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)