Ta có: \(S_{m-n}=\frac{\left(\sqrt{2}+1\right)^m}{\left(\sqrt{2}+1\right)^n}+\frac{\left(\sqrt{2}-1\right)^m}{\left(\sqrt{2}-1\right)^n}\)
\(=\left(\sqrt{2}+1\right)^m\cdot\left(\sqrt{2}-1\right)^n+\left(\sqrt{2}-1\right)^m\left(\sqrt{2}+1\right)^n\)
Do đó:
\(S_{m+n}+S_{m-n}=\left(\sqrt{2}+1\right)^{m+n}+\left(\sqrt{2}-1\right)^{m+n}+\left(\sqrt{2}+1\right)^m\cdot\left(\sqrt{2}-1\right)^n+\left(\sqrt{2}-1\right)^m\cdot\left(\sqrt{2}+1\right)^n\)
\(=\left(\sqrt{2}+1\right)^m\left[\left(\sqrt{2}+1\right)^n+\left(\sqrt{2}-1\right)^n\right]+\left(\sqrt{2}-1\right)^m\cdot\left[\left(\sqrt{2}-1\right)^n+\left(\sqrt{2}+1\right)^n\right]\)
\(=\left[\left(\sqrt{2}+1\right)^n+\left(\sqrt{2}-1\right)^n\right]\cdot\left[\left(\sqrt{2}+1\right)^m+\left(\sqrt{2}-1\right)^m\right]\)
\(=S_m\cdot S_n\)(đpcm)
