B=1/3+1/3^2+1/3^3+....+1/3^2004+1/3^2005
Những câu hỏi liên quan
B=1/3+1/3^2+1/3^3+...+1/3^2004+1/3^2005 CMR B<1/2
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 = \(\frac{1}{2}-\frac{1}{3^{2005}.2}< \frac{1}{2}\left(\text{ĐPCM}\right)\)
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+........+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\)
\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+........+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\)
\(\Rightarrow3B-B=1-\frac{1}{3^{2005}}\)
\(\Rightarrow2B=1-\frac{1}{3^{2005}}\)\(\Rightarrow B=\frac{1-\frac{1}{3^{2005}}}{2}\)
Vì \(1-\frac{1}{3^{2005}}< 1\)\(\Rightarrow\frac{1-\frac{1}{3^{2005}}}{2}< \frac{1}{2}\)
hay \(B< \frac{1}{2}\)( đpcm )
(1/2003+1/2004-1/2005)/(5/2003+5/2004-5/2005)-(2/2002+2/2003-2/2004)/(3/2002+3/2003-3/2004)
B=1/3+1/3^2+1/3^3+...+1/3^2004+1/3^2005 chứng minh rằng B<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<\(\frac{1}{2}\)
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Cho B=1/3+1/3^2+1/3^3+...+1/3^2004+1/3^2005.Chứng minh B<1/2
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\)
\(\Leftrightarrow2B=3\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\right)\)
\(\Leftrightarrow2B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\)
\(\Leftrightarrow2B-B=\left(1+\frac{1}{3}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\right)\)
\(\Leftrightarrow B=1-\frac{1}{3^{2005}}\)
\(\Leftrightarrow B=1-\frac{1}{3^{2005}}< \frac{1}{2}\)
Vậy \(B< \frac{1}{2}\) (Đpcm)
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\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+..+\dfrac{1}{3^{2004}}+\dfrac{1}{3^{2005}}\\ \)
\(C=3B=1+\dfrac{1}{3}+..+\dfrac{1}{3^{2004}}\)
\(C-B=1-\dfrac{1}{3^{3005}}\)
\(B=\dfrac{1}{2}-\dfrac{1}{2.3^{2005}}< \dfrac{1}{2}\)
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\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2005}}\)
\(3B=3\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2005}}\right)\)
\(3B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2004}}\)
\(3B-B=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2004}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2005}}\right)\)
\(2B=1-\dfrac{1}{3^{2005}}\)
\(B=\dfrac{1-\dfrac{1}{3^{2005}}}{2}\\ \)
\(\text{Mà }1-\dfrac{1}{3^{2005}}< 1\\ \Rightarrow\dfrac{1-\dfrac{1}{3^{2005}}}{2}< \dfrac{1}{2}\\ \Rightarrow B< \dfrac{1}{2}\left(ĐPCM\right)\)
Vậy \(B< \dfrac{1}{2}\)
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Chứng tỏ rằng: B=1/3+1/3^2+1/3^3+1/3^4+...+1/3^2004+1/3^2005<1/2
Ta có :
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\)
\(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\right)\)
\(2B=1-\frac{1}{3^{2005}}< 1\)
\(\Rightarrow\frac{2B}{2}=\frac{1-\frac{1}{3^{2005}}}{2}< \frac{1}{2}\)
\(\Rightarrow B< \frac{1}{2}\)
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B=1/3+1/3^2+1/3^3+...+1/3^2004+1/3^2005 cmr 4/9<B<1/2
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bài 1 : (4đ) 1) Tính : A = 1 phần 2003 + 1 phần 2004 - 1 phần 2005 : 5 phần 2003 + 5 phần 2004 - 5 phần 2005 - ( qua phân số khác rồi nhé ) 2/2002 + 2/2003 - 2/2004 : 3/2002 + 3/2003 - 3/2004 2) Cho B = 1/3+1/3 mũ 2 + 1/3 mũ 3 + 1/3 mũ 4 + ... +1/3 mũ 2015 + 1/3 mũ 2016 . Chứng minh ràng B<1/2
Thuc hien phep tinh:
B=1/3+1/3^2+1/3^3+...+1/3^2004+1/3^2005
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\)
\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\)
\(\Rightarrow3B-B=1-\frac{1}{3^{2005}}\)
\(\Rightarrow2B=1-\frac{1}{3^{2005}}\)
\(\Rightarrow B=\frac{1-\frac{1}{3^{2005}}}{2}\)
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Thuc hien phep tinh:
B=1/3+1/3^2+1/3^3+...+1/3^2004+1/3^2005
Giải:
\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2004}}+\dfrac{1}{3^{2005}}\)
\(\Leftrightarrow\dfrac{1}{3}B=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}...+\dfrac{1}{3^{2005}}+\dfrac{1}{3^{2006}}\)
\(\Leftrightarrow B-\dfrac{1}{3}B=\dfrac{1}{3}-\dfrac{1}{3^{2006}}\)
\(\Leftrightarrow\dfrac{2}{3}B=\dfrac{1}{3}-\dfrac{1}{3^{2006}}\)
\(\Leftrightarrow B=\dfrac{\dfrac{1}{3}-\dfrac{1}{3^{2006}}}{\dfrac{2}{3}}\)
\(\Leftrightarrow B=\dfrac{1-\dfrac{1}{3^{2005}}}{2}\)
\(\Leftrightarrow B=\dfrac{\dfrac{3^{2005}-1}{3^{2005}}}{2}\)
\(\Leftrightarrow B=\dfrac{3^{2005}-1}{2.3^{2005}}\)
Vậy ...
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