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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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\(S_{n^3}\) có vẻ là ghi sai đề, \(S_n^3\) mới đúng
Đặt \(\left\{{}\begin{matrix}a=\left(2-\sqrt{3}\right)^n\\b=\left(2+\sqrt{3}\right)^n\end{matrix}\right.\) \(\Rightarrow ab=\left[a=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)\right]^n=1^n=1\)
\(S_n^3=\left(a+b\right)^3\)
\(S_{3n}+3S_n=a^3+b^3+3\left(a+b\right)=a^3+b^3+3.1.\left(a+b\right)\)
\(=a^3+b^3+3ab\left(a+b\right)=\left(a+b\right)^3=S_n^3\)
b/ Thay trực tiếp vào casio và bấm, hoặc nếu giải kiểu tổng quát thì:
\(S_1=2-\sqrt{3}+2+\sqrt{3}=4\) ; \(S_2=7-4\sqrt{3}+7+4\sqrt{3}=14\)
\(\Rightarrow S_3+3S_1=S_1^3\Rightarrow S_3=S_1^3-3S_1=4^3-3.4=52\)
Đặt \(\left\{{}\begin{matrix}2-\sqrt{3}=x\\2+\sqrt{3}=y\end{matrix}\right.\) \(\Rightarrow xy=1\)
\(S_1=x+y=4\) ; \(S_3=x^3+y^3\)
\(S_1S_3=\left(x+y\right)\left(x^3+y^3\right)=x^4+y^4+x^3y+y^3x\)
\(\Rightarrow S_1S_3=x^4+y^4+xy\left(x^2+y^2\right)=S_4+S_2\)
\(\Rightarrow S_4=S_1S_3-S_2=194\)
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)
a) \(u_n=\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\sqrt{\frac{n^2\left(n+1\right)^2+\left(n+1\right)^2+n^2}{n^2\left(n+1\right)^2}}\)
\(=\sqrt{\frac{\left[n\left(n+1\right)\right]^2+2n^2+2n+1}{\left[n\left(n+1\right)\right]^2}}=\sqrt{\frac{\left[n\left(n+1\right)\right]^2+2n\left(n+1\right)+1}{\left[n\left(n+1\right)\right]^2}}\)
\(=\sqrt{\frac{\left[n\left(n+1\right)+1\right]^2}{\left[n\left(n+1\right)\right]^2}}=\frac{n\left(n+1\right)+1}{n\left(n+1\right)}\in Q\)
b) \(u_n=\frac{n\left(n+1\right)+1}{n\left(n+1\right)}=1+\frac{1}{n\left(n+1\right)}=1+\frac{1}{n}-\frac{1}{n+1}\)
Vậy \(S_{2021}=u_1+u_2+...+u_{2021}=1+\frac{1}{1}-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+...+1+\frac{1}{2021}-\frac{1}{2022}\)
\(=2022-\frac{1}{2022}=\frac{2022^2-1}{2022}\)
Ta có:
\(S_2=\dfrac{\sqrt{3}+1}{1-\sqrt{3}}=\dfrac{\left(\sqrt{3}+1\right)^2}{-2}=-2-\sqrt{3}\)
\(S_3=\dfrac{\sqrt{3}-2-\sqrt{3}}{1+\sqrt{3}\left(2+\sqrt{3}\right)}=\dfrac{-2}{4+2\sqrt{3}}=\sqrt{3}-2\)
\(S_4=\dfrac{\sqrt{3}+\sqrt{3}-2}{1-\sqrt{3}\left(\sqrt{3}-2\right)}=1\)
Tới đây thì đã thấy được vòng lặp của dãy số rồi thì đơn giản rồi ha.
Với \(n\)lẻ: \(n=2k-1\)
\(S_n=1-2+3-...+\left(-1\right)^{n-1}n=1+\left(3-2\right)+...+\left[\left(-1\right)^{n-1}n-\left(-1\right)^{n-2}\left(n-1\right)\right]\)
\(=1+1+...+1=k\)
Với \(n\)chẵn: \(n=2k\)
\(S_n=1-2+3-...+\left(-1\right)^{n-1}n=\left(1-2\right)+\left(3-4\right)+...+\left[\left(-1\right)^{n-1}n-\left(-1\right)^{n-2}\left(n-1\right)\right]\)
\(=-1-1-...-1=-k\)
Áp dụng:
\(D=S_{35}+S_{60}+S_{100}=18-30-50=-62\)
Em cảm ơn thầy nhiều ạ !