bạn ơi hình như bạn viết sai đề thì phải có lẽ là 1/77 chứ

 bài làm 1/4 +1/28+1/77+ .....+1/10300

           = 1/1x4 +1/4x7 +1/7x11+......+1/100x103

           =1/3 x (3/1x4+3/4x7+3/7x11+.......+3/100X103)

           =1/3x(1-1/4+1/4-1/7+1/7-1/11+.........+1/100x1/103)

           =1/3x(1-1/103)

          =1/3x102/103

         =34/103

           CHÚC BẠN HOK TỐT

Đặt :

\(M=\dfrac{1}{4}+\dfrac{1}{28}+\dfrac{1}{70}+................+\dfrac{1}{550}\)

\(\Rightarrow M=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+..................+\dfrac{1}{22.25}\)

\(\Rightarrow3M=\dfrac{3}{1.4}+\dfrac{3}{4.7}+.................+\dfrac{3}{22.25}\)

\(\Rightarrow3M=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+.........+\dfrac{1}{22}-\dfrac{1}{25}\)

\(\Rightarrow3M=1-\dfrac{1}{25}\)

\(\Rightarrow3M=\dfrac{24}{25}\)

\(\Rightarrow M=\dfrac{24}{75}\)

Đặt:

\(A=\dfrac{1}{4}+\dfrac{1}{28}+\dfrac{1}{70}+.....+\dfrac{1}{550}\)

\(A=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+.....+\dfrac{1}{22.25}\)

\(A=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+.....+\dfrac{1}{22}-\dfrac{1}{25}\right)\)

\(A=\dfrac{1}{3}\left(1-\dfrac{1}{25}\right)=\dfrac{1}{3}.\dfrac{24}{25}=\dfrac{25}{72}\)

\(3M=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\)

\(3M=\frac{4-1}{1.4}+\frac{7-4}{4.7}+...+\frac{100-97}{97.100}\)

\(3M=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\)

\(3M=1-\frac{1}{100}\)

\(3M=\frac{99}{100}\)

\(M=\frac{33}{100}\)

$\dfrac14+\dfrac1{28}+\dfrac1{70}+\cdots+\dfrac1{9700}$

$=\dfrac1{2\cdot2}+\dfrac1{4\cdot7}+\dfrac1{7\cdot10}+\cdots+\dfrac1{97\cdot100}$

$=\dfrac13\left(\dfrac12-\dfrac15\right)+\dfrac13\left(\dfrac14-\dfrac17\right)+\dfrac13\left(\dfrac17-\dfrac1{10}\right)+\cdots+\dfrac13\left(\dfrac1{97}-\dfrac1{100}\right)$

$=\dfrac13\left(\dfrac12-\dfrac15+\dfrac14-\dfrac17+\dfrac17-\dfrac1{10}+\cdots+\dfrac1{97}-\dfrac1{100}\right)$

$=\dfrac13\left(\dfrac12+\dfrac14-\dfrac15-\dfrac1{100}\right)$

$=\dfrac13\left(\dfrac{50+25-20-1}{100}\right)$

$=\dfrac13\cdot\dfrac{54}{100}$

$=\dfrac{9}{50}$

Theo đề bài:

$\dfrac{9}{50}=\dfrac{0,33x}{2009}$

$\Leftrightarrow\dfrac{9}{50}=\dfrac{33x}{100\cdot2009}$

$\Leftrightarrow 9\cdot100\cdot2009=50\cdot33x$

$\Leftrightarrow x=\dfrac{9\cdot100\cdot2009}{50\cdot33}$

$\Leftrightarrow x=\dfrac{6\cdot2009}{11}$

$\Leftrightarrowx=\dfrac{12054}{11}$.

M=1/4+1/4.7+1/7.10+1/10.13+.............+1/88.91

3M=3/4+3/4.7+3/7.10+3/10.13+........+3/88.91

3M= 3/4+1/4-1/7+1/7-1/10+1/10-1/13+......+1/88-1/91

3M=3/4+1/4-1/91=1-1/91=90/91 

----->M= 30/91

$M=\dfrac14+\dfrac1{28}+\dfrac1{70}+\dfrac1{130}+\cdots+\dfrac1{8008}$

$=\dfrac1{2\cdot2}+\dfrac1{4\cdot7}+\dfrac1{7\cdot10}+\dfrac1{10\cdot13}+\cdots+\dfrac1{88\cdot91}$

$=\dfrac14+\dfrac13\left(\dfrac14-\dfrac17\right)+\dfrac13\left(\dfrac17-\dfrac1{10}\right)+\dfrac13\left(\dfrac1{10}-\dfrac1{13}\right)+\cdots+\dfrac13\left(\dfrac1{88}-\dfrac1{91}\right)$

$=\dfrac14+\dfrac13\left(\dfrac14-\dfrac17+\dfrac17-\dfrac1{10}+\dfrac1{10}-\dfrac1{13}+\cdots+\dfrac1{88}-\dfrac1{91}\right)$

$=\dfrac14+\dfrac13\left(\dfrac14-\dfrac1{91}\right)$

$=\dfrac14+\dfrac{29}{364}$

$=\dfrac{91+29}{364}$

$=\dfrac{120}{364}$

$=\dfrac{30}{91}$.

Phần nào không hiểu bạn có thể nhắn hỏi mình nhe

Ta có : mẫu số 1 : 4 . 1

mẫu số hai : 4.7

... mẫu số thứ 96 = 100.103 = 10300

=> Số số hạng y là 100

Ta có :

\((y+..+y) + (\frac{3}{1.4} + \frac{3}{4.7} + ...+ \frac{3}{100.103})\)

\(= ( y+...+y) + [1. (\frac{1}{1.4} + \frac{1}{4.7} + ..+ \frac{1}{100.103})]\)

\(= (y+...y) + [1.(\frac{1}{1} - \frac{1}{4} + \frac{1}{4} - \frac{1}{7} + ...+ \frac{1}{100} - \frac{1}{103}) ]\)

\(= (y+...+y) + (1 - \frac{1}{103})\)

\(= (y+...+y) + \frac{102}{103}\)

\(=> (y+...+y) = \frac{308}{103} - \frac{102}{103} = \frac{206}{103}\)

\(=> y = \frac{206}{103} : 100 = \frac{206}{10300} = \frac{103}{5150}\) ( Chia 100 vì có 100 số hạng y)

Vậy \(y = \frac{103}{5150}\)

\(4-\dfrac{1}{4}-\dfrac{1}{28}-\dfrac{1}{70}-.....-\dfrac{1}{2002.2005}\)

\(=4-\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+.....+\dfrac{1}{2002.2005}\right)\)

\(=4-\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+.....+\dfrac{1}{2002}-\dfrac{1}{2005}\right)\)\(=4-\left(1-\dfrac{1}{2005}\right)\)

\(=4-1+\dfrac{1}{2005}=3+\dfrac{1}{2005}=\dfrac{6016}{2005}\)

\(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+\frac{1}{208}\)

\(=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}\)

\(=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}\right)\)

\(=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\right)\)

\(=\frac{1}{3}.\left(1-\frac{1}{16}\right)\)

\(=\frac{1}{3}.\frac{15}{16}=\frac{5}{16}\)

1/1.4+1/4.7+1/7.10+..+1/13.16

1-1/4+1/4-1/7+..+1/13-1/16

1-1/16

15/16