2/1.3 + 2/3.5 + 2/5.7 + ... + 2/2009.2011

= 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/2009 - 1/2011

= 1 - 1/2011

= 2010/2011

 A = 1/1x3 + 1/3x5 + 1/5x7 +.........+ 1/2009x2011

    = 1/1-1 +1/3-5 + 1/5-7 + .......+ 1/2009-2011

    = 1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 +........+ 1/2009 -1/2011

    = 1/1 - 1/2011

    = 2010/2011

Ta có;\(\frac{4}{1\times3}+\frac{4}{3\times5}+\frac{4}{5\times7}+....+\frac{4}{19\times21}\)

\(=2\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+....+\frac{2}{19\times21}\right)\)

\(=2\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)\)

\(=2\times\left(1-\frac{1}{21}\right)=2\times\frac{20}{21}=\frac{40}{21}\)

4/1 x 3 + 4/ 3 x 5 + 4/ 5 x 7 + ....+ 4/ 17 x 19 + 4/ 19 x 21

= 2 x ( 2/ 1 x 3 + 2/ 3 x 5 + 2/ 5 x 7 + ...+ 2/ 17 x 19 + 2/ 19 x 21 ) 

= 2 x ( 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ...+ 1/17 - 1/19 + 1/19 - 1/21 ) 

= 2 x ( 1 - 1/21 ) 

= 2 x  20/21

= 40/21 

Chúc bạn học giỏi !!! 

Đặt \(K=\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{17.19}+\frac{4}{19.21}\)

\(\Leftrightarrow K=\left(\frac{4}{1}-\frac{4}{3}\right)+\left(\frac{4}{3}-\frac{4}{5}\right)+\left(\frac{4}{5}-\frac{4}{7}\right)+\left(\frac{4}{7}-\frac{4}{9}\right)+...+\left(\frac{4}{19}-\frac{4}{21}\right)\)

\(\Leftrightarrow K=\frac{4}{1}-\frac{4}{3}+\frac{4}{3}-\frac{4}{5}+\frac{4}{5}-\frac{4}{7}+\frac{4}{7}-\frac{4}{9}+...+\frac{4}{19}-\frac{4}{21}\)

\(\Leftrightarrow K=\frac{4}{1}-\frac{4}{21}=\frac{84}{21}-\frac{4}{21}=\frac{80}{21}\)

Lời giải:

Gọi tổng trên là $A$
$A=2\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{25.27}\right)$

$=2\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{27-25}{25.27}\right)$

$=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{25}-\frac{1}{27}\right)$

$=2\left(1-\frac{1}{27})=\frac{52}{27}$

\(K=\dfrac{4}{1\times3}+\dfrac{4}{3\times5}+...+\dfrac{4}{299\times301}\)

\(=2\times\left(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+...+\dfrac{2}{299\times301}\right)\)

\(=2\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{299}-\dfrac{1}{301}\right)\)

\(=2\times\left(1-\dfrac{1}{301}\right)=2\times\dfrac{300}{301}=\dfrac{600}{301}\)

\(K=\dfrac{4}{1\cdot3}+\dfrac{4}{3\cdot5}+...+\dfrac{4}{299\cdot301}\)

\(=2\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{299}-\dfrac{1}{301}\right)\)

\(=2\cdot\dfrac{300}{301}=\dfrac{600}{301}\)

a=1/1x2+1/2x3+....+1/99x100

a=1-1/2+1/2-1/3+....+1/99-1/100

a=1-1/100

a=99/100

b=4/1x3+4/3x5+.....+4/51x53

b=2x(2/1x3+2/3x5+....+2/51x53)

b=2x(1-1/3+1/3-1/5+...+1/51-1/53)

b=2x(1-1/53)

b=2x52/53

b=104/53

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Bài này cũng dễ mà

\(S=\frac{4}{1\times3}+\frac{16}{3\times5}+\frac{36}{5\times7}+...+\frac{2500}{49\times51}\)

\(=\frac{1\times3+1}{1\times3}+\frac{3\times5+1}{3\times5}+\frac{5\times7+1}{5\times7}+...+\frac{49\times51+1}{49\times51}\)

\(=\frac{1\times3}{1\times3}+\frac{1}{1\times3}+\frac{3\times5}{3\times5}+\frac{1}{3\times5}+\frac{5\times7}{5\times7}+\frac{1}{5\times7}+...+\frac{49\times51}{49\times51}+\frac{1}{49\times51}\)

\(=1+\frac{1}{1\times3}+1+\frac{1}{3\times5}+1+\frac{1}{5\times7}+...+\frac{1}{49\times51}\) (  Có : \(\left(51-3\right)\div2+1=25\)chữ số 1 )

\(=25+\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{49\times51}\)

\(=25+\frac{1}{2}\times\left(1-\frac{1}{3}\right)+\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}\times\left(\frac{1}{5}-\frac{1}{7}\right)+...+\frac{1}{2}\times\left(\frac{1}{49}-\frac{1}{51}\right)\)

\(=25+\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(=25+\frac{1}{2}\times\left(1-\frac{1}{51}\right)\)

\(=25+\frac{1}{2}\times\frac{50}{51}\)

\(=25+\frac{25}{51}\)

\(=\frac{1300}{51}\)

\(S=\frac{4}{1.3}+\frac{16}{3.5}+\frac{36}{5.7}+...+\frac{2500}{49.51}\)

\(=\frac{4}{3}+\frac{16}{15}+\frac{36}{35}+...+\frac{2500}{2499}\)

\(=1+\frac{1}{3}+1+\frac{1}{15}+1+\frac{1}{35}+...+1+\frac{1}{2499}\)

\(=\left(1+1+1+...+1\right)+\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2500}\right)\)

\(=25+\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{35}+...+\frac{1}{2499}\right)\)

Đặt \(A=\frac{1}{3}+\frac{1}{5}+\frac{1}{35}+...+\frac{1}{2499}\)

\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)

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

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

\(\Rightarrow S=25+\frac{50}{51}=\frac{1325}{51}\)

Vậy S=\(\frac{1325}{51}\)

Bài 1:

 A  = \(\dfrac{1}{1\times3}\) + \(\dfrac{1}{3\times5}\) + \(\dfrac{1}{5\times7}\) +...+ \(\dfrac{1}{2019\times2021}\)

A =   \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{2}{1\times3}\) + \(\dfrac{2}{3\times5}\) + \(\dfrac{2}{5\times7}\)+...+ \(\dfrac{2}{2019\times2021}\))

A = \(\dfrac{1}{2}\) \(\times\)( \(\dfrac{1}{1}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\)+...+ \(\dfrac{1}{2019}\) - \(\dfrac{1}{2021}\))

A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{1}{1}\) - \(\dfrac{1}{2021}\))

A = \(\dfrac{1010}{2021}\)

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Bài 2:

B = \(\dfrac{4}{11\times16}\) + \(\dfrac{4}{16\times21}\)+ \(\dfrac{4}{21\times26}\)+...+ \(\dfrac{4}{61\times66}\)
B = \(\dfrac{4}{5}\) \(\times\) ( \(\dfrac{5}{11\times16}\)+ \(\dfrac{5}{16\times21}\) + \(\dfrac{5}{21\times26}\)+...+ \(\dfrac{5}{61\times66}\))

B = \(\dfrac{4}{5}\) \(\times\) ( \(\dfrac{1}{11}\) - \(\dfrac{1}{16}\) + \(\dfrac{1}{16}\) - \(\dfrac{1}{21}\) + \(\dfrac{1}{21}\) - \(\dfrac{1}{26}\)+...+ \(\dfrac{1}{61}\) - \(\dfrac{1}{66}\))

B = \(\dfrac{4}{5}\) \(\times\)( \(\dfrac{1}{11}\) - \(\dfrac{1}{66}\))

B = \(\dfrac{4}{5}\) \(\times\)  \(\dfrac{5}{66}\)

B = \(\dfrac{2}{33}\)