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

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

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

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

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

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

Vậy \(B< \frac{1}{2}\) (Đpcm)

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

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

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

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

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

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

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

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

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

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

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

Vậy \(B< \dfrac{1}{2}\)

Ta có:3B\(\frac{1}{3}+\frac{1}{3}^2+\frac{1}{3}^3+...+\frac{1}{3}^{2003}+\frac{1}{3}^{2004}\)

B=\(\frac{1}{3}+\frac{1}{3}^2+\frac{1}{3}^3+..+\frac{1}{3}^{2003}+\frac{1}{3}^{2004}+\frac{1}{3}^{2005}\)

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

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

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

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

a)ta có 3B=1+1/3+1/3^2+........+1/3^2003+1/3^2004

             B=    1/3+1/3^2+........+1/3^2003+1/3^2004+1/3^2005

suy ra 2B=1-1/3^2005

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

suy ra B=1/2-1/3^2005/2 bé hơn 1/2

từ đấy suy ra B bé hơn 1/2

làm ơn tách ra giùm mk

nguyên một hàng mk đọc ko hỉu????????????

không hiểu......>><

khó hiểu quá

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

\(\Rightarrow3B-B=\left(3+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2005}}\right)\)

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

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

3B=1+1/3+1/3²+...+1/3²⁰⁰⁴

3B-B=1-1/3²⁰⁰⁵

2B=1-1/3²⁰⁰⁵

B=1/2-1/(3²⁰⁰⁵.2)

Vậy B <1/2

Hùng ơi sai rồi

3B=1+1/3+1/3^2+...+1/3^2004 chứ

Thay số 3 thành 1 vì 1/3*3=1 ko phải bằng 3

Ta có :

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

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

\(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\right)\)

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

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

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

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 + 1/3^3 + ..... + 1/3^2005 )

                  = 1 - 1/3^2005 < 1

=> B < 1 : 2 = 1/2

=> ĐPCM

Tk mk nha

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

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

\(\Rightarrow3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2005}}\right)\)

\(\Rightarrow2B=1-\frac{1}{3^{2005}}< 1\)

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

Kết bạn LQM ko

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