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Ta có: \(a_n=1+\frac{2^n\left[1.3.5...\left(2n-1\right)\right]}{\left(n+5\right)\left(n+6\right)...\left(2n\right)}\)
\(=1+\frac{2^n\left(2n\right)!}{\left[2.4.6..\left(2n\right)\right]\left[\left(n+5\right)\left(n+6\right)..\left(2n\right)\right]}\)
\(=1+\frac{\left(2n\right)!}{n!\left(n+5\right)\left(n+6\right)...\left(2n\right)}\)
\(=1+\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)\)
mặt khác \(1+\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)=\left(n^2+5n+5\right)^2\)
do đó an luôn là SCP
Bài cuối có Max nữa nhé, cần thì ib mình làm cho.
Giả sử \(c=min\left\{a;b;c\right\}\Rightarrow c\le1< 2\Rightarrow2-c>0\)
Ta có:\(P=ab+bc+ca-\frac{1}{2}abc=\frac{ab}{2}\left(2-c\right)+bc+ca\ge0\)
Đẳng thức xảy ra tại \(a=3;b=0;c=0\) và các hoán vị
Nhận xét: \(\left(n+1\right)\sqrt{n}=\sqrt{\left(n+1\right)^2n}=\sqrt{\left(n+1\right)n\left(n+1\right)};n\sqrt{n+1}=\sqrt{n^2\left(n+1\right)}=\sqrt{n.n\left(n+1\right)}\)
=> \(\left(n+1\right)\sqrt{n}>n\sqrt{n+1}\) => \(2.\left(n+1\right)\sqrt{n}>\left(n+1\right)\sqrt{n}+n\sqrt{n+1}\)
=> \(\frac{2}{2.\left(n+1\right)\sqrt{n}}<\frac{2}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{2}{\sqrt{n\left(n+1\right)}.\left(\sqrt{n+1}+\sqrt{n}\right)}\)
=> \(\frac{1}{\left(n+1\right)\sqrt{n}}<\frac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{n\left(n+1\right)}.\left(\left(\sqrt{n+1}\right)^2-\left(\sqrt{n}\right)^2\right)}=2.\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Áp dụng ta có:
\(\frac{1}{2\sqrt{1}}<2.\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}\right)\)
....
\(\frac{1}{3\sqrt{2}}<2.\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)\)
\(\frac{1}{\left(n+1\right)\sqrt{n}}<2.\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
=> A < \(2\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)=2\left(1-\frac{1}{\sqrt{n+1}}\right)<2\)
Vậy A < 2
Ta có:
\(\frac{1}{\left(n-1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)\(=\left(1+\frac{\sqrt{n}}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)<2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Dễ dàng giải tiếp bài toán