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thích làm mỗi bài 3 vi các bai khac vua de, vua dai viet mệt
3) 3n : 3n-1 = 3n-n+1 = 3
Số nguyên n thỏa mãn đẳng thức -81/(-3)^n =-243 <=> (-3)^n x (-243) = -81 <=> (-3)^n x (-3)^5 = (-3)^4
<=> (-3)^n = (-3)^4 : (-3)^5 <=> (-3)^n = (-3)^4-5 <=> (-3)^n = (-3)^(-1) => n=-1.
a/ \(\frac{1}{n\left(n-1\right)\left(n+1\right)}=\frac{1}{n^3-n}>\frac{1}{n^3}\)
b/ \(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n^3+3n^2+2n}< \frac{1}{n^3}\)
c/ Ap dụng câu b ta được
\(\frac{1}{2^3}+\frac{1}{3^3}+...+\frac{1}{2006^3}>\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2006.2007.2008}\)
\(=\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2006.2007}-\frac{1}{2007.2008}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{2007.2008}\right)>\frac{1}{12}>\frac{1}{15}\)
\(\text{a)}A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(=\frac{1}{1}-\frac{1}{n}=1-\frac{1}{n}<1\left(\text{vì n}\ge2\text{ hay n dương}\right)\)
Vậy A<1
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Đặt \(A=1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+.........+\frac{1}{1+2+....+n}\)
Ta có: \(1+2=\frac{2.3}{2}\); \(1+2+3=\frac{3.4}{2}\); \(1+2+3+4=\frac{4.5}{2}\); .......... ; \(1+2+.......+n=\frac{n\left(n+1\right)}{2}\)
\(\Rightarrow A=1+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+.......+\frac{1}{\frac{n\left(n+1\right)}{2}}\)
\(=1+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+.......+\frac{2}{n\left(n+1\right)}\)
\(=1+2.\left[\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+........+\frac{1}{n\left(n+1\right)}\right]\)
\(=1+2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+........+\frac{1}{n}-\frac{1}{n+1}\right)\)
\(=1+2.\left(\frac{1}{2}-\frac{1}{n+1}\right)=1+1-\frac{2}{n+1}=2-\frac{2}{n+1}\)
Để A có GTNN thì \(\frac{2}{n+1}\)phải có GTLN \(\Rightarrow n+1\)phải có GTNN
mà \(n>1\)\(\Rightarrow n+1>2\)\(\Rightarrow min\left(n+1\right)=3\)\(\Leftrightarrow n=2\)
\(\Rightarrow A=2-\frac{2}{1+2}=2-\frac{2}{3}=\frac{4}{3}\)
Vậy \(minA=\frac{4}{3}\Leftrightarrow n=2\)