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\(S=\frac{1}{3}+\frac{1}{9}+\frac{1}{18}+\frac{1}{30}+\frac{1}{45}+...+\frac{1}{14850}\)
\(\Rightarrow\frac{3}{2}S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(S=\frac{1}{3}+\frac{1}{9}+\frac{1}{30}+\frac{1}{45}+...+\frac{1}{14850}\)
\(\Rightarrow\frac{3}{2}S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
Vậy S = \(\frac{99}{100}:\frac{3}{2}\) = \(\frac{33}{50}\)
\(F=\left(\frac{3}{1.8}+\frac{3}{8.15}+\frac{3}{15.22}+...+\frac{3}{106.113}\right)\)\(-\)\(\left(\frac{25}{50.55}+\frac{25}{55.60}+\frac{25}{60.65}+...+\frac{25}{95.100}\right)\)
\(=\frac{3}{7}\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{15}+...+\frac{1}{106}-\frac{1}{113}\right)\) - \(5\left(\frac{1}{50}-\frac{1}{55}+\frac{1}{55}-\frac{1}{60}+...+\frac{1}{95}-\frac{1}{100}\right)\)
\(=\frac{3}{7}\left(\frac{1}{3}-\frac{1}{113}\right)-5\left(\frac{1}{50}-\frac{1}{100}\right)\)
\(=\frac{3}{7}.\frac{110}{339}-5.\frac{1}{100}\)
\(=\frac{1}{7}-\frac{1}{20}=\frac{13}{140}\)
= \(\frac{3}{7}\left(\frac{7}{1.8}+\frac{7}{8.15}+...+\frac{7}{106.103}\right)-5\left(\frac{5}{50.55}+\frac{5}{55.60}+...+\frac{5}{95.100}\right)\)
=\(\frac{3}{7}\left(1-\frac{1}{8}+\frac{1}{8}-\frac{1}{15}+...+\frac{1}{106}-\frac{1}{113}\right)-5\left(\frac{1}{50}-\frac{1}{55}+\frac{1}{55}-\frac{1}{60}+...+\frac{1}{95}-\frac{1}{100}\right)\)
=\(\frac{3}{7}\left(1-\frac{1}{113}\right)-5\left(\frac{1}{50}-\frac{1}{100}\right)\)
=\(\frac{3}{7}\cdot\frac{112}{113}-5\cdot\frac{1}{100}\)
=\(\frac{48}{113}-\frac{1}{20}\)
=\(\frac{847}{2260}\)
Như vậy ta sẽ so sánh 1 và 1/3 + 1/6 + 1/10 +......+ 1/45
Ta có : 1/3 + 1/6 + 1/10 + .....+ 1/45 < 1/10 + 1/10 + 1/10 +......+ 1/10
Mà 1/10 + 1/10 + 1/10 + ....+ 1/10 = 8/10 < 1
Vậy S <2
ta có 1/3=10/30
1/21+1/22+...+1/30 có 10 p/số
mà 1/21>1/30
1/22>1/30
....
1/29>1/30
1/30=1/30
=>1/21+..1/30>1/30+....1/30 có 10 phân số
=>1/21+...1/30>1/3
Ta có
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
..............
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
=> S < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
S < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(S< 1-\dfrac{1}{100}< 1\)(do 1/100 >0)
ĐPcm
Giải:
\(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}\)
Ta có:
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}=\dfrac{1}{4.4}< \dfrac{1}{3.4}\)
\(...\)
\(\dfrac{1}{99^2}=\dfrac{1}{99.99}< \dfrac{1}{98.99}\)
\(\dfrac{1}{100^2}=\dfrac{1}{100.100}< \dfrac{1}{99.100}\)
\(\Rightarrow S< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\)
\(\Rightarrow S< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow S< \dfrac{1}{1}-\dfrac{1}{100}< 1\)
\(\Rightarrow S< 1\)
Vậy S < 1.
S=1/5+1/52+1/53+...+1/58
5S=1+1/5+1/52+...+1/57
4S=1-1/58<1
\(\Rightarrow\)S<1
gium nha
S=1/5+1/52+1/53+...+1/58
5S=1+1/5+1/52+...+1/57
4S=1-1/58<1
=> S<1
\(S=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{2014}{5^{2014}}\)
\(5S=1+\frac{2}{5}+\frac{3}{5^2}+...+\frac{2014}{5^{2013}}\)
\(\Rightarrow5S-S=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2013}}-\frac{2014}{5^{2014}}\)
\(S=\frac{1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2013}}-\frac{2014}{5^{2014}}}{4}\)
Xét \(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2013}}\)
\(5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2012}}\)
\(5A-A=1-\frac{1}{5^{2013}}\Leftrightarrow A=\frac{1-\frac{1}{5^{2013}}}{4}=\frac{1}{4}-\frac{1}{4.5^{2013}}\)
\(\Rightarrow S=\frac{1+\frac{1}{4}-\left(\frac{1}{4.5^{2013}}+\frac{2014}{5^{2014}}\right)}{4}=\frac{5}{16}-\frac{\frac{1}{4.5^{2013}}+\frac{2014}{5^{2014}}}{4}< \frac{1}{3}\)
a,Ta có: \(\frac{3}{10}=\frac{3}{10};\frac{3}{11}< \frac{3}{10};\frac{3}{12}< \frac{3}{10};\frac{3}{13}< \frac{3}{10};\frac{3}{14}< \frac{3}{10}\)
\(\Rightarrow S< \frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}=\frac{15}{10}=\frac{3}{2}=1,5\left(1\right)\)
Lại có: \(\frac{3}{10}>\frac{3}{15};\frac{3}{11}>\frac{3}{15};\frac{3}{12}>\frac{3}{15};\frac{3}{13}>\frac{3}{15};\frac{3}{14}>\frac{3}{15}\)
\(\Rightarrow S>\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}=\frac{15}{15}=1\left(2\right)\)
Từ (1) và (2) => 1 < S < 1,5
Vậy...
b, \(A=\frac{1}{61}+\frac{1}{62}+...+\frac{1}{100}\)
\(=\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}\right)+\left(\frac{1}{81}+\frac{1}{82}+...+\frac{1}{100}\right)\)
Ta có: \(\frac{1}{61}>\frac{1}{80};\frac{1}{62}>\frac{1}{80};...;\frac{1}{80}=\frac{1}{80}\)
\(\Rightarrow\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}>\frac{1}{80}+\frac{1}{80}+...+\frac{1}{80}=\frac{20}{80}=\frac{1}{4}\left(1\right)\)
Lại có: \(\frac{1}{81}>\frac{1}{100};\frac{1}{82}>\frac{1}{100};...;\frac{1}{100}=\frac{1}{100}\)
\(\Rightarrow\frac{1}{81}+\frac{1}{82}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{20}{100}=\frac{1}{5}\left(2\right)\)
Từ (1) và (2) => \(A>\frac{1}{4}+\frac{1}{5}=\frac{9}{20}\)
Vậy...
S=\(\frac{1}{3}.\left(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{4950}\right)\)
S=\(\frac{1}{3}.2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\right)\)
S=\(\frac{2}{3}.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
S=\(\frac{2}{3}.\left(1-\frac{1}{100}\right)=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)
\(\frac{33}{50}>\frac{30}{50}=\frac{3}{5}->S>\frac{3}{5}\)