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A=2.(1/1.3 + 1/3.5 + 1/5.7 +.......+1/99.101)
=2.(1/1 + 1/3 + 1/5 + 1/5 + 1/7 +...+1/99 + 1/101)
=2.(1-1/101)
=2.(101/101-1/101)
=2.100/101
200/101
B=2.(1/1.3+1/3.5+1/3.1+....+1/99.101)
=2.(1/1+1/3+1/3+1/5+1/3+1/7+....+1/99+1/101)
=2.(1/1+1/101)
=2.(101/101+1/101)
=2.102/101
=204/101
Ta đặt
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{49.51}=A\)
\(\Rightarrow2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+..+\dfrac{2}{49.51}\)
\(2A=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\)
\(\Rightarrow2A=\dfrac{1}{1}-\dfrac{1}{51}=\dfrac{50}{51}\)
\(A=\dfrac{50}{51}:2=\dfrac{25}{51}\)
Vậy : \(\dfrac{1}{3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{49.51}=\dfrac{25}{51}\)
a)
\(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{24.25}\)
\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{24}-\dfrac{1}{25}\)
\(=\dfrac{1}{5}-\dfrac{1}{25}\)
\(=\dfrac{4}{25}\)
b)
\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{99.101}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(=1-\dfrac{1}{101}\)
\(=\dfrac{100}{101}\)
a) \(\dfrac{1}{5.6}=\dfrac{1}{5}-\dfrac{1}{6}\)
⇒ \(\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{24.25}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}=\dfrac{1}{5}-\dfrac{1}{25}=\dfrac{4}{25}\)b) \(\dfrac{2}{1.3}=1-\dfrac{1}{3}\)
tương tự
\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}=1-\dfrac{1}{101}=\dfrac{100}{101}\)
Giải:
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2009.2011}.\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right).\)
\(=\dfrac{1}{2}\left[\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+...+\left(\dfrac{1}{2009}-\dfrac{1}{2009}\right)+\left(1-\dfrac{1}{2011}\right)\right].\)
\(=\dfrac{1}{2}\left[0+0+0+...+\left(1-\dfrac{1}{2011}\right)\right].\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{2011}\right).\)
\(=\dfrac{1}{2}.\dfrac{2010}{2011}=\dfrac{2010}{4022}=\dfrac{1005}{2011}.\)
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1\1-1\3+1\3-1\5+1\5-1\7+...+ 1\2009- 1\2011
=1- 1\2011
=2010\2011
dấu \ là 1 trên 
\(B=\dfrac{1}{1.3}\dfrac{1}{3.5}+\dfrac{1}{5.7}+.....+\dfrac{1}{2003.2005}\\ =\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2003.2005}\right)\\ =\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-...+\dfrac{1}{2003}-\dfrac{1}{2005}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{2005}\right)\\ =\dfrac{1}{2}.\dfrac{2004}{2005}\\ =\dfrac{1002}{2005}\)
A = \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
A=\(\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{50}{100}-\dfrac{1}{100}=\dfrac{49}{100}\)
B = \(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{49.51}\)
B = \(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{49}-\dfrac{1}{51}\)
B = \(\dfrac{1}{2}-\dfrac{1}{51}=\dfrac{51}{102}-\dfrac{2}{102}=\dfrac{49}{102}\)
Tính
\(\dfrac{1}{1.3}\)+ \(\dfrac{1}{3.5}\) + \(\dfrac{1}{5.7}\) + .......+ \(\dfrac{1}{2009.2011}\)
Lời giải:
Ta có: \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2009.2011}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{2009.2011}\)
\(2A=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+....+\frac{2011-2009}{2009.2011}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-....+\frac{1}{2009}-\frac{1}{2011}\)
\(2A=1-\frac{1}{2011}=\frac{2010}{2011}\Rightarrow A=\frac{1005}{2011}\)
\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...\dfrac{1}{99}-\dfrac{1}{101}\right)-\dfrac{1}{101}\)
\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{101}\right)-\dfrac{1}{101}\)
\(A=\dfrac{1}{2}.\left(\dfrac{100}{101}\right)-\dfrac{1}{101}\)
\(A=\dfrac{50}{101}-\dfrac{1}{101}=\dfrac{49}{101}\)
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{99.101}-\dfrac{1}{101}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}\right)-\dfrac{1}{101}\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)-\dfrac{1}{101}\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{101}\right)-\dfrac{1}{101}\)
\(=\dfrac{1}{2}.\dfrac{100}{101}-\dfrac{1}{101}=\dfrac{50}{101}-\dfrac{1}{101}=\dfrac{49}{101}\)
Vậy...
\(2H=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{49.51}\)
\(2H=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+...+\dfrac{51-49}{49.51}\)
\(2H=\dfrac{3}{1.3}-\dfrac{1}{1.3}+\dfrac{5}{3.5}-\dfrac{3}{3.5}+...+\dfrac{51}{49.51}-\dfrac{49}{49.51}\)
\(2H=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{49}-\dfrac{1}{51}\)
\(2H=1-\dfrac{1}{51}\)
\(2H=\dfrac{50}{51}\)
\(H=\dfrac{25}{51}\)