a: \(A=\dfrac{5}{3\cdot7}+\dfrac{5}{7\cdot11}+...+\dfrac{5}{83\cdot87}\)
\(=\dfrac{5}{4}\left(\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+...+\dfrac{4}{83\cdot87}\right)\)
\(=\dfrac{5}{4}\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{83}-\dfrac{1}{87}\right)\)
\(=\dfrac{5}{4}\left(\dfrac{1}{3}-\dfrac{1}{87}\right)=\dfrac{5}{4}\cdot\dfrac{28}{87}=\dfrac{35}{87}\)
b: \(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}=1-\dfrac{1}{2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)
...
\(\dfrac{1}{50^2}< \dfrac{1}{49\cdot50}=\dfrac{1}{49}-\dfrac{1}{50}\)
Do đó: \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
=>\(\dfrac{1}{1^2}+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}< 1-\dfrac{1}{50}+1=2-\dfrac{1}{50}\)
=>\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2^2}\left(2-\dfrac{1}{50}\right)=\dfrac{1}{2}-\dfrac{1}{200}\)
=>\(S< \dfrac{1}{2}\)
c: \(\dfrac{1}{51}>\dfrac{1}{100}\)
\(\dfrac{1}{52}>\dfrac{1}{100}\)
...
\(\dfrac{1}{99}>\dfrac{1}{100};\dfrac{1}{100}=\dfrac{1}{100}\)
Do đó: \(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}>\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{50}{100}=\dfrac{1}{2}\)
\(\dfrac{1}{51}< \dfrac{1}{50}\)
\(\dfrac{1}{52}< \dfrac{1}{50}\)
...
\(\dfrac{1}{99}< \dfrac{1}{50};\dfrac{1}{100}< \dfrac{1}{50}\)
Do đó: \(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}< \dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}=\dfrac{50}{50}=1\)
Do đó: \(\dfrac{1}{2}< \dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}< 1\)
6a.
$A=5(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{83.87})$
$4A=5(\frac{7-3}{3.7}+\frac{11-7}{7.11}+\frac{15-11}{11.15}+...+\frac{87-83}{83.87})$
$=5(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+....+\frac{1}{83}-\frac{1}{87})$
$=5(\frac{1}{3}-\frac{1}{87})=\frac{140}{87}$
$\Rightarrow A=\frac{140}{87}:4=\frac{35}{87}$
6b.
$S=\frac{1}{4}(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2})$
$<\frac{1}{4}(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50})$
$=\frac{1}{4}(1+\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50})$
$=\frac{1}{4}(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50})$
$=\frac{1}{4}(2-\frac{1}{50})=\frac{1}{2}-\frac{1}{200}< \frac{1}{2}$
6c.
$\frac{1}{51}< \frac{1}{50}$
$\frac{1}{52}< \frac{1}{50}$
...............
$\frac{1}{100}< \frac{1}{50}$
$\Rightarrow \frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}< \underbrace{\frac{1}{50}+\frac{1}{50}+....+\frac{1}{50}}_{50}=\frac{1}{50}.50=1(*)$
Lại có:
$\frac{1}{51}+\frac{1}{52}+..+\frac{1}{100}> \underbrace{\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}}_{50}=\frac{50}{100}=\frac{1}{2}(**)$
Từ $(*); (**)$ ta có đpcm.