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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)
Ta có: \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{\left(n+1\right)n}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)\)
\(=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=\left(1+\frac{\sqrt{n}}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(< \left(1+\frac{\sqrt{n+1}}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Áp dụng vào bài toán ta được
\(\frac{1}{2}+\frac{1}{3\sqrt{2}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}< 2\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=2\left(1-\frac{1}{\sqrt{n+1}}\right)< 2\)
\(\RightarrowĐPCM\)
Ta có S m-n = (√2 + 1)m /(√2 + 1)n + (√2 - 1)m /(√2 - 1)n = (√2 + 1)m (√2 - 1)n + (√2 - 1)m (√2 + 1)n
Từ đó
S m+n + S m-n = (√2 + 1)m+n + (√2 - 1)m+n +(√2 + 1)m (√2 - 1)n + (√2 - 1)m (√2 + 1)n
= (√2 + 1)m [(√2 + 1)n + (√2 -1)n] + (√2 - 1)m [(√2 - 1)n + (√2 + 1)n]
= [(√2 + 1)n + (√2 - 1)n] [(√2 + 1)m + (√2 - 1)m]
= S m .S n
sorry mk ko bít!!! ^^
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