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Ta có:
\(\left(a^2+b^2\right)+\left(b^2+c^2\right)+\left(c^2+a^2\right)\ge2\left(ab+bc+ca\right)\)
\(\Leftrightarrow ab+bc+ca\le a^2+b^2+c^2\)
\(\Leftrightarrow3\left(ab+bc+ca\right)\le\left(a+b+c\right)^2=1\)
\(\Leftrightarrow ab+bc+ca\le\frac{1}{3}< \frac{1}{2}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\)
\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(=1-\frac{1}{n}\)
\(< 1\)
\(\sqrt{n+1}-\sqrt{n}=\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}\)
\(\sqrt{n+1}+\sqrt{n}>2\sqrt{n}\Leftrightarrow\frac{1}{\sqrt{n+1}+\sqrt{n}}< \frac{1}{2\sqrt{n}}\)
\(\sqrt{n+1}+\sqrt{n}< 2\sqrt{n+1}\Leftrightarrow\frac{1}{\sqrt{n+1}+\sqrt{n}}>\frac{1}{2\sqrt{n+1}}\)
Do đó ta có đpcm.
Với \(k\in N;k>0\) Ta có :
\(\frac{1}{k\left(k+1\right)\left(k+2\right)}=\frac{1}{2}.\frac{\left(k+2\right)-k}{k\left(k+1\right)\left(k+2\right)}=\frac{1}{2}\left(\frac{1}{k\left(k+1\right)}-\frac{1}{\left(k+1\right)\left(k+2\right)}\right)\)
Áp dụng ta có :
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+.....+\frac{1}{\left(n-1\right)n\left(n+1\right)}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{n\left(n+1\right)}\right)=\frac{1}{2}.\frac{n\left(n+1\right)-2}{2n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{4n\left(n+1\right)}\)(đpcm)
Ta có :
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{\left(n-1\right)n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{4n\left(n+1\right)}\)
\(\Leftrightarrow\)\(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{\left(n-1\right)n\left(n+1\right)}=\frac{2\left(n-1\right)\left(n+2\right)}{4n\left(n+1\right)}\)
\(\Leftrightarrow\)\(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}=\frac{n\left(n-1\right)+2\left(n-1\right)}{2n\left(n+1\right)}\)
\(\Leftrightarrow\)\(\frac{1}{2}-\frac{1}{n\left(n+1\right)}=\frac{n^2-n+2n-2}{2n^2+2n}\)
\(\Leftrightarrow\)\(\frac{n\left(n+1\right)}{2n\left(n+1\right)}-\frac{2}{2n\left(n+1\right)}=\frac{n^2+n-2}{2n^2+2n}\)
\(\Leftrightarrow\)\(\frac{n^2+n-2}{2n^2+2n}=\frac{n^2+n-2}{2n^2+2n}\) với \(n\ge2\)
Vậy ...
vì bài dài quá nên mình làm từng bài 1 nhé
1. Ta thấy : \(\frac{1}{n^3}< \frac{1}{n^3-n}=\frac{1}{\left(n-1\right)n\left(n+1\right)}=\frac{1}{2}.\frac{\left(n+1\right)-\left(n-1\right)}{\left(n-1\right)n\left(n+1\right)}=\frac{1}{2}.\left[\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right]\)
Do đó :
\(B< \frac{1}{2}.\left[\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right]< \frac{1}{2}.\frac{1}{6}=\frac{1}{12}\)
2.
Nhận xét : \(1+\frac{1}{n\left(n+2\right)}=\frac{\left(n+1\right)^2}{n\left(n+2\right)}\)
Do đó :
\(A=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{\left(n+1\right)^2}{n\left(n+2\right)}=\frac{2.3...\left(n+1\right)}{1.2...n}.\frac{2.3...\left(n+1\right)}{3.4...\left(n+2\right)}=\frac{n+1}{1}.\frac{2}{n+2}< 2\)
Ta có :
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}=1-\frac{1}{n}< 1\) ( đpcm )
Vậy ...
Bạn tham khảo nhé mình mới lớp 7
bạn vào đây nè https://olm.vn/hoi-dap/question/62675.html