Tính P = \(\dfrac{1^2}{3.5}+\dfrac{2^2}{3.5}+\dfrac{3^2}{5.7}+....+\dfrac{1004^2}{2007.2009}+\dfrac{1005^2}{2009.2011}\)
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8 tháng 11 2017
\(A=\frac{1^2}{1.3}+\frac{2^2}{3.5}+...+\frac{1006^2}{2011.2013}\)
\(\Leftrightarrow4A=\frac{2^2.1^2}{2^2-1}+\frac{2^2.2^2}{4^2-1}+...+\frac{2^2.1006^2}{2012^2-1}\)
\(=1006+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2011.2013}\right)\)
\(=1006+\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2011}-\frac{1}{2013}\right)\)
\(=1006+\frac{1}{2}\left(1-\frac{1}{2013}\right)=\frac{2026084}{2013}\)
\(\Rightarrow A=\frac{506521}{2013}\)
8 tháng 2 2022
I: Để 3n+4/n+2 là số nguyên thì \(3n+4⋮n+2\)
\(\Leftrightarrow3n+6-2⋮n+2\)
\(\Leftrightarrow n+2\in\left\{1;-1;2;-2\right\}\)
hay \(n\in\left\{-1;-3;0;-4\right\}\)
II: \(D=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2007}-\dfrac{1}{2009}\right)\)
\(D=2\cdot\left(1-\dfrac{1}{2009}\right)=2\cdot\dfrac{2008}{2009}=\dfrac{4016}{2009}\)
OY
22 tháng 7 2021
Đặt A=\(\dfrac{2}{3.5}.\dfrac{2}{7.9}.....\dfrac{2}{99.101}\)
A=\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
A=\(\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}\)
22 tháng 7 2021
Ta có: \(P=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{1}{15}\)
\(=\dfrac{4}{15}\)
LB
15 tháng 11 2017
\(A=\dfrac{1}{2}+\dfrac{3-2}{3.2}+\dfrac{4-3}{3.4}+...+\dfrac{100-99}{100.99}\)
\(A=\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=1-\dfrac{1}{100}\)
\(A=\dfrac{99}{100}\)
LB
15 tháng 11 2017
\(2B=\dfrac{2}{1.3}+\dfrac{2}{3.5}+....+\dfrac{2}{2007.2009}+\dfrac{2}{2009..2011}\)
\(2B=\dfrac{3-1}{1.3}+\dfrac{5-3}{3,5}+...+\dfrac{2009-2007}{2009.2007}+\dfrac{2011-2009}{2011.2009}\)
\(2B=\dfrac{3}{3}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2007}-\dfrac{1}{2009}+\dfrac{1}{2009}-\dfrac{1}{2011}\)
\(2B=1-\dfrac{1}{2011}\)
\(2B=\dfrac{2010}{2011}\)
\(B=\dfrac{2010}{4022}\)
QT
1
TA
17 tháng 1 2018
2S=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2007.2009}\)
=\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-...+\dfrac{1}{2007}-\dfrac{1}{2009}\)
= 1- \(\dfrac{1}{2009}\)
= \(\dfrac{2008}{2009}\)
=> S=\(\dfrac{1004}{2009}\)
LS
2
26 tháng 4 2022
\(B=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{99.100}\)
\(B=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99.100}+\dfrac{1}{99.100}\)
\(B=\dfrac{1}{3}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{100}\)
\(B=\dfrac{1}{3}-\dfrac{2}{100}+\dfrac{1}{99}\)
\(B=\dfrac{1}{3}-\dfrac{1}{50}+\dfrac{1}{99}\)
Đến đây thì hết tính hợp lý được rồi:v
\(B=\dfrac{34}{99}-\dfrac{1}{50}\)
\(B=\dfrac{1601}{4950}\)
H9
HT.Phong (9A5)
CTVHS
15 tháng 6 2023
\(1-\dfrac{2}{3.5}-\dfrac{2}{5.7}-...-\dfrac{2}{61.63}-\dfrac{2}{63.65}\)
\(=1-\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{63}-\dfrac{1}{65}\right)\)
\(=1-\left(\dfrac{1}{3}-\dfrac{1}{65}\right)\)
\(=1-\dfrac{62}{195}\)
\(=\dfrac{133}{195}\)
H
5 tháng 4 2021
Trước hết ta chứng minh (a-1)(a+1) + 1 = a^2 (*)
Thật vậy VT = (a-1)(a+1)+1=(a-1)a + a-1 +1 = a^2-a+a=a^2 =VP
Áp dụng (*) ta có:
\(A=\dfrac{1\cdot3+2}{2^2}+\dfrac{2\cdot4+2}{3^2}+...+\dfrac{2009\cdot2011+2}{2010^2}\\ =\dfrac{2^2+1}{2^2}+\dfrac{3^2+1}{3^2}+...+\dfrac{2010^2+1}{2010^2}=2009+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2010^2}\\ < 2009+\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2009\cdot2010}\\ =2009+\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{2009}-\dfrac{1}{2010}=2010-\dfrac{1}{2010}< 2020< 2011\)
\(P=\dfrac{1^2}{1.3}+\dfrac{2^2}{3.5}+...+\dfrac{1005^2}{2009.2011}\)
\(\Leftrightarrow4P=\dfrac{4.1^2}{1.3}+\dfrac{4.2^2}{3.5}+...+\dfrac{4.1005^2}{2009.2011}\)
\(=\dfrac{2^2}{2^2-1}+\dfrac{4^2}{4^2-1}+...+\dfrac{2010^2}{2010^2-1}\)
\(=2009+\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2009.2011}\right)\)
\(=2009+\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\)
\(=2009+\dfrac{1}{2}\left(1-\dfrac{1}{2011}\right)=2009+\dfrac{1005}{2011}\)
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