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Chứng minh 122+132+142+...+120222<1\frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{{2022}^2}}} < 1.

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Lời giải

Hướng dẫn giải:

Ta có: 122<11.2\frac{1}{{{2^2}}} < \frac{1}{{1.2}}; 132<12.3\frac{1}{{{3^2}}} < \frac{1}{{2.3}}; 142<13.4\frac{1}{{{4^2}}} < \frac{1}{{3.4}};…; 120212<12020.2021\frac{1}{{{{2021}^2}}} < \frac{1}{{2020.2021}}.

Đặt A=122+132+142+...+120222A = \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{{2022}^2}}}

=122+132+142+...+120222<11.2+12.3+13.4+...+12020.2021 = \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}}

A<112+1213+1314+...+1202012021 \Rightarrow A < 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{{2020}} - \frac{1}{{2021}}

A<112021 \Rightarrow A < 1 - \frac{1}{{2021}}

A<20202021 \Rightarrow A < \frac{{2020}}{{2021}}.

2020<20212020 < 2021 nên 20202021<1\frac{{2020}}{{2021}} < 1.

Do đó A<20202021<1A < \frac{{2020}}{{2021}} < 1.

Vậy 122+132+142+...+120222<1\frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{{2022}^2}}} < 1 (đpcm).

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