6-я олимпиада им. Шалтая Смагулова, 6-7 класс, 3 (командный) тур

Теңсіздікті дәлелдеңіз: $2023 < \frac{2}{1} + \frac{3}{2} + \ldots + \frac{{2023}}{{2022}} + \frac{{2024}}{{2023}} < 2282$.

Комментарий/решение

$2023<\dfrac{2}{1}+\dfrac{3}{2}+\cdots+\dfrac{2023}{2022}+\dfrac{2024}{2023}<2282$

$2023<1+\dfrac{1}{1}+1+\dfrac{1}{2}+\cdots+1+\dfrac{1}{2022}+1+\dfrac{1}{2023}<2282$

$0<\dfrac{1}{1}+\dfrac{1}{2}+\cdots+\dfrac{1}{2022}+\dfrac{1}{2023}<259$

$\dfrac{1}{8}>\dfrac{1}{2023}>0$

$\dfrac{1}{8}>\dfrac{1}{2022}>0$

$\cdots$

$\dfrac{1}{8}>\dfrac{1}{9}>0$

$\dfrac{1}{7}>\dfrac{1}{8}>0$

$\cdots$

$1>\dfrac{1}{2}>0$

$1,5>1>0$

$0<\dfrac{1}{1}+\dfrac{1}{2}+\cdots+\dfrac{1}{2022}+\dfrac{1}{2023}<\dfrac{2023-8}{8}+\dfrac{1}{7}+\cdots+1+1,5$

$0<\dfrac{1}{1}+\dfrac{1}{2}+\cdots+\dfrac{1}{2022}+\dfrac{1}{2023}<251\dfrac{7}{8}+\dfrac{6+7}{42}+\dfrac{5+4}{20}+\dfrac{3+2}{6}+2,5$

$0<\dfrac{1}{1}+\dfrac{1}{2}+\cdots+\dfrac{1}{2022}+\dfrac{1}{2023}<254\dfrac{3}{8}+\dfrac{13+35}{42}+\dfrac{5+4}{20}$

$0<\dfrac{1}{1}+\dfrac{1}{2}+\cdots+\dfrac{1}{2022}+\dfrac{1}{2023}<254\dfrac{3}{8}+\dfrac{480+189}{420}$

$0<\dfrac{1}{1}+\dfrac{1}{2}+\cdots+\dfrac{1}{2022}+\dfrac{1}{2023}<254\dfrac{3}{8}+1\dfrac{249}{420}$

$0<\dfrac{1}{1}+\dfrac{1}{2}+\cdots+\dfrac{1}{2022}+\dfrac{1}{2023}<254\dfrac{3}{8}+1\dfrac{83}{140}$

$0<\dfrac{1}{1}+\dfrac{1}{2}+\cdots+\dfrac{1}{2022}+\dfrac{1}{2023}<255\dfrac{105+166}{280}$

$0<\dfrac{1}{1}+\dfrac{1}{2}+\cdots+\dfrac{1}{2022}+\dfrac{1}{2023}<255\dfrac{271}{280}<259$