a) =1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101

    =1-1/101

    =100/101

b) =(2/1.3+2/3.5+2/5.7+...+2/99.101).2,5

    =(1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101).2,5

    =(1-1/101).2,5

    =100/101.2,5

    =250/101

c) =(2/2.4+2/4.6+2/6.8+...+2/2008-2/2010).2

    =(1/2-1/4+1/4-1/6+1/6-1/8+...+1/2008-1/2010).2

    =(1/2-1/2010).2

    =1004/1005

\(S=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}-\left(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+\dfrac{1}{8\cdot10}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}\right)-\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+\dfrac{2}{8\cdot10}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{9}\right)-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{10}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{8}{9}-\dfrac{1}{2}\cdot\dfrac{2}{5}\)

\(=\dfrac{4}{9}-\dfrac{1}{5}\)

\(=\dfrac{11}{45}\)

A=\(\dfrac{2}{1.3}-\dfrac{2}{3.5}-\dfrac{2}{5.7}-.....-\dfrac{2}{23.25}-\dfrac{1}{27}\)

A=\(\dfrac{2}{3}-\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+....+\dfrac{2}{23.25}\right)-\dfrac{1}{27}\)

A=\(\dfrac{2}{3}-\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+......+\dfrac{1}{23}-\dfrac{1}{25}\right)-\dfrac{1}{27}\)

A=\(\dfrac{2}{3}-\left(\dfrac{1}{3}-\dfrac{1}{25}\right)-\dfrac{1}{27}\)

A=\(\dfrac{2}{3}-\dfrac{22}{75}-\dfrac{1}{27}\)

A=\(\dfrac{227}{675}\)

bằng 2588

cách làm như thế nào ạ

\(C=1.3+3.5+5.7+...+23.25\)

\(6C=1.3.\left(1+5\right)+3.5.\left(7-1\right)+5.7.\left(9-3\right)+...+23.25.\left(27-21\right)\)

\(6C=1.3+1.3.5+3.5.7-1.3.5+5.7.9-3.5.7+...+23.25.27-21.23.25\)

\(6C=3+23.25.27\)

 \(6C=3+15525\)

\(6C=15528\)

\(C=\frac{15528}{6}=2588\)

\(\frac{4}{1.3}\)+\(\frac{4}{3.5}\)+\(\frac{4}{5.7}\)+\(\frac{4}{7.9}\)+...+\(\frac{4}{2011.2013}\)

= 1+\(\frac{1}{3}\)-\(\frac{1}{3}\)+\(\frac{1}{5}\)-\(\frac{1}{5}\)+\(\frac{1}{7}\)-\(\frac{1}{7}\)+\(\frac{1}{9}\)+...+\(\frac{1}{2011}\)+\(\frac{1}{2013}\)

=1+       0          +        0        +        0         +...+        0          +         \(\frac{1}{2013}\)

=1+\(\frac{1}{2013}\)

=\(\frac{2014}{2013}\)

k dùm nha

\(\frac{4}{1\cdot3}+\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+...+\frac{4}{2011\cdot2013}\)

\(=2\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{2011\cdot2013}\right)\)

\(=2\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2011}-\frac{1}{2013}\right)\)

\(=2\cdot\left(1-\frac{1}{2013}\right)\)

\(=2\cdot\frac{2012}{2013}\)

\(=\frac{4024}{2013}\)

Đặt A ta có : \(A=\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{2011.2013}\)

\(2A=4\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}...+\frac{1}{2011.2013}\right)\)

\(2A=4\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2011}-\frac{1}{2013}\right)\)

\(2A=4\left(1-\frac{1}{2013}\right)\)

\(2A=4.\frac{2012}{2013}\)

\(2A=\frac{8048}{2013}\)

\(\Rightarrow A=\frac{4024}{2013}\)

\(B=\dfrac{2^{24}\cdot3^5-2^{24}\cdot3^4}{2^{24}\cdot3^5}+1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{301}-\dfrac{1}{303}\)

\(=\dfrac{2^{24}\cdot3^4\left(3-1\right)}{2^{24}\cdot3^5}+\dfrac{302}{303}\)

\(=\dfrac{2}{3}+\dfrac{302}{303}=\dfrac{202+302}{303}=\dfrac{504}{303}\)

=168/101

(a+\(\dfrac{1}{1.3}\))+(a+\(\dfrac{1}{3.5}\))+(a+\(\dfrac{1}{5.7}\))+..+(a+\(\dfrac{1}{23.25}\))=11.a+(\(\dfrac{1}{3}\)+\(\dfrac{1}{9}\)+\(\dfrac{1}{27}\)+\(\dfrac{1}{81}\)+\(\dfrac{1}{243}\))

(a+a+..+a)+(\(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+...+\(\dfrac{1}{23.25}\)) = 11.a+ \(\dfrac{1}{3}\)+\(\dfrac{1}{9}\)+\(\dfrac{1}{27}\)+\(\dfrac{1}{81}\)+\(\dfrac{1}{243}\))

Đặt A =(a+a+..+a) + \(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+...+\(\dfrac{1}{23.25}\)

Xét dãy số 1; 3; 5;...;25 Dãy số trên là dãy số cách đều với khoảng cách là: 3-1 = 2

Dãy số trên có số số hạng là: (25 - 1): 2 + 1  = 13

Vậy A = a\(\times\)13 + \(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+...+\(\dfrac{1}{23.25}\)

A = a\(\times\)13 + \(\dfrac{1}{2}\) \(\times\)(\(\dfrac{2}{1.3}\)+\(\dfrac{2}{3.5}\)+\(\dfrac{2}{5.7}\)+...+\(\dfrac{2}{23.25}\))

A = a \(\times\) 13 + \(\dfrac{1}{2}\times\)( \(\dfrac{1}{1}-\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)- \(\dfrac{1}{7}\)+...+\(\dfrac{1}{23}\) - \(\dfrac{1}{25}\))

A = a\(\times\)13 + \(\dfrac{1}{2}\) \(\times\) \(\dfrac{24}{25}\)

A = a\(\times\)13 + \(\dfrac{12}{25}\) (1)

Đặt B =    \(\dfrac{1}{3}\) + \(\dfrac{1}{9}\)+ \(\dfrac{1}{27}\)+\(\dfrac{1}{81}\)+\(\dfrac{1}{243}\)

B\(\times\)3 =1 + \(\dfrac{1}{3}\)+\(\dfrac{1}{9}\)+\(\dfrac{1}{27}\)+\(\dfrac{1}{81}\)

B\(\times\)3 - B = 1 - \(\dfrac{1}{243}\) = \(\dfrac{242}{243}\)

2B = \(\dfrac{242}{243}\)

B = \(\dfrac{242}{243}\): 2

B = \(\dfrac{121}{243}\)

11a + B = 11a + \(\dfrac{121}{243}\) (2)

Từ (1) và(2) ta có:

a\(\times\)13  + \(\dfrac{12}{25}\) = 11\(\times\) a + \(\dfrac{121}{143}\)

a \(\times\) 13 + \(\dfrac{12}{25}\) - 11 \(\times\)a = \(\dfrac{121}{143}\) 

\(a\times\)(13 - 11) + \(\dfrac{12}{25}\) = \(\dfrac{121}{143}\)

a \(\times\) 2 + \(\dfrac{12}{25}\) = \(\dfrac{121}{243}\)

a \(\times\) 2 = \(\dfrac{121}{243}\) - \(\dfrac{12}{25}\)

a \(\times\) 2 = \(\dfrac{109}{6075}\)

a = \(\dfrac{109}{6075}\): 2

a = \(\dfrac{109}{12150}\)

\(\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.7}+...+\dfrac{4}{99.101}\\ =\dfrac{4}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\ =2.\left(1-\dfrac{1}{101}\right)\\ =2.\dfrac{100}{101}\\ =\dfrac{200}{101}\)

`4/1.3+4/3.5+4/5.7+...+4/99.101`

`=2(2/1.3+2/3.5+2/5.7+...+2/99.101)`

`=2(1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101)`

`=2(1-1/101)`

`=2. 100/101`

`=200/101`