] và \[B = \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{{13}}{{60}}\]
b] \[C = \frac{{2019}}{{2020}}.\frac{{2021}}{{2022}}\] và \[D = \frac{{2020 + 2022}}{{2019 + 2021}}.\frac{3}{2}\]
Phương pháp giải - Xem chi tiết
Rút gọn rồi so sánh hai kết quả thu được.
Lời giải chi tiết
a] Ta có:
\[\begin{array}{l}A = \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{2020}}}} + \frac{1}{{{2^{2021}}}}\\ \Rightarrow 2A = 2.\left[ {\frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{2020}}}} + \frac{1}{{{2^{2021}}}}} \right]\\ = 1 + \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{2019}}}} + \frac{1}{{{2^{2020}}}}\\ \Rightarrow A = \left[ {1 + \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{2019}}}} + \frac{1}{{{2^{2020}}}}} \right] - \left[ {\frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{2020}}}} + \frac{1}{{{2^{2021}}}}} \right]\\ \Rightarrow A = 1 - \frac{1}{{{2^{2021}}}} < 1\end{array}\]
Và
\[\begin{array}{l}B = \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{{13}}{{60}} = \frac{{1.20}}{{3.20}} + \frac{{1.15}}{{4.15}} + \frac{{1.12}}{{5.12}} + \frac{{13}}{{60}}\\ = \frac{{20}}{{60}} + \frac{{15}}{{60}} + \frac{{12}}{{60}} + \frac{{13}}{{60}} = \frac{{20 + 15 + 12 + 13}}{{60}} = \frac{{60}}{{60}} = 1\end{array}\]
Vậy A < B.
b] \[C = \frac{{2019}}{{2020}}.\frac{{2021}}{{2022}}\] và \[D = \frac{{2020 + 2022}}{{2019 + 2021}}.\frac{3}{2}\]
Ta có:
\[C = \frac{{2019}}{{2020}}.\frac{{2021}}{{2022}} = \frac{{2019.2021}}{{2020.2022}}\]
Mà: \[2019.2021 < 2020.2022\] nên \[C < 1\]
\[D = \frac{{2020 + 2022}}{{2019 + 2021}}.\frac{3}{2}\]
Mà \[2020>2019; 2022>2021\] nên \[2020+2022 > 2019+2021\]. Do đó \[\frac{{2020 + 2022}}{{2019 + 2021}} > 1 \Rightarrow D > 1.\frac{3}{2} = \frac{3}{2} > 1\]