\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}=\dfrac{49}{50}\)

\(=\dfrac{49}{50}\)

\(A=\) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=1-\frac{1}{50}\)

\(A=\frac{49}{50}\)

\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.50}\)

A= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)

A = \(\frac{1}{1}-\frac{1}{51}=\frac{50}{51}\)

Ta có: A = 1.2 + 2.3 + 3.4 + … + n.(n + 1)

   =>  3A = 1.2.(3-0) + 2.3.(4-1) + .... + n.(n+1).(n+2 - n+1)

   => 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + .... + n.(n+1).(n+2)

  =>  3A = n.(n+1).(n+2)

  = > A = \(\frac{\text{n.(n+1).(n+2)}}{3}\)

\(A=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)

1.50+2.49+3.48+...+49.2+50.1=

= (1.50+2.50+3.50+...+50.1)-(1.2+2.3+3.4+...+49.50)

= (2500+50).50:2-41650

= 63750-41650=22100

2, 

A = 1.2 + 2.3 + 3.4 + ... + 2011.2012

3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 2011.2012.3

3A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 2011.2012.(2013 - 2010)

3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 2011.2012.2013 - 2010.2011.2012

3A = 2011.2012.2013

A = 2011.2012.2013 : 3 

A = 2714954572

1)A=22100

2)B=60236

Ta có A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{19.20}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}=1-\frac{1}{20}\)

\(=\frac{19}{20}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{19.20}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}...+\frac{1}{19}-\frac{1}{20}\)

\(=\frac{1}{1}-\frac{1}{20}\)

\(=\frac{19}{20}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{19.20}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\)

\(=1-\frac{1}{20}=\frac{19}{20}\)

\(=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}+\dfrac{8-7}{7.8}+\dfrac{9-8}{8.9}+\dfrac{10-9}{9.10}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\\ =1-\dfrac{1}{10}\\ =\dfrac{10-1}{10}=\dfrac{9}{10}\)

1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10

=2-1/1.2+3-2/2.3+4-3/3.4+...+10-9/9.10

=1-1/2+1/2-1/3+1/3-1/4+....+1/9-1/10

=1-1/10

=9/10

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=1-\dfrac{1}{10}\)

\(=\dfrac{9}{10}\)

3A=1.2.(3-0)+2.3.(4-1)+...+n(n+1)[(n-1)(n+2)]

3A=1.2.3-0.1.2+2.3.4-1.2.3+...n.(n+1)(n+2)-(n-1)n(n+1)

  A=n(n+1)(n+2):3
 

Ta thấy:\(\frac{1}{1.2}=1-\frac{1}{2},\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3},...,\frac{1}{49.50}=\frac{1}{49}-\frac{1}{50}\)

=>\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

=>\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

=>\(A=1-\frac{1}{50}\)

=>\(A=\frac{49}{50}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(\Rightarrow A=1-\frac{1}{50}\)

\(\Rightarrow A=\frac{49}{50}\)

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=\frac{1}{1}-\frac{1}{50}\)

\(A=\frac{50}{50}-\frac{1}{50}\)

\(A=\frac{49}{50}\)

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