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S=1/5.6+1/10.9+1/15.12+...+1/3350.2013
=(1/5).(1/3).(1/1.2+1/2.3+1/3.4+...+1/670.671)
=(1/15). (1-1/2+1/2-1/3+...+1/670-1/671)
=(1/15). (1-1/671)
=1/15.670/671
=134/2013
Đặt tổng trên là A ta có
\(2A=\frac{2}{10.12}+\frac{2}{12.14}+\frac{2}{14.16}+...+\frac{2}{48.52}\)
\(2A=\frac{12-10}{10.12}+\frac{14-12}{12.14}+\frac{16-14}{14.16}+...+\frac{50-48}{48.50}\)
\(2A=\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+\frac{1}{14}-\frac{1}{16}+...+\frac{1}{48}-\frac{1}{50}=\frac{1}{10}-\frac{1}{50}=\frac{2}{25}\)
\(\Rightarrow A=\frac{2A}{2}=\frac{1}{25}\)
\(S=\dfrac{2}{10\cdot12}+\dfrac{2}{12\cdot14}+...+\dfrac{2}{98\cdot100}\)
\(S=\dfrac{2}{10}-\dfrac{2}{12}+\dfrac{2}{12}-\dfrac{2}{14}+...+\dfrac{2}{98}-\dfrac{2}{100}\)
\(S=\dfrac{2}{10}-\dfrac{2}{100}=\dfrac{9}{50}=0,18\)
Vậy \(S>\dfrac{1}{10}\)
\(S=\dfrac{2}{10\cdot12}+\dfrac{2}{12\cdot14}+\dfrac{2}{14\cdot16}+...+\dfrac{2}{98\cdot100}\)
\(S=\dfrac{2}{10}-\dfrac{2}{12}+\dfrac{2}{12}-\dfrac{2}{14}+...+\dfrac{2}{98}-\dfrac{2}{100}\)
\(S=\dfrac{2}{10}-\dfrac{2}{100}\)
\(S=\dfrac{20}{100}-\dfrac{2}{100}\)
\(S=\dfrac{18}{100}=\dfrac{9}{50}=0,18\)
\(\dfrac{1}{10}=0,1\), mà \(0,1< 0,18\)
\(\Rightarrow S>\dfrac{1}{10}\left(đpcm\right)\)
Đặt \(A=\frac{2}{10\cdot12}+\frac{2}{12\cdot14}+\frac{2}{14\cdot16}+...+\frac{2}{48\cdot50}\)
\(A=\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+...+\frac{1}{48}-\frac{1}{50}\)
\(A=\frac{1}{10}-\frac{1}{50}=\frac{5}{50}-\frac{1}{50}=\frac{4}{50}=\frac{2}{25}\)
Vậy \(A=\frac{2}{25}\)
= \(\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+...+\frac{1}{48}-\frac{1}{50}\)
= \(\frac{1}{10}-\frac{1}{50}\)= \(\frac{2}{25}\)
\(\frac{8^2}{7.9}.\frac{9^2}{8.10}...\frac{14^2}{13.15}\)
\(\frac{8.8}{7.9}.\frac{9.9}{8.10}...\frac{14.14}{13.15}\)
\(\frac{8.9...14}{7.8...13}.\frac{8.9...14}{9.10...15}\)
\(\frac{14}{7}.\frac{8}{15}\)
\(2.\frac{8}{15}\)
\(\frac{16}{15}\)
(8.9.10.11.12.13.14)(8.9.10.11.12.13.14)/7.8.9.10.11.12.13).(9.10.11.12.13.14.15)
=14.8/7.15
=16/15
k cho mình nhá
Ta có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}.\)
Mà\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
\(=1-\frac{1}{9}=\frac{8}{9}\)
\(\Rightarrow S< \frac{8}{9}\)
Và \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{9.10}\)
Mà \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
\(\Rightarrow S>\frac{2}{5}\)
Vậy: \(\frac{2}{5}< S< \frac{8}{9}\)
a)Ta có: \(\frac{3}{1.4}=\frac{4-1}{1.4}=1-\frac{1}{4}\)
\(\frac{3}{4.7}=\frac{7-4}{4.7}=\frac{1}{4}-\frac{1}{7}\)
... . . . .
\(\frac{3}{n\left(n+3\right)}=\frac{1}{n}-\frac{1}{n+3}\)
\(\Leftrightarrow S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+3}< 1^{\left(đpcm\right)}\)
b) Ta có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
Suy ra \(\frac{2}{5}< S\) (1)
Ta lại có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\)
Từ đó suy ra S < 8/9
Từ (1) và (2) suy ra đpcm
S=\(\frac{2}{10.12}+\frac{2}{12.14}+\frac{2}{14.16}+.....+\frac{2}{98.100}\)
S=\(\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+........+\frac{1}{98}-\frac{1}{100}\)
S=\(\frac{1}{10}-\frac{1}{100}\)
S=\(\frac{9}{100}\)<\(\frac{1}{10}\)
tính S = cánh tính sai phân