Cho \(S=\frac{4}{50}+\frac{4}{51}+\frac{4}{52}+.......+\frac{4}{99}\)
Chứng minh rằng 2<S<4
K
Khách
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19 tháng 5 2021
Giải:
\(S=\dfrac{1}{50}+\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{98}+\dfrac{1}{99}\)
\(S=\left(\dfrac{1}{50}+\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{74}\right)+\left(\dfrac{1}{75}+...+\dfrac{1}{98}+\dfrac{1}{99}\right)\)
\(\Rightarrow S>\left(\dfrac{1}{50}+\dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}\right)+\left(\dfrac{1}{75}+...+\dfrac{1}{75}+\dfrac{1}{75}\right)\)
\(\Rightarrow S>\dfrac{1}{2}+\dfrac{1}{3}>\dfrac{1}{2}\)
\(\Rightarrow S>\dfrac{1}{2}\left(đpcm\right)\)
26 tháng 2 2019
Ta có: \(\frac{1}{50}\)>\(\frac{1}{100}\)
\(\frac{1}{51}\)>\(\frac{1}{100}\)
\(\frac{1}{52}\)>\(\frac{1}{100}\)
..................
\(\frac{1}{99}\)>\(\frac{1}{100}\)
=>\(\frac{1}{50}\)+\(\frac{1}{51}\)+.............+\(\frac{1}{99}\)>\(\frac{1}{100}\).50=\(\frac{1}{2}\)(50 là số số hạng của S nha)
=>S>\(\frac{1}{2}\)
PH
Cho \(S=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
50 mũ 2 nhé
Chứng minh rằng S<\(\frac{3}{4}\)
1
7 tháng 4 2017
\(S=\frac{1}{4}+\left(\frac{1}{3^2}+\frac{1}{4^2}+..+\frac{1}{50^2}\right)\)
Xét \(A=\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\)
\(A< \frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A< \frac{1}{2}-\frac{1}{50}< \frac{1}{2}\)
\(=>A< \frac{1}{2}\)
=>\(S=\frac{1}{4}+A< \frac{1}{4}+\frac{1}{2}=\frac{3}{4}\)
vậy S<3/4
N
6 tháng 4 2019
\(S=\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{98}+\frac{1}{99}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\)(có 50 số hạng)\(=\frac{50}{100}=\frac{1}{2}\)
Vậy \(S>\frac{1}{2}\) .
6 tháng 4 2019
Có: \(\frac{1}{50}>\frac{1}{100}\\ \frac{1}{51}>\frac{1}{100}\\ \frac{1}{52}>\frac{1}{100}\\ .\\ .\\ .\\ \frac{1}{98}>\frac{1}{100}\\ \frac{1}{99}>\frac{1}{100}\)
\(\Rightarrow\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{98}+\frac{1}{99}>\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}+\frac{1}{100}\)(có 50 số hạng \(\frac{1}{100}\))
\(\Rightarrow S>\frac{1}{100}\cdot50\)
\(\Rightarrow S>\frac{50}{100}\)
\(\Rightarrow S>\frac{1}{2}\left(đpcm\right)\)
KN
1 tháng 5 2019
\(M=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}\)
\(\Rightarrow M< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}\)
\(\Rightarrow M< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}\)
\(\Rightarrow M< 1-\frac{1}{99}< 1\)
Dễ thấy M > 0 nên 0 < M < 1
Vậy M không là số tự nhiên.
KN
1 tháng 5 2019
\(S=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)
\(\Rightarrow S>\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\) (50 số hạng \(\frac{1}{100}\))
\(\Rightarrow S>\frac{1}{100}.50=\frac{1}{2}\)
Vậy \(S>\frac{1}{2}\left(đpcm\right)\)
11 tháng 3 2022
Xét \(4S=1+\dfrac{2}{4}+\dfrac{3}{4^2}+\dfrac{4}{4^3}+...+\dfrac{2014}{4^{2013}}\)
=> \(3S=4S-S=\left(1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2014}{4^{2013}}\right)-\left(\dfrac{1}{4}+\dfrac{2}{4^2}+...+\dfrac{2014}{4^{2014}}\right)\)
=> \(3S=1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2013}}-\dfrac{2014}{4^{2014}}< 1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2013}}\)
Đặt \(A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2013}}\)
=> \(4A=4+1+\dfrac{1}{4}+...+\dfrac{1}{4^{2012}}\)
=> \(3A=4A-A=\left(4+1+\dfrac{1}{4}+...+\dfrac{1}{4^{2012}}\right)-\left(1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2013}}\right)\)
=> \(3A=4-\dfrac{1}{4^{2013}}< 4\)
=> \(A< \dfrac{4}{3}\)
=> \(3S< \dfrac{4}{3}\)
=> \(S< \dfrac{4}{9}< \dfrac{1}{2}\)
11 tháng 3 2022
\(4S=1+\frac{2}{4}+\frac{3}{4^2}+....+\frac{2014}{4^{2013}}\)
\(4S-S=3S=1+\frac{2}{4}+\frac{3}{4^2}+....+\frac{2014}{4^{2013}}-\left(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+....+\frac{2014}{4^{2014}}\right)\)
\(3S=1+\left(\frac{2}{4}-\frac{1}{4}\right)+\left(\frac{3}{4^2}-\frac{2}{4^2}\right)+......+\left(\frac{2014}{4^{2013}}-\frac{2013}{4^{2013}}\right)-\frac{2014}{4^{2014}}\)
\(3S=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+.....+\frac{1}{4^{2013}}-\frac{2014}{4^{2014}}\)
đặt \(A=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+....+\frac{1}{4^{2023}}\)
\(4A-A=4+1+\frac{1}{4}+\frac{1}{4^2}+.....+\frac{1}{4^{2022}}-\left(1+\frac{1}{4}+\frac{1}{4^2}+....+\frac{1}{4^{2023}}\right)\)
\(3A=4-\frac{1}{4^{2023}}\)
\(A=\frac{4}{3}-\frac{1}{3.4^{2023}}\)
\(\Rightarrow3S=\frac{4}{3}-\frac{1}{3.4^{2023}}-\frac{2014}{4^{2024}}\)
\(\Rightarrow S=\frac{4}{9}-\frac{1}{9.4^{2023}}-\frac{2014}{3.4^{2024}}\)
do \(\frac{4}{9}< \frac{4}{8}=\frac{1}{2}\)
\(\Rightarrow S=\frac{4}{9}-\frac{1}{9.4^{2023}}-\frac{2014}{3.4^{2024}}< \frac{4}{8}=\frac{1}{2}\left(đpcm\right)\)