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

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

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

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

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

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

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

1/2A=1/2(6/4+6/28+6/70+6/130+6/208)

       = 3/4+3/28+3/70+3/130+3/208

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

       =1-1/16

       =15/16 => A=15/8

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$=\dfrac{120}{364}$

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

4 = 1.4 =1.(1+3)

28=4.7=4.(4+3)

70=7.10=7.(7+3)

130=10.13=10.(10+3)

208=13.16=13.(13+3)

 suy ra n(n+3)

\(B=\frac{3}{4}+\frac{3}{28}+\frac{3}{70}+\frac{3}{130}+\frac{3}{208}+\frac{3}{304}\)

\(\Rightarrow B=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}+\frac{3}{16.19}\)

\(\Rightarrow B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}\)

\(\Rightarrow B=1-\frac{1}{19}=\frac{18}{19}\)

Vậy \(B=\frac{18}{19}\)

Chúc bn học tốt

B = \(\frac{3}{4}+\frac{3}{28}+\frac{3}{70}+\frac{3}{130}+\frac{3}{208}+\frac{3}{304}\)

= \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}+\frac{3}{16.19}\)

= \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-...+\frac{1}{13}-\frac{1}{16}\)

= \(1-\frac{1}{16}\)

= \(\frac{15}{16}\)

Vào câu hỏi tương tự nha bạn !

$M=\dfrac14+\dfrac1{28}+\dfrac1{70}+\dfrac1{130}+\cdots$ (có $30$ số hạng)

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

Với mọi số hạng từ số hạng thứ hai:

$\dfrac1{a(a+3)}=\dfrac13\left(\dfrac1a-\dfrac1{a+3}\right)$

Do đó:

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

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

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

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

$=\dfrac{120}{364}$

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