\(A=\frac{1}{\sqrt{2001}+\sqrt{2003}}+\frac{1}{\sqrt{2003}+\sqrt{2005}}+...+\frac{1}{\sqrt{2015}+\sqrt{2017}}\)
Ta có công thức:
\(\frac{1}{\sqrt{n}+\sqrt{n+1}}=\frac{\sqrt{n+1}-\sqrt{n}}{\left(\sqrt{n}+\sqrt{n+1}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}=\sqrt{n+1}-\sqrt{n}\)
Áp dụng vào công thức ta có:
\(A=\frac{1}{\sqrt{2001}+\sqrt{2003}}+\frac{1}{\sqrt{2003}+\sqrt{2005}}+...+\frac{1}{\sqrt{2015}+\sqrt{2017}}\)
\(A=\sqrt{2003}-\sqrt{2001}+\sqrt{2005}-\sqrt{2003}+...+\sqrt{2017}-\sqrt{2015}\)
\(A=\sqrt{2017}-\sqrt{2001}\approx0,17848\)
\(A=\frac{1}{\sqrt{2001}+\sqrt{2003}}+\frac{1}{\sqrt{2003}+\sqrt{2005}}+....+\frac{1}{\sqrt{2015}+\sqrt{2017}}\)
\(=\sqrt{2003}-\sqrt{2001}+\sqrt{2005}-\sqrt{2003}+....+\sqrt{2017}-\sqrt{2015}\)
\(=-\sqrt{2001}+\sqrt{2017}\)
Sai hết kìa ; thắng nghuyễn cx sai nốt :((
\(\frac{1}{\sqrt{n}+\sqrt{n+2}}=\frac{\sqrt{n+2}-\sqrt{n}}{\left(\sqrt{n+2}+\sqrt{n}\right)\left(\sqrt{n+2}-\sqrt{n}\right)}=\frac{\sqrt{n+2}-\sqrt{n}}{\left(n+2\right)-n}\)
\(=\frac{\sqrt{n+2}-\sqrt{n}}{2}\)
\(\Rightarrow A=\frac{\sqrt{2003}-\sqrt{2001}}{2}+\frac{\sqrt{2005}-\sqrt{2003}}{2}+...+\frac{\sqrt{2017}-\sqrt{2015}}{2}\)
\(=\frac{\sqrt{2017}-\sqrt{2001}}{2}\)
