Question

Cho B=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+......+\frac{1}{3^{2005}}\)

CMR   B < \(\frac{1}{2}\)

Answer 1

ta có      \(B=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}.\)

      \(\Rightarrow3B=1+\frac{1}{3}+...+\frac{1}{3^{2004}}\)

    \(\Leftrightarrow3B-B=1+\frac{1}{3}-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^2}+...+\frac{1}{3^{2004}}-\frac{1}{3^{2004}}-\frac{1}{3^{2005}}\)

 \(\Leftrightarrow2B=1-\frac{1}{3^{2005}}\)   \(\Rightarrow B=\frac{1-\frac{1}{3^{2005}}}{2}< \frac{1}{2}\left(đpcm\right)\)

Answer 2

Có : 

3B = 1  +1/3 + 1/3^2 + ...... + 1/3^2004

2B = 3B - B = ( 1 + 1/3 + 1/3^2 + ....... + 1/3^2004 ) - ( 1/3 + 1/3^2 + ...... + 1/3^2004 )

     = 1 - 1/3^2004 < 1

=> B < 1/2

Tk mk nha

Answer 3

Bạn tham khảo nhé :) 

Ta có : 

\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2005}}\)

\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\)

\(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2005}}\right)\) ( loại bỏ các phân số đối nhau ) 

\(2B=1-\frac{1}{3^{2005}}< 1\)

\(B< \frac{1}{2}\)

Vậy \(B< \frac{1}{2}\)

Chúc bạn học tốt ~