Câu hỏi:
51 lượt xemLời giải
Hướng dẫn giải:
Ta có: \(\frac{1}{{{2^2}}} < \frac{1}{{1.2}}\); \(\frac{1}{{{3^2}}} < \frac{1}{{2.3}}\); \(\frac{1}{{{4^2}}} < \frac{1}{{3.4}}\);…; \(\frac{1}{{{{2021}^2}}} < \frac{1}{{2020.2021}}\).
Đặt \(A = \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{{2022}^2}}}\)
\( = \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{{2022}^2}}} < \frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + ... + \frac{1}{{2020.2021}}\)
\( \Rightarrow A < 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{{2020}} - \frac{1}{{2021}}\)
\( \Rightarrow A < 1 - \frac{1}{{2021}}\)
\( \Rightarrow A < \frac{{2020}}{{2021}}\).
Vì \[2020 < 2021\] nên \(\frac{{2020}}{{2021}} < 1\).
Do đó \(A < \frac{{2020}}{{2021}} < 1\).
Vậy \(\frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{{2022}^2}}} < 1\) (đpcm).
1. Thực hiện phép tính (tính nhanh nếu có thể):
a) ; b) .
2. Tìm :
a) ; b) .
