\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.....+\frac{1}{3^{2011}}+\frac{1}{3^{2012}}\)

\(\Rightarrow3Á=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{2010}}+\frac{1}{3^{2011}}\)

\(\Rightarrow3A-A=2A=1-\frac{1}{3^{2012}}\)

\(\Rightarrow A=\frac{1-\frac{1}{3^{2012}}}{2}< \frac{1}{2}\)

Vậy \(A< \frac{1}{2}\)

Xét biểu thức A 

A= 1+(1+2) +....... +(1+2+3+...+2012)

A = 1+1+2+1+2+3+...+1+2+3+...+2012

 A có 2012  số 1

      có 2011  số 2

         ...

        có 1 số 2012

A = 1 x2012 +2x2011+...+2012x1

 mà B = 1 x2012 +2x2011+...+2012x1

nên A=B

\(A=1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2012\right)\)

\(=\left(1+1+1+...+1\right)+\left(2+2+...+2\right)+...+2012\)

\(=1\times2012+2\times2011+...+2012\times1\)

\(=B\)

\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2011}}+\dfrac{1}{3^{2012}}\)

\(3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2010}}+\dfrac{1}{3^{2011}}\)

\(3A-A=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2010}}+\dfrac{1}{3^{2011}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2011}}+\dfrac{1}{3^{2012}}\right)\)

\(2A=1-\dfrac{1}{3^{2012}}\Leftrightarrow A=\dfrac{1}{2}-\dfrac{1}{3^{2012}.2}< \dfrac{1}{2}\)

B<3\4 là đúng

khó thế

\(\frac{1}{2.2}< \frac{1}{1.2}\)

\(\frac{1}{3.3}< \frac{1}{2.3}\)

......

\(\frac{1}{100.100}< \frac{1}{99.100}\)

\(\Rightarrow\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{100.100}< \frac{1}{1.2}+..+\frac{1}{99.100}\)

\(\Rightarrow\frac{1}{2.2}+..+\frac{1}{100.100}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow\frac{1}{2.2}+..+\frac{1}{100.100}< 1-\frac{1}{100}< 1\).Suy ra điều phải chứng minh. câu b tương tự. bấm đúng cho mình nha

a hon b nhe thanh ha