\(S-\frac{1}{3}S=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}-\frac{2008}{3^{2009}}<\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\)(3)

\(\Rightarrow\frac{2}{3}S<\frac{1}{2}\Rightarrow S<\frac{3}{4}\)(ĐPCM)

Ta có: Xét thừa số tổng quát. Với mọi n là số thực dương thì: \(\dfrac{1}{n^3}< \dfrac{1}{n^3-n}=\dfrac{1}{n\left(n^2-1\right)}=\dfrac{1}{\left(n-1\right)n\left(n+1\right)}\) Áp dụng vào bài toán: \(NL=\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{2008^3}< \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2007.2008.2009}=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{2007.2008}-\dfrac{1}{2008.2009}\right)=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2008.2009}\right)=\dfrac{1}{4}-\dfrac{1}{2008.2009.2}< \dfrac{1}{4}\left(đpcm\right)\)

giải dùm mk vs đi

Ta có:

\(3^1+3^2+3^3+3^4+...+3^{2009}+3^{2010}\)

\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{2008}+3^{2009}+3^{2010}\right)\)

\(=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{2008}\left(1+3+3^2\right)\)

\(=\left(1+3+3^2\right)\left(3+3^4+...+3^{2008}\right)\)

\(=13\left(3+3^4+...+3^{2008}\right)⋮13\)

=> ĐPCM