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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:
\(\frac{3}{2}-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}+\frac{19}{90}\)
\(=\frac{3}{1.2}-\frac{5}{2.3}+\frac{7}{3.4}-\frac{9}{4.5}+\frac{11}{5.6}-\frac{13}{6.7}+\frac{15}{7.8}-\frac{17}{8.9}+\frac{19}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
\(A=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
TA có :\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{50^2}< \frac{1}{49.50}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1\Rightarrow1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1+1=2\)
\(A=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)< \frac{1}{2^2}.2=\frac{1}{2}\left(đpcm\right)\)