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a,1/102+1/112+1/122+...+1/1002<1/9.10+1/10.11+1/11.12+...+1/99.100=1/9-1/10+1/10-1/11+...+1/99-1/100
=1/9-1/100=91/900<3/4
Vậy 1/102+1/112+1/122+...+1/1002<3/4
b,1/22+1/32+1/42+...+1/1002<1/1.2+1/2.3+1/3.4+...+1/99.100=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
=1-1/100=99/100
Vậy 1/22+1/32+1/42+...+1/1002<99/100
c,1/22+1/32+1/42+...+1/1002<1/22+(1/2.3+1/3.3+...+1/99.100)=1/4+(1/2-1/3+1/3-1/4+...+1/99-1/100)
=1/4+(1/2-1/100)=1/4+49/100=74/100<3/4=75/100
Vậy 1/22+1/32+1/42+...+1/1002<3/4
1/ Tính:
\(\frac{3}{2}-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}+\frac{19}{90}\)
\(=\frac{3}{1.2}-\frac{5}{2.3}+\frac{7}{3.4}-\frac{9}{4.5}+\frac{11}{5.6}-\frac{13}{6.7}+\frac{15}{7.8}-\frac{17}{8.9}+\frac{19}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
\(\frac{1}{11^2}+\frac{1}{12^2}+\frac{1}{13^2}+\frac{1}{14^2}+...+\frac{1}{100^2}\)
\(=\frac{1}{11.11}+\frac{1}{12.12}+\frac{1}{13.13}+\frac{1}{14.14}+...+\frac{1}{100.100}\)
\(< \frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}+...+\frac{1}{99.100}\)
\(=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{10}-\frac{1}{100}\)
Vì \(\frac{1}{100}>0\Rightarrow\frac{1}{10}-\frac{1}{100}< \frac{1}{10}\)
\(\RightarrowĐPCM\)
theo mình tình thi \(\frac{1}{11^2}+\frac{1}{12^2}+......+\frac{1}{100^2}=0,08521616902\)
mà \(\frac{1}{10}=0,1\)
\(\Rightarrow0,08521515902< 0,1\)