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\(\dfrac{1}{5^2}< \dfrac{1}{4.5}\)
\(\dfrac{1}{6^2}< \dfrac{1}{5.6}\)
......
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(\Rightarrow\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\)
Ta có: \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\)
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
\(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{2019^2}>\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+...+\dfrac{1}{2019\cdot2020}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{2019}-\dfrac{1}{2020}=\dfrac{1}{5}-\dfrac{1}{2020}=\dfrac{404-1}{2020}=\dfrac{403}{2020}>\dfrac{40}{2020}=\dfrac{20}{101}\left(1\right)\) \(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{2019^2}< \dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+...+\dfrac{1}{2018\cdot2019}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{2018}-\dfrac{1}{2019}=\dfrac{1}{4}-\dfrac{1}{2019}=\dfrac{2019-4}{4\cdot2019}=\dfrac{2015}{4\cdot2019}< \dfrac{2019}{4\cdot2019}=\dfrac{1}{4}\left(2\right)\) Từ (1) và (2) \(\Rightarrow\dfrac{20}{101}< A< \dfrac{1}{4}\)
Giải:
Ta có:
\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{99.100}\)
Đặt \(A=\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
\(B=\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)
\(A=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=\dfrac{1}{4}-\dfrac{1}{100}\)
\(A=\dfrac{6}{25}\)
Mà \(\dfrac{1}{6}< \dfrac{6}{25}< \dfrac{1}{4}\)
Ta lại có \(A< \dfrac{6}{25}\)
Vậy \(\dfrac{1}{6}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}\)
1/5^2< 1/4.5=1/4-1/5
1/6^2<1/5.6=1/5-1/6
..
1/99^2<1/98.99=1/98-1/99
1/100^2<1/99.100=1/99-1/100
Cộng vế theo vế, đơn giản:
=> 1/5^2+1/6^2+...+1/100^2< 1/4 -1/100<1/4
**
1/5^2> 1/5.6=1/5-1/6
1/6^2>1/6.7=1/6-1/7
..
1/99^2>1/99.100=1/99-1/100
1/100^2>1/100.101=1/100-1/101
Cộng vế theo vế, đơn giản:
=> 1/5^2+1/6^2+...+1/100^2>1/5 -1/101=96/505>1/6
Vậy:
1/6<1/5^2+1/6^2+...+1/100^2<1/4
2,
\(M=\dfrac{\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{11}}\) =\(\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\)
\(=\dfrac{3}{4}\)
`S=1/19+1/19^2+1/19^3+........+1/19^20`
`=>19S=1+1/19+1/19^2+.....+1/19^19`
`=>19S-S=18S=1-1/19^20<1`
`=>S<1/18(đpcm)`
Giải:
S=\(\dfrac{1}{19}+\dfrac{1}{19^2}+\dfrac{1}{19^3}+...+\dfrac{1}{19^{10}}\)
19S=\(1+\dfrac{1}{19}+\dfrac{1}{19^2}+...+\dfrac{1}{19^9}\)
19S-S=\(\left(1+\dfrac{1}{19}+\dfrac{1}{19^2}+...+\dfrac{1}{19^9}\right)-\left(\dfrac{1}{19}+\dfrac{1}{19^2}+\dfrac{1}{19^3}+...+\dfrac{1}{19^{10}}\right)\)
18S=1-\(\dfrac{1}{19^{10}}\)
S=(1-\(\dfrac{1}{19^{10}}\) ):18
S=\(1:18-\dfrac{1}{19^{10}}:18\)
S=\(\dfrac{1}{18}-\dfrac{1}{19^{10}.18}\)
⇒S<\(\dfrac{1}{18}\) (đpcm)
Chúc bạn học tốt!
\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{20}-\dfrac{1}{20}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{20}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{10}\right)\)
\(=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}\) (đpcm)
Ta có \(\dfrac{6}{15}>\dfrac{6}{16}>...>\dfrac{6}{19}\) nên \(S< \dfrac{6}{15}.5=2\).
Lại có \(S>\dfrac{6}{19}.5>1\) nên \(1< S< 2\)
\(\dfrac{1}{19}< \dfrac{1}{5}\) ;\(\dfrac{1}{18}< \dfrac{1}{5}\) ;\(\dfrac{1}{17}< \dfrac{1}{5}\) ;\(\dfrac{1}{16}< \dfrac{1}{5}\) ;\(\dfrac{1}{15}< \dfrac{1}{5}\) ;\(\dfrac{1}{14}< \dfrac{1}{5}\)
\(\dfrac{1}{13}< \dfrac{1}{5}\) ;\(\dfrac{1}{12}< \dfrac{1}{5}\) ;\(\dfrac{1}{11}< \dfrac{1}{5}\); \(\dfrac{1}{10}< \dfrac{1}{5}\) ;\(\dfrac{1}{9}< \dfrac{1}{5}\) ;\(\dfrac{1}{8}< \dfrac{1}{5}\)
\(\dfrac{1}{7}< \dfrac{1}{5}\) ;\(\dfrac{1}{6}< \dfrac{1}{5}\)
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?)