Question

Cho B=1/3+1/3^2+1/3^3+...+1/3^2004+1/3^2005.Chứng minh B<1/2

Answer 1

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

\(\Leftrightarrow2B=3\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\right)\)

\(\Leftrightarrow2B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\)

\(\Leftrightarrow2B-B=\left(1+\frac{1}{3}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\right)\)

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

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

Vậy \(B< \frac{1}{2}\) (Đpcm)

Answer 2

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

\(C=3B=1+\dfrac{1}{3}+..+\dfrac{1}{3^{2004}}\)

\(C-B=1-\dfrac{1}{3^{3005}}\)

\(B=\dfrac{1}{2}-\dfrac{1}{2.3^{2005}}< \dfrac{1}{2}\)

Answer 3

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

\(3B=3\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2005}}\right)\)

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

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

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

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

\(\text{Mà }1-\dfrac{1}{3^{2005}}< 1\\ \Rightarrow\dfrac{1-\dfrac{1}{3^{2005}}}{2}< \dfrac{1}{2}\\ \Rightarrow B< \dfrac{1}{2}\left(ĐPCM\right)\)

Vậy \(B< \dfrac{1}{2}\)