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Tổng quát: \(1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}=1+\frac{\left(n+1\right)^2+n^2}{n^2\left(n+1\right)^2}=1+\frac{1}{n^2\left(n+1\right)^2}+\frac{2}{n\left(n+1\right)}\)
\(=\left(1+\frac{1}{n\left(n+1\right)}\right)^2=\left(1+\frac{1}{n}-\frac{1}{n+1}\right)^2\)
\(\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\left|1+\frac{1}{n}-\frac{1}{n+1}\right|=1+\frac{1}{n}-\frac{1}{n+1}\)
Áp dụng ta được:
\(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{2009^2}+\frac{1}{2010^2}}\)
\(=1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+...+1+\frac{1}{2009}-\frac{1}{2010}\)
\(=2008+\frac{1}{2}-\frac{1}{2010}\)
\(=2008\frac{502}{1005}\)
Câu a:
Có dạng tổng quát:\(\frac{1}{\left(k+1\right)\sqrt{k}+k\sqrt{x+1}}=\frac{1}{\sqrt{\left(k+1\right)k}\left(\sqrt{k+1}+\sqrt{k}\right)}=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{\left(k+1\right)k}}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k-1}}\)
Áp dụng kết quả trên suy ra câu a
Xét với n là số tự nhiên không nhỏ hơn 1 , ta có
\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\)
\(=\left(1+\frac{\sqrt{n}}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Áp dụng điều trên :
\(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2010\sqrt{2009}}< \)
\(< 2\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2009}}-\frac{1}{\sqrt{2010}}\right)=2\left(1-\frac{1}{\sqrt{2010}}\right)< \)
\(< 2\left(1-\frac{1}{\sqrt{2025}}\right)=\frac{88}{45}\)
Tổng quát \(n\in N\text{*};n\ge2\) ta có \(\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}+\frac{2\left(n+1-n-1\right)}{n\left(n+1\right)}}\)
\(=\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}+2\cdot1\cdot\frac{1}{n}-2\cdot1\cdot\frac{1}{n+1}-2\cdot\frac{1}{n}\cdot\frac{1}{n+1}}\)
\(=\sqrt{\left(1+\frac{1}{n}-\frac{1}{n-1}\right)^2}=1+\frac{1}{n}-\frac{1}{n-1}\).Áp dụng vào ta có:
\(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{2008^2}+\frac{1}{2009^2}}=1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+...+1+\frac{1}{2008}-\frac{1}{2009}\)
\(=\left(1+1+...+1\right)+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2008}-\frac{1}{2009}\right)\)
Super dễ nhé !! Cho bn xử nốt
\(\text{Ta thấy: }\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}=1+\frac{1}{1}-\frac{1}{2}\)
\(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}=1+\frac{1}{2}-\frac{1}{3}\)
\(...................\)
\(\sqrt{1+\frac{1}{2009^2}+\frac{1}{2010^2}}\)
\(\text{Suy ra: }\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{2009^2}+\frac{1}{2010^2}}\)
\(=1+\frac{1}{1}-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+...+1+\frac{1}{2009}-\frac{1}{2010}\)
\(=2009+\frac{1}{1}-\frac{1}{2010}=2010-\frac{1}{2010}\)