chứng minh rằng:
\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{19}< 2\)
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Những câu hỏi liên quan
28 tháng 1 2018
\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+........+\dfrac{1}{100^2}\)
Ta có :
\(\dfrac{1}{5^2}< \dfrac{1}{4.5}\)
\(\dfrac{1}{6^2}< \dfrac{1}{5.6}\)
...................
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(\Leftrightarrow\dfrac{1}{5^2}+\dfrac{1}{6^2}+....+\dfrac{1}{100^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+.......+\dfrac{1}{99.100}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+......+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{4}-\dfrac{1}{100}=\dfrac{6}{25}\)
Mà \(\dfrac{1}{6}< \dfrac{5}{26}< \dfrac{1}{4}\)
Mà \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+.........+\dfrac{1}{100^2}< \dfrac{6}{25}\)
\(\Leftrightarrow\dfrac{1}{6}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+.......+\dfrac{1}{100^2}< \dfrac{1}{4}\left(đpcm\right)\) \(\left(1\right)\)
2 tháng 7 2021
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)
\(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+...+\dfrac{19}{81.100}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{81}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}< 1\left(dpcm\right)\)
30 tháng 9 2021
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)
\(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+...+\dfrac{19}{81.100}\)\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{81}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}< 1\)
NP
2
6 tháng 9 2021
\(\dfrac{1}{5^2}< \dfrac{1}{4.5}\)
\(\dfrac{1}{6^2}< \dfrac{1}{5.6}\)
......
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(\Rightarrow\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\)
6 tháng 9 2021
Ta có: \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\)
TG
11 tháng 4 2022
1/4^2 + 1/5^2 +... + 1/100^2 < 1/3.4 + 1/4.5 +...+ 1/99.100
A=1/3 - 1/4 + 1/4 - 1/5 +...+ 1/99 - 1/100
=1/3 - 1/100 < 1/3
\(\dfrac{1}{19}< \dfrac{1}{5}\) ;\(\dfrac{1}{18}< \dfrac{1}{5}\) ;\(\dfrac{1}{17}< \dfrac{1}{5}\) ;\(\dfrac{1}{16}< \dfrac{1}{5}\) ;\(\dfrac{1}{15}< \dfrac{1}{5}\) ;\(\dfrac{1}{14}< \dfrac{1}{5}\)
\(\dfrac{1}{13}< \dfrac{1}{5}\) ;\(\dfrac{1}{12}< \dfrac{1}{5}\) ;\(\dfrac{1}{11}< \dfrac{1}{5}\); \(\dfrac{1}{10}< \dfrac{1}{5}\) ;\(\dfrac{1}{9}< \dfrac{1}{5}\) ;\(\dfrac{1}{8}< \dfrac{1}{5}\)
\(\dfrac{1}{7}< \dfrac{1}{5}\) ;\(\dfrac{1}{6}< \dfrac{1}{5}\)
Ta có :
(19 - 5) : 1 + 1= 15 (số hạng)
\(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{7}+\dfrac{1}{6}\) < \(\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{5}+...+\dfrac{1}{5}\)
(14 số hạng) (14 số hạng)
\(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{7}+\dfrac{1}{6}+\dfrac{1}{5}\) < \(\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{5}+...+\dfrac{1}{5}+\dfrac{1}{5}\)
(15 số hạng) (15 số hạng)
= > A < 3
= > A < 2
(sao cứ thấy có vấn đề nhỉ?)