Tính \(\frac{1^2}{1.3}+\frac{2^2}{3.5}+...+\frac{2005^2}{2009.2011}+\frac{2006^2}{2011.2013}\)
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8 tháng 11 2017
\(A=\frac{1^2}{1.3}+\frac{2^2}{3.5}+...+\frac{1006^2}{2011.2013}\)
\(\Leftrightarrow4A=\frac{2^2.1^2}{2^2-1}+\frac{2^2.2^2}{4^2-1}+...+\frac{2^2.1006^2}{2012^2-1}\)
\(=1006+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2011.2013}\right)\)
\(=1006+\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2011}-\frac{1}{2013}\right)\)
\(=1006+\frac{1}{2}\left(1-\frac{1}{2013}\right)=\frac{2026084}{2013}\)
\(\Rightarrow A=\frac{506521}{2013}\)
PV
3
28 tháng 4 2019
\(A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2011.2013}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{2013}\right)\)
\(A=\frac{1}{2}.\frac{2012}{2013}\)
\(A=\frac{1006}{2013}\)
28 tháng 4 2019
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2011.2013}\)
\(A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2011}-\frac{1}{2013}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{2013}\right)\)
\(A=\frac{1}{2}.\frac{2012}{2013}\)
\(A=\frac{1006}{2013}\)
TM
23 tháng 10 2016
Theo quy luật mà mình nhận thấy thì 20112 phải sửa thành 20122 bạn ạ!
Đặt \(A=\frac{1.3+2}{2^2}+\frac{2.4+2}{3^2}+\frac{3.5+2}{4^2}+...+\frac{2011.2013+2}{2012^2}\)
\(\Leftrightarrow A=\frac{2^2+1}{2^2}+\frac{3^2+1}{3^2}+\frac{4^2+1}{4^2}+...+\frac{2012^2+1}{2012^2}\)
\(\Leftrightarrow A=1+\frac{1}{2^2}+1+\frac{1}{3^2}+1+\frac{1}{4^2}+...+1+\frac{1}{2012^2}\)
\(\Leftrightarrow A=\left(1+1+1+...+1\right)+\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2012^2}\right)\)
\(\Leftrightarrow A=2011+\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2012^2}\right)\)
Đặt \(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2012^2}\)
Có: \(B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2011.2012}\)
\(\Leftrightarrow B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}\)
\(\Leftrightarrow B< 1-\frac{1}{2012}\)
\(\Rightarrow A=2011+B< 2011+1-\frac{1}{2012}\)
\(\Rightarrow A< 2012-\frac{1}{2012}< 2013\)
Ta có đpcm
TC
2
10 tháng 4 2018
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2009.2011}\)
=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\)
=\(1-\frac{1}{2011}\)
=\(\frac{2010}{2011}\)
10 tháng 4 2018
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}\frac{1}{5\cdot7}+...+\frac{1}{2009\cdot2011}\)
\(=\frac{1\cdot2}{2\cdot1\cdot3}+\frac{1\cdot2}{2\cdot3\cdot5}+\frac{1\cdot2}{2\cdot5\cdot7}+...+\frac{1\cdot2}{2\cdot2009\cdot2011}\)
\(=\frac{1}{2}\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{2009\cdot2011}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{2009\cdot2011}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{1}{1}-\frac{1}{2011}\right)\)= .......
Mình không chắc là đúng đâu nha
GC
3
KN
28 tháng 4 2019
\(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2011.2013}\)
\(\Rightarrow2S=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2011.2013}\)
\(\Rightarrow2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2011}-\frac{1}{2013}\)
\(\Rightarrow2S=1-\frac{1}{2013}\)
\(\Rightarrow2S=\frac{2012}{2013}\)
\(\Rightarrow S=\frac{2012}{2013}\div2\)
\(\Rightarrow S=\frac{1006}{2013}\)
BD
28 tháng 4 2019
\(2S=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{2011\cdot2013}\)
\(2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2011}-\frac{1}{2013}\)
\(2S=1-\frac{1}{2013}\)
\(2S=\frac{2012}{2013}\)
\(S=\frac{2012}{2013}\div2=\frac{1006}{2013}\)
#Louis
CB
2
20 tháng 7 2019
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{2009.2011}=(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2009.2011}):2\)
\(=\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2009}-\frac{1}{2011}\right):2=\left(1-\frac{1}{2011}\right):2=\frac{1}{2}-\frac{1}{4022}=...\)
20 tháng 7 2019
\(\frac{1}{2}\cdot\left(\frac{2}{1\cdot3}+\cdot\cdot\cdot+\frac{2}{2009\cdot2011}\right)\)
\(=\frac{1}{2}\cdot\left(1-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{2009}-\frac{1}{2011}\right)\)
\(=\frac{1}{2}\cdot\left(1-\frac{1}{2011}\right)\)
\(=\frac{1}{2}\cdot\frac{2010}{2011}\)
\(=\frac{1005}{2011}\)