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\(\dfrac{1}{2}\)+\(\dfrac{1}{2^2}\) +...+\(\dfrac{1}{2^{2005}}\)
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TC
5 tháng 5 2022
a)\(=\left(\dfrac{2}{2}+\dfrac{1}{2}\right)\times\left(\dfrac{3}{3}+\dfrac{1}{3}\right)\times...\times\left(\dfrac{2005}{2005}+\dfrac{1}{2005}\right)\)
\(=\dfrac{3}{2}\times\dfrac{4}{3}\times...\times\dfrac{2006}{2005}=\dfrac{2006}{2}=1003\)
b)\(=\left(\dfrac{2}{3}+\dfrac{1}{3}\right)\times\dfrac{1}{2}=\dfrac{3}{3}\times\dfrac{1}{2}=\dfrac{1}{2}\)
NK
1
17 tháng 4 2018
ta thấy : \(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}=1-\dfrac{1}{2}\)
tương tự: \(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\)
....
\(\dfrac{1}{2005^2}=\dfrac{1}{2005.2005}< \dfrac{1}{2004.2005}=\dfrac{1}{2004}-\dfrac{1}{2005}\)
cộng vế theo vé các BĐT trên, ta có:
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2005^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2004}-\dfrac{1}{2005}=1-\dfrac{1}{2005}=\dfrac{2004}{2005}\)=> đpcm
NK
2
LK
17 tháng 4 2018
\(A=\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{2005^2}\)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2004.2005}\)
\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2004}-\frac{1}{20055}\)
\(A< 1-\frac{1}{2005}=\frac{2004}{2005}\)
\(\Rightarrow A< \frac{2004}{2005}\left(đpcm\right)\)
DK
17 tháng 4 2018
Đặt M=1/2^2+1/3^2+1/4^2+...+1/2005^2
M<1/1.2+1/2.3+1/3.4+...+1/2004.2005
M<1-1/2+1/2-1/3+1/3-1/4+...+1/2004-1/2005
M<1-1/2005=2004/2005(đpcm)
24 tháng 4 2021
Ta có: D\(=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2005}\right)\)
\(\Leftrightarrow D=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2004}{2005}=\dfrac{1.2.3...2004}{2.3.4...2005}=\dfrac{1}{2005}\)
Ta có: \(E=\dfrac{1^2}{1.3}.\dfrac{2^2}{2.4}.\dfrac{3^2}{3.5}...\dfrac{999^2}{999.1000}.\dfrac{1000^2}{1000.1001}=\dfrac{\left(1.2.3.4...1000\right)\left(1.2.3.4...1000\right)}{\left(1.2.3....1000\right)\left(3.4.5....1001\right)}=\dfrac{2}{1001}\)
TT
1 tháng 12 2017
Chữa lại đề.Bạn xem lại đề xem đúng chưa nhé!
\(D=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}+\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}+\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}+\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}+\dfrac{3}{2004}}\)
\(D=\dfrac{1.\left(\dfrac{1}{2003}+\dfrac{1}{2004}+\dfrac{1}{2005}\right)}{5.\left(\dfrac{1}{2003}+\dfrac{1}{2004}+\dfrac{1}{2005}\right)}-\dfrac{2.\left(\dfrac{1}{2002}+\dfrac{1}{2003}+\dfrac{1}{2004}\right)}{3\left(\dfrac{1}{2002}+\dfrac{1}{2003}+\dfrac{1}{2004}\right)}\)
\(D=\dfrac{1}{5}-\dfrac{2}{3}\)
\(D=-\dfrac{7}{15}\)
Cái này học lâu rồi.Bạn xem lại xem mình làm đúng chưa nhé!
Đặt: \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\)
\(2A=2\cdot\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\right)\)
\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2004}}\)
\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2004}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\right)\)
\(A=1-\dfrac{1}{2^{2005}}\)