Question

B=1/3+1/3^2+1/3^3+...+1/3^2004+1/3^2005 CMR B<1/2

Answer 1

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

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

Khi đó 3B - B = \(\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2005}}\right)\)

=> 2B = \(1-\frac{1}{3^{2005}}\)

=> B = \(\frac{1}{2}-\frac{1}{3^{2005}.2}< \frac{1}{2}\left(\text{ĐPCM}\right)\)

Answer 2

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

\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+........+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\)

\(\Rightarrow3B-B=1-\frac{1}{3^{2005}}\)

\(\Rightarrow2B=1-\frac{1}{3^{2005}}\)\(\Rightarrow B=\frac{1-\frac{1}{3^{2005}}}{2}\)

Vì \(1-\frac{1}{3^{2005}}< 1\)\(\Rightarrow\frac{1-\frac{1}{3^{2005}}}{2}< \frac{1}{2}\)

hay \(B< \frac{1}{2}\)( đpcm )