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Ta có:
\(\frac{1}{\left(k+1\right)\sqrt{k}}< 2\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)\)
\(\Leftrightarrow\frac{1}{\left(k+1\right)\sqrt{k}}-2\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)< 0\)
\(\Leftrightarrow\frac{1-2k-2+2\sqrt{k\left(k+1\right)}}{\sqrt{k}\left(k+1\right)}< 0\)
Lại có: \(k>0\)
\(\Rightarrow k+1>0\)
\(\Rightarrow\sqrt{k}\left(k+1\right)>0\)
\(\Rightarrow-1-2k+2\sqrt{k\left(k+1\right)}< 0\)
Áp dụng BĐT Cô-si ta có:
\(k+\left(k+1\right)\ge2\sqrt{k\left(k+1\right)}\)
\(\Leftrightarrow2k+1\ge2\sqrt{k\left(k+1\right)}\)
\(\Leftrightarrow2\sqrt{k\left(k+1\right)}-2k-1\le0\forall k>0\)
Vậy \(\frac{1}{\left(k+1\right)\sqrt{k}}< 2\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)\)
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).
Đặt \(\sqrt{2}+1=a\Rightarrow\sqrt{2}-1=\frac{1}{a}\)
\(\Rightarrow S_k=a^k+\frac{1}{a^k}\) ; \(S_{k+1}=a^{k+1}+\frac{1}{a^{k+1}}\) ;
\(S_1=a+\frac{1}{a}=\sqrt{2}+1+\sqrt{2}-1=2\sqrt{2}\)
\(\Rightarrow S_k.S_{k+1}=\left(a^k+\frac{1}{a^k}\right)\left(a^{k+1}+\frac{1}{a^{k+1}}\right)\)
\(=a^k.a^{k+1}+\frac{a^k}{a^{k+1}}+\frac{a^{k+1}}{a^k}+\frac{1}{a^k.a^{k+1}}\)
\(=a^{2k+1}+\frac{1}{a^{2k+1}}+a+\frac{1}{a}\)
\(=S_{2k+1}+S_1=S_{2k+1}+2\sqrt{2}\)
\(\Rightarrow S_k.S_{k+1}-S_{2k+1}=2\sqrt{2}\)
Thay \(k=2009\) vào ta được:
\(S_{2009}.S_{2010}-S_{4019}=2\sqrt{2}\) (đpcm)