\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{1998.1999.2000}\)
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Những câu hỏi liên quan
QD
26 tháng 1 2017
a)A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{2009.2010}\)
A=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{2009}-\frac{1}{2010}\)
A=1-\(\frac{1}{2010}\)=\(\frac{2009}{2010}\)
c)C=\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+......+\frac{1}{2006.2008}\)
C=\(\frac{1}{2}\).(\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+..+\frac{1}{2006}-\frac{1}{2008}\))
C=\(\frac{1}{2}\).(\(\frac{1}{2}-\frac{1}{2008}\))
C=\(\frac{1}{2}\).\(\frac{1003}{2008}\)=\(\frac{1003}{4016}\)
Câu b mình chưa nghĩ ra
Chúc bạn học tốt!
26 tháng 1 2017
a) A = \(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) + \(\frac{1}{3.4}\) + ...+ \(\frac{1}{2009.2000}\)
= 1 - \(\frac{1}{2}\) + \(\frac{1}{2}\) - \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{4}\) + ... + \(\frac{1}{2009}\) - \(\frac{1}{2000}\)
= 1 - \(\frac{1}{2000}\) = \(\frac{1999}{2000}\)
b) B = \(\frac{1}{1.2.3}\) + \(\frac{1}{2.3.4}\) + \(\frac{1}{3.4.5}\) + ... + \(\frac{1}{1998.1999.2000}\)
= \(\frac{1}{2}\) ( \(\frac{2}{1.2.3}\) + \(\frac{2}{2.3.4}\) + \(\frac{2}{3.4.5}\) + ... + \(\frac{2}{1998.1999.2000}\))
= \(\frac{1}{2}\) (\(\frac{1}{1.2}\) - \(\frac{1}{2.3}\) + \(\frac{1}{2.3}\) - \(\frac{1}{3.4}\) + \(\frac{1}{3.4}\) - \(\frac{1}{4.5}\) + ... + \(\frac{1}{1998.1999}\) - \(\frac{1}{1999.2000}\))
= \(\frac{1}{2}\) (\(\frac{1}{1.2}\) - \(\frac{1}{1999.2000}\))
= \(\frac{1}{2}\) (\(\frac{1}{2}\) - \(\frac{1}{3998000}\))
= \(\frac{1}{4}\) - \(\frac{1}{7996000}\) = ?
c) C = \(\frac{1}{2.4}\) + \(\frac{1}{4.6}\) + \(\frac{1}{6.8}\) + ... + \(\frac{1}{2006.2008}\)
= \(\frac{1}{2}\) (\(\frac{1}{2}\) - \(\frac{1}{4}\)) + \(\frac{1}{2}\)(\(\frac{1}{4}\) - \(\frac{1}{6}\)) + ... + \(\frac{1}{2}\)(\(\frac{1}{2006}\) - \(\frac{1}{2008}\))
= \(\frac{1}{2}\)(\(\frac{1}{2}\) - \(\frac{1}{4}\) + \(\frac{1}{4}\) - \(\frac{1}{6}\) + ... + \(\frac{1}{2006}\) - \(\frac{1}{2008}\))
= \(\frac{1}{2}\)(\(\frac{1}{2}\) - \(\frac{1}{2008}\))
= \(\frac{1}{2}\) . \(\frac{1003}{2008}\) = \(\frac{1003}{4016}\).
KQ
1
7 tháng 5 2019
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\\ =\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2450}\right)\)
\(=\frac{1}{2}.\frac{612}{1225}\\ =\frac{306}{1225}\)(mà đây là toán 6 mà :V)
NN
2
24 tháng 6 2016
Ta có nhận xét:
\(\frac{2}{n.\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
Áp dụng công thức trên vào bài tập, ta có:
B=\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\)
\(\Rightarrow B=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{37.38.39}\right)\)
\(\Rightarrow B=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)
\(\Rightarrow B=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{38.39}\right)\)
\(\Rightarrow B=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{1482}\right)\)
\(\Rightarrow B=\frac{1}{2}.\frac{370}{741}=\frac{185}{741}\)
Vậy \(B=\frac{185}{741}\)
NC
5
10 tháng 6 2016
Ta có nhận xét:
\(\frac{2}{n.\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
Áp dụng công thức trên vào bài tập, ta có:
\(\Rightarrow B=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{37.38.39}\right)\)
\(\Rightarrow B=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)
\(\Rightarrow B=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{38.39}\right)\)
\(\Rightarrow B=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{1482}\right)\)
\(\Rightarrow B=\frac{1}{2}.\frac{370}{741}=\frac{185}{741}\)
10 tháng 6 2016
\(\Rightarrow B=\frac{1}{2}\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{39-37}{37.38.39}\right)\)
\(\Rightarrow B=\frac{1}{2}\left(\frac{3}{1.2.3}-\frac{1}{1.2.3}+\frac{4}{2.3.4}-\frac{2}{2.3.4}+\frac{5}{3.4.5}-\frac{3}{3.4.5}+...+\frac{39}{37.38.39}-\frac{37}{37.38.39}\right)\)
\(\Rightarrow B=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)
\(\Rightarrow B=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{38.29}\right)\)
\(\Rightarrow B=\frac{1}{2}.\frac{370}{741}=\frac{185}{741}\)
H
4
14 tháng 6 2016
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}...+\frac{2}{37.38.39}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{38.39}\right)=\frac{185}{741}\)
HN
2
K
1 tháng 7 2015
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\)
\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{37.38.39}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{38.39}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{1482}\right)\)
\(=\frac{1}{2}.\frac{370}{741}\)
\(=\frac{1}{2}.\frac{370}{741}\)
\(=\frac{185}{741}\)
26 tháng 7 2019
Đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{37.38.39}\)
\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{37.38.39}\)
\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{37.38}-\frac{1}{38.39}\)
\(2A=\frac{1}{1.2}-\frac{1}{38.39}=\frac{370}{741}\)
\(A=\frac{185}{741}\)
Chúc bn hc tốt <3
T
3
29 tháng 5 2015
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{37.38.39}=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{37.38.39}\right)=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{38.39}\right)=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{1482}\right)=\frac{1}{2}.\frac{370}{741}=\frac{185}{741}\)
29 tháng 5 2015
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{37\cdot38\cdot39}=\frac{1}{2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{37\cdot38}-\frac{1}{38\cdot39}\)
\(=\left(\frac{1}{2}-\frac{1}{38\cdot39}\right)+\left(\frac{1}{2\cdot3}-\frac{1}{2\cdot3}\right)+...+\left(\frac{1}{37\cdot38}-\frac{1}{37\cdot38}\right)=\left(\frac{741}{1482}-\frac{1}{1482}\right)+0+...+0=\frac{740}{1482}=\frac{370}{741}\)Chúc bạn học tốt!^_^
UT
2
14 tháng 3 2016
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...........+\frac{1}{37.38.39}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+..........+\frac{1}{37.38}-\frac{1}{38.39}\)
\(=\frac{1}{1.2}-\frac{1}{38.39}\)
\(=\frac{1}{2}-\frac{1}{1482}\)
\(=\frac{741}{1482}-\frac{1}{1482}\)
\(=\frac{370}{741}\)
Đặt tổng trên là A.Ta có:
A=\(\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{1998.1999.2000}\right)\)
A=\(\frac{1}{2}\left[\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\left(\frac{1}{1998.1999}-\frac{1}{1999.2000}\right)\right]\)
A=\(\frac{1}{2}\left(\frac{1}{2}-\frac{1}{1999.2000}\right)\)
A=\(\frac{1}{4}-\frac{1}{1999.2000.2}\)
\(\frac{1}{5678}\)
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