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Ta co:
\(\frac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{n+1+n}< \frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n+1}.\sqrt{n}}=\frac{1}{2}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Ap vào bài toan được
\(S_n=\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+...+\frac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}\)
\(< \frac{1}{2}\left(\frac{1}{1}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{\sqrt{n+1}}\right)< \frac{1}{2}\)
Đáp án của bạn ở đây: https://dethihsg.com/de-thi-hoc-sinh-gioi-toan-9-phong-gddt-cam-thuy-2011-2012/amp/
\(f\left(n\right)=\dfrac{2n-1+2n+1+\sqrt{\left(2n+1\right)\left(2n+1\right)}}{\sqrt{2n+1}+\sqrt{2n-1}}\\ f\left(n\right)=\dfrac{\left(\sqrt{2n+1}-\sqrt{2n-1}\right)\left(2n-1+2n+1+\sqrt{\left(2n+1\right)\left(2n+1\right)}\right)}{2n+1-2n+1}\\ f\left(n\right)=\dfrac{\left(\sqrt{2n+1}\right)^3-\left(\sqrt{2n+1}\right)^3}{2}=\dfrac{\left(2n+1\right)\sqrt{2n+1}-\left(2n-1\right)\sqrt{2n+1}}{2}\)
\(\Leftrightarrow f\left(1\right)+f\left(2\right)+...+f\left(40\right)=\dfrac{3\sqrt{3}-1\sqrt{1}+5\sqrt{5}-3\sqrt{3}+...+81\sqrt{81}-79\sqrt{79}}{2}\\ =\dfrac{81\sqrt{81}-1\sqrt{1}}{2}=\dfrac{9^3-1}{2}=364\)
