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2.+ \(\left(2n+1\right)^2=4n^2+4n+1>4n^2+4n\)
\(\Rightarrow2n+1>\sqrt{4n\left(n+1\right)}=2\sqrt{n\left(n+1\right)}\)
+ \(\frac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(2n+1\right)\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{2n+1}< \frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n\left(n+1\right)}}=\frac{1}{2}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Do đó : \(A< \frac{1}{2}\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{48}}-\frac{1}{\sqrt{49}}\right)\)
\(\Rightarrow A< \frac{1}{2}\)
1. + \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\left(n+1\right)-n}{\left(n+1\right)\sqrt{n}}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(n+1\right)\sqrt{n}}\)
\(< \frac{\left(\sqrt{n+1}-\sqrt{n}\right)\cdot2\sqrt{n+1}}{\sqrt{n}\left(n+1\right)}=2\cdot\frac{n+1-\sqrt{n\left(n+1\right)}}{\left(n+1\right)\sqrt{n}}=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Do đó : \(A< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\right)\)
\(\Rightarrow A< 2\)
Bài 2 tạm thời chưa nghĩ ra :))
Đặt B là tên biểu thức
Với mọi n thuộc N*, 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 (*), ta được:
\(B< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{2011}}-\frac{1}{\sqrt{2012}}+\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\right)\)
\(=2\left(1-\frac{1}{\sqrt{2013}}\right)=2-\frac{1}{\sqrt{2013}}< 2\)
Ta có: \(16a^4+4=16a^4+2.4a^2.2+4-16a^2\)
\(=\left(4a+2\right)^2-16a^2\)
\(=\left(4a+2\right)^2-16a^2\)
\(=\left(4a^2-4a+2\right).\left(4a^2+4a+2\right)\)
\(=\left[\left(2a-1\right)^2+1\right].\left[\left(2a+1\right)^2+1\right]\) ( a \(\in\) N* )
Do đó: \(16a^4+4=\left[\left(2a-1\right)^2+1\right].\left[\left(2a+1\right)^2+1\right]\) ( * )
Thay a lần lượt bằng 1, 2, 3, ..., 2014, ta có:
\(16.1^4+4=\left[\left(2.1-1\right)^2+1\right].\left[\left(2.1+1\right)^2+1\right]=\left(1^2+1\right).\left(3^2+1\right)\)
\(16.2^4+4=\left[\left(2.2-1\right)^2+1\right].\left[\left(2.2+1\right)^2+1\right]=\left(3^2+1\right).\left(5^2+1\right)\)
\(16.3^4+4=\left[\left(2.3-1\right)^2+1\right].\left[\left(2.3+1\right)^2+1\right]=\left(5^2+1\right).\left(7^2+1\right)\)
\(16.4^4+4=\left[\left(2.4-1\right)^2+1\right].\left[\left(2.4+1\right)^2+1\right]=\left(7^2+1\right).\left(9^2+1\right)\)
\(......\)
\(16.2005^4+4=\left[\left(2.2005-1\right)^2+1\right].\left[\left(2.2005+1\right)^2+1\right]=\left(4009^2+1\right).\left(4011^2+1\right)\)
\(16.2006^4+4=\left[\left(2.2006-1\right)^2+1\right].\left[\left(2.2006+1\right)^2+1\right]=\left(4011^2+1\right).\left(4013^2+1\right)\)
Đặt \(T=\dfrac{\left(1^4+\dfrac{1}{4}\right).\left(3^4+\dfrac{1}{4}\right)...\left(2005^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right).\left(4^4+\dfrac{1}{4}\right)...\left(2006^4+\dfrac{1}{4}\right)}\)
\(\Leftrightarrow T=\dfrac{16.\left(1^4+\dfrac{1}{4}\right).16\left(3^4+\dfrac{1}{4}\right)...16\left(2005^4+\dfrac{1}{4}\right)}{16.\left(2^4+\dfrac{1}{4}\right).16\left(4^4+\dfrac{1}{4}\right)...16\left(2006^4+\dfrac{1}{4}\right)}\)
\(\Leftrightarrow T=\dfrac{\left(16.1^4+4\right).\left(16.3^4+4\right)...\left(16.2005^4+4\right)}{\left(16.2^4+4\right).\left(16.4^4+4\right)...\left(16.2006^4+4\right)}\)
\(\Leftrightarrow T=\dfrac{\left(1^2+1\right).\left(3^2+1\right).\left(5^2+1\right)...\left(4009^2+1\right).\left(4011^2+1\right)}{\left(3^2+1\right).\left(5^2+1\right).\left(7^2+1\right)...\left(4011^2+1\right).\left(4013^2+1\right)}\)
\(\Leftrightarrow T=\dfrac{1^2+1}{4013^2+1}\)
\(\Leftrightarrow T=\dfrac{2}{4013^2+1}\)
