CMR: \(\frac{1}{2\sqrt{k}}< \frac{1}{\sqrt{k-1}+\sqrt{k}}\)
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Ta thấy: k thuộc N* nên \(\sqrt{k+1}>\sqrt{k}\)
\(\Rightarrow\frac{1}{\left(k+1\right)\sqrt{k}}=\frac{2}{\left(2\sqrt{k+1}\right).\left(\sqrt{k+1}.\sqrt{k}\right)}< \frac{2}{\left(\sqrt{k+1}.\sqrt{k}\right).\left(\sqrt{k+1}+\sqrt{k}\right)}\)
\(=\frac{2\left(\sqrt{k+1}-\sqrt{k}\right)}{\left(\sqrt{k+1}.\sqrt{k}\right)\left(k+1-k\right)}=2\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)\)
\(\Rightarrow\frac{1}{\left(k+1\right)\sqrt{k}}< 2\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)\)(đpcm).
Với mọi \(k\ge2\) thì \(\frac{2k+\sqrt{k^2-1}}{\sqrt{k-1}+\sqrt{k+1}}=\frac{\left[\left(\sqrt{k-1}\right)^2+\left(\sqrt{k+1}\right)^2+\sqrt{\left(k-1\right)\left(k+1\right)}\right]\left(\sqrt{k+1}-\sqrt{k-1}\right)}{\left(\sqrt{k-1}+\sqrt{k+1}\right)\left(\sqrt{k+1}-\sqrt{k-1}\right)}\)
\(=\frac{\sqrt{\left(k+1\right)^3}-\sqrt{\left(k-1\right)^3}}{2}\)
Suy ra tổng đã cho có thể viết là :
\(A=\frac{1}{2}\left[\sqrt{3^3}-\sqrt{1^3}+\sqrt{4^3}-\sqrt{2^3}+\sqrt{5^3}-\sqrt{3^3}+\sqrt{6^3}-\sqrt{4^3}+...+\sqrt{101^3}-\sqrt{99^3}\right]\)
\(=\frac{1}{2}\left[-1-\sqrt{2^3}+\sqrt{101^3}+\sqrt{100^3}\right]\)
\(=\frac{999+\sqrt{101^3}-\sqrt{8}}{2}\)
Ta có:
\(\frac{1}{\sqrt{k}}=\frac{2}{2\sqrt{k}}=\frac{2}{\sqrt{k}+\sqrt{k}}< \frac{2}{\sqrt{k}+\sqrt{k-1}}=\frac{2\left(\sqrt{k}-\sqrt{k-1}\right)}{\left(\sqrt{k}-\sqrt{k-1}\right)\sqrt{k}+\sqrt{k-1}}\)
\(=\frac{2\left(\sqrt{k}-\sqrt{k-1}\right)}{k-\left(k-1\right)}=2\left(\sqrt{k}-\sqrt{k-1}\right)\)
\(\frac{1}{\sqrt{k}\left(k+1\right)}=\frac{1}{\sqrt{k+1}}.\frac{1}{\sqrt{k}\sqrt{k+1}}=\frac{1}{\sqrt{k+1}}.\frac{k+1-k}{\sqrt{k\left(k+1\right)}}=\frac{1}{\sqrt{k+1}}\left(\frac{\left(\sqrt{k+1}-\sqrt{k}\right)\left(\sqrt{k+1}-\sqrt{k}\right)}{\sqrt{k}\sqrt{k+1}}\right)\)
\(=\frac{\left(\sqrt{k+1}-\sqrt{k}\right)}{\sqrt{k}\sqrt{k+1}}.\frac{\left(\sqrt{k+1}+\sqrt{k}\right)}{\sqrt{k+1}}<\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k}\sqrt{k+1}}.2\)
Đề đúng sory nhé
\(2\frac{1998}{1999}\)là hỗn số hay \(2.\frac{1998}{1999}\)hả bạn?
\(\sqrt{1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}}=\sqrt{\frac{k^2\left(k+1\right)^2+\left(k+1\right)^2+k^2}{k^2\left(k+1\right)^2}}=\sqrt{\frac{k^2\left(k+1\right)^2+2k\left(k+1\right)+1}{k^2\left(k+1\right)^2}}\)
\(=\sqrt{\frac{\left[k\left(k+1\right)+1\right]^2}{k^2\left(k+1\right)^2}}=\frac{k\left(k+1\right)+1}{k\left(k+1\right)}=1+\frac{1}{k\left(k+1\right)}\)
\(\Rightarrow A=1+\frac{1}{2.3}+1+\frac{1}{3.4}+...+1+\frac{1}{k\left(k+1\right)}\)
\(=k-1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{k}-\frac{1}{k+1}\)
\(=k-1+\frac{1}{2}-\frac{1}{k+1}=...\)
với \(a>0,b>0\)ta có \(\sqrt{a}.\sqrt{b}\le\frac{a+b}{2}\Rightarrow\frac{1}{\sqrt{a}.\sqrt{b}}\ge\frac{2}{a+b}\)
từ đó ta có : \(\frac{1}{\sqrt{k\left(2016-k\right)}}\ge\frac{2}{k+2016-k}\ge\frac{2}{2016}=\frac{1}{1008},\)với mọi \(k\in N^{\cdot}\)
Suy ra \(S_k\)\(\ge k.\frac{1}{1008}>k.\frac{1}{1018}\)(đpcm).
\(\frac{1}{\sqrt{k-1}+\sqrt{k}}>\frac{1}{\sqrt{k}+\sqrt{k}}=\frac{1}{2\sqrt{k}}\)