\(A=4\left(\frac{1}{20}-\frac{1}{80}\right)=4.\frac{3}{80}=60\)

\(A=\frac{16}{20\cdot24}+\frac{16}{24\cdot28}+\frac{16}{28\cdot32}+...+\frac{16}{76\cdot80}\)

\(A=4\left[\frac{4}{20\cdot24}+\frac{4}{24\cdot28}+\frac{4}{28\cdot32}+...+\frac{4}{76\cdot80}\right]\)

\(A=4\left[\frac{1}{20}-\frac{1}{24}+...+\frac{1}{76}-\frac{1}{80}\right]\)

\(A=4\left[\frac{1}{20}-\frac{1}{80}\right]\)

\(A=4\left[\frac{4}{80}-\frac{1}{80}\right]=4\cdot\frac{3}{80}=\frac{4\cdot3}{80}=\frac{1\cdot3}{20}=\frac{3}{20}\)

\(A=\frac{16}{20\cdot24}+\frac{16}{24\cdot28}+...+\frac{16}{76\cdot80}\)

\(A=\frac{4^2}{\left(5\cdot6\right)\cdot4^2}+\frac{4^2}{\left(6\cdot7\right)\cdot4^2}+...+\frac{4^2}{\left(19\cdot20\right)\cdot4^2}\)

\(A=\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{19\cdot20}\)

\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{20}=\frac{1}{5}-\frac{1}{20}=\frac{3}{20}\)

Lời giải:

$D=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+......+\frac{2018}{4^{2018}}+\frac{2019}{4^{2019}}$

$4D=1+\frac{2}{4}+\frac{3}{4^2}+....+\frac{2018}{4^{2017}}+\frac{2019}{4^{2018}}$

Trừ theo vế:

\(3D=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+....+\frac{1}{4^{2018}}-\frac{2019}{4^{2019}}\)

\(\Rightarrow 12D=4+1+\frac{1}{4}+\frac{1}{4^2}+....+\frac{1}{4^{2017}}-\frac{2019}{4^{2018}}\)

Trừ theo vế:
$9D=4-\frac{2019}{4^{2018}}+\frac{2019}{4^{2019}}-\frac{1}{4^{2018}}$

$=4-\frac{6061}{4^{2019}}< 4$

$\Rightarrow D< \frac{4}{9}<\frac{4}{8}$ hay $D< \frac{1}{2}$ (đpcm)

15135454

 Ta có:\(\frac{1}{2^2}=\frac{1}{4};\frac{1}{3^2}< \frac{1}{2\cdot3}=\frac{1}{2}-\frac{1}{3};\frac{1}{3^2}< \frac{1}{3\cdot4}=\frac{1}{3}-\frac{1}{4};.....;\frac{1}{100^2}< \frac{1}{99\cdot100}=\frac{1}{99}-\frac{1}{100}\)

\(A=\frac{1}{4}+\frac{1}{2}-\frac{1}{100}< \frac{3}{4}\left(đpcm\right)\)

Gọi \(D=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}+\frac{1}{100^2}< \frac{3}{4}\)

Vì \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{100^2}< \frac{1}{99.100}\)

Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}< \frac{3}{4}\)

\(\Rightarrow D< \frac{3}{4}\left(đpcm\right)\)

Thôi, cho phép mình góp ý bài mình đã làm bằng cách đơn giản hơn nha ^^.

Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}\) ta có:

\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{2010^2}< \frac{1}{2009.2010}\)

\(=A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)

\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{2009}-\frac{1}{2010}\)

\(\Rightarrow A< 1-\frac{1}{2010}\)

\(\Rightarrow A< 1\)

\(\Rightarrow A< \frac{3}{4}\)

Có: \(\frac{1}{2^2}< \frac{1}{1.2}\); \(\frac{1}{3^2}< \frac{1}{2.3}\);...;\(\frac{1}{2010^2}< \frac{1}{2009.2010}\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2010^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2009.2010}=1-\frac{1}{2010}=\frac{2009}{2010}\)Mà \(\frac{2009}{2010}>\frac{3}{4}\) -> Sai đề

Với mọi k ta luôn có \(k^2\ge k^2-1=\left(k-1\right)\left(k+1\right)\)

\(\Rightarrow\frac{1}{k^2}\le\frac{1}{\left(k-1\right)\left(k+1\right)}=\frac{1}{2}.\left(\frac{1}{k-1}-\frac{1}{k+1}\right)\)

Áp dụng vào ta suy ra

\(2A\le\frac{1}{2}+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+...+\left(\frac{1}{2009}-\frac{1}{2011}\right)=\frac{1}{2}+\frac{1}{2}+\frac{1}{3}-\frac{1}{2010}-\frac{1}{2011}< \frac{3}{2}\)