\(C=1-\frac{2}{2.3}+1-\frac{2}{3.4}+...+1-\frac{2}{2019.2020}\)

\(=2018-2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\right)\)

\(=2018-2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\right)\)

\(=2018-2\left(\frac{1}{2}-\frac{1}{2020}\right)\)

\(=2018-2.\frac{1009}{2020}\)

\(=2018-\frac{1009}{1010}\)

\(=\frac{2037171}{1010}\)

Cách lớp 7 nà:)

\(\frac{1}{n.\left(n+1\right)^2}=\frac{1}{n.\left(n+1\right).\left(n+1\right)}< \frac{1}{n.n\left(n+1\right)}< \frac{1}{\left(n-1\right)n\left(n+1\right)}\) (n>=2_

\(\text{Suy ra }VT< \frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{\left(n-1\right)n\left(n+1\right)}\)

Mặt khác ta có công thức \(\frac{1}{\left(n-1\right)n\left(n+1\right)}=\frac{\left[\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right]}{2}\) (n>= 2)

Suy ra \(VT< \frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{n\left(n+1\right)}\right)< \frac{1}{2}.\frac{1}{2}=\frac{1}{4}\left(\text{do }\frac{1}{n\left(n+1\right)}>0\right)\)

Vậy ta có đpcm

Gắt chưa??? :>> Dương Bá Gia Bảo

kinh!

oOo Hello the world oOo, làm được không?

Hình như đề là thế này :

\(\frac{1}{\sqrt{1}+\sqrt{2}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}=9\)

= \(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}=\sqrt{100}-1=10-1=9\)

ta có \(\frac{1}{\sqrt{1.2}}khác\frac{1}{\sqrt{1}+\sqrt{2}}\)

................................

 \(\frac{1}{\sqrt{99.100}}khấc\frac{1}{\sqrt{99}+\sqrt{100}}\)

\(\sqrt{1+\frac{8n^2-1}{\left(2n-1\right)^2\left(2n+1\right)^2}}=\sqrt{1+\frac{8n^2-1}{\left(4n^2-1\right)^2}}=\sqrt{\frac{\left(4n^2-1\right)^2+8n^2-1}{\left(4n^2-1\right)^2}}\)

\(=\sqrt{\frac{16n^4-8n^2+1+8n^2-1}{\left(4n^2-1\right)^2}}=\frac{4n^2}{4n^2-1}=1+\frac{1}{4n^2-1}=1+\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)\)

\(\Rightarrow S=1009+\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)

\(=1009+\frac{1}{2}\left(1-\frac{1}{2019}\right)=...\)