A = \(\frac{1}{1.4}\)+ \(\frac{1}{4.7}\)+\(\frac{1}{7.10}\)+...+ \(\frac{1}{2014.2017}\)
3A = \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{2014.2017}\)
3A = \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{2014}-\frac{1}{2017}\)
3A= 1 - \(\frac{1}{2017}\)
A = \(\frac{1}{3}-\frac{1}{2017.3}\)
A = \(\frac{672}{2017}\)

Ta có \(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2014.2017}\)

\(\Rightarrow A=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2014}-\frac{1}{2017}\right)\)

\(\Rightarrow A=\frac{1}{3}.\left(1-\frac{1}{2017}\right)\)

\(\Rightarrow A=\frac{1}{3}.\frac{2016}{2017}=\frac{672}{2017}\)

Vậy \(A=\frac{672}{2017}\)

~ Học tốt

# Chiyuki Fujito

672/2017

\(B=\dfrac{4}{1\cdot4}+\dfrac{4}{4\cdot7}+...+\dfrac{4}{2014\cdot2017}\)

\(=\dfrac{4}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{2014\cdot2017}\right)\)

\(=\dfrac{4}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{2014}-\dfrac{1}{2017}\right)\)

\(=\dfrac{4}{3}\left(1-\dfrac{1}{2017}\right)=\dfrac{4}{3}\cdot\dfrac{2016}{2017}=\dfrac{8064}{6051}\)

\(a,A=\dfrac{101}{100}+\dfrac{102}{100}+\dfrac{103}{100}+...+\dfrac{199}{100}\)

\(A=\dfrac{101+102+103+...+109}{100}\)

Xét tử số : \(101+102+103+...+199\)

Có : \(\left(199-101\right):1+1=99\) (số hạng)

\(\Rightarrow\) Tử số bằng \(:\left(199+101\right).99:2=14850\)

\(\Rightarrow A=\dfrac{14850}{100}=\dfrac{297}{2}\)

\(b,B=\dfrac{10002}{10000}+\dfrac{10004}{10000}+\dfrac{10006}{10000}+...+\dfrac{12014}{10000}\)

\(B=\dfrac{10002+10004+10006+...+12014}{10000}\)

\(B=\dfrac{10002+10004+10006+...+12014}{10000}\)

Xét tử số : \(10002+10004+10006+...+12014\)

Có : \(\left(12014-10002\right):2+1=1007\) (số hạng)

\(\Rightarrow\) Tử số bằng : \(\left(12014+10002\right).1007:2=11085056\)

\(\Rightarrow B=\dfrac{11085056}{10000}\)

Bạn tự làm câu C nha

\(D=\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{2014.2015}\)

\(\Rightarrow D=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{2014}-\dfrac{1}{2015}\)

\(\Rightarrow D=\dfrac{1}{5}-\dfrac{1}{2015}=\dfrac{402}{2015}\)

\(E=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{2014.2017}\)

\(\Rightarrow3E=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{2014.2017}\)

\(\Rightarrow3E=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{2014}-\dfrac{1}{2017}\)

\(\Rightarrow3E=1-\dfrac{1}{2017}=\dfrac{2016}{2017}\)

\(\Rightarrow E=\dfrac{2016}{2017}:3=\dfrac{672}{2017}\)

D = \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\) + \(\dfrac{1}{7.8}\) +...+ \(\dfrac{1}{2014.2015}\)
D = \(\dfrac{1}{5}\) - \(\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)+...+ \(\dfrac{1}{2014}-\dfrac{1}{2015}\)
D = \(\left(\dfrac{1}{5}-\dfrac{1}{2015}\right)\)
D = \(\dfrac{403}{2015}-\dfrac{1}{2015}\)
D = \(\dfrac{402}{2015}\)

E = \(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}\)+ ...+ \(\dfrac{1}{2014.2017}\)
E = \(\dfrac{1.3}{3.1.4}+\dfrac{1.3}{3.4.7}+\dfrac{1.3}{3.7.10}+...+\dfrac{1.3}{3.2014.2017}\)
E = \(\dfrac{1}{3}\) .( \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{2014.2017}\) )
E = \(\dfrac{1}{3}\).\(\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{2014}+\dfrac{1}{2015}\right)\)
E = \(\dfrac{1}{3}\) . \(\left(1-\dfrac{1}{2015}\right)\)
E = \(\dfrac{1}{3}\) . \(\left(\dfrac{2015}{2015}-\dfrac{1}{2015}\right)\)
E = \(\dfrac{1}{3}\) . \(\dfrac{2014}{2015}\)
E = \(\dfrac{2014}{6045}\)

Ta có : B = \(\frac{4}{1.4}+\frac{4}{4.7}+\frac{4}{7.10}+......+\frac{4}{2014.2017}\)

\(=\frac{4}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{2014.2017}\right)\)

\(=\frac{4}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+......+\frac{1}{2014}-\frac{1}{2017}\right)\)

\(=\frac{4}{3}.\left(1-\frac{1}{2017}\right)\)

\(=\frac{4}{3}.\frac{2016}{2017}=\frac{2688}{2017}\)

thank

sai zùi

`#3107.101107`

1.

a)

`1/(1*4) + 1/(4*7) + 1/(7*10) + ... + 1/(100*103)`

`= 1/3 * (3/(1*4) + 3/(4*7) + 3/(7*10) + ... + 3/(100*103) )`

`= 1/3 * (1 - 1/4 + 1/4 - 1/7 + ... + 1/100 - 1/103)`

`= 1/3* (1 - 1/103)`

`= 1/3*102/103`

`= 34/103`

b)

`-1/3 + (-1/15) + (-1/35) + (-1/63) + ... + (-1/9999)`

`= - 1/3 - 1/15 - 1/35 - 1/63 - ... - 1/9999`

`= - (1/3 + 1/15 + 1/35 + ... + 1/9999)`

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

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

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

`= -1/2 * (1 - 1/101)`

`= -1/2*100/101`

`= -50/101`

2.

`3/(1*4) + 3/(4*7) + ... + 3/(94*97) + 3/(97*100)`

`= 1 - 1/4 + 1/4 - 1/7 + ... + 1/94 - 1/97 + 1/97 - 1/100`

`= 1-1/100`

`= 99/100`

Đặt \(A=\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+...+\frac{1}{2014\cdot2017}\)

\(\Rightarrow A=\frac{1}{3}\cdot\left(\frac{3}{1\cdot3}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{2014\cdot2017}\right)\)

\(\Rightarrow A=\frac{1}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2014}-\frac{1}{2017}\right)\)

\(\Rightarrow A=\frac{1}{3}\cdot\left(1-\frac{1}{2017}\right)=\frac{1}{3}-\frac{1}{6051}< \frac{1}{3}\)

\(\Rightarrow A< \frac{1}{3}\left(ĐPCM\right)\)

Ta có :

\(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{2014.2017}\)

\(=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{2014.2017}\right)\)

\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{2014}-\frac{1}{2017}\right)\)

\(=\frac{1}{3}\left(1-\frac{1}{2017}\right)\)

\(=\frac{1}{3}.\frac{2016}{2017}< \frac{1}{3}\left(đpcm\right)\)

\(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{2014.2017}\) 

\(=\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2014}-\frac{1}{2017}\right)\) 

\(=\frac{1}{3}\left(1-\frac{1}{2017}\right)=\frac{1}{3}\cdot\frac{2016}{2017}\)

\(=\frac{672}{2017}< \frac{1}{3}\) 

\(\RightarrowĐPCM\) 

S = 1.4 + 4.7 + 7.10 + 10.13 + ... + 61.64

1.4.9 = 1.4.(7 + 2) = 1.4.7 + 1.4.2

4.7.9 = 4.7.(10 - 1) = 4.7.10 - 1.4.7

7.10.9 = 7.10.(13 - 4) = 7.10.13 - 4.7.10

10.13.9 = 10.13.(16 - 7) = 10.13.16 - 7.10.13

.......................................................................

61.64.9 = 61.64.(67 - 58) = 61.64.67 - 58.61.64

Cộng vế với vế ta có:

1.4.9 + 4.7.9 + 7.10.9 +...+ 61.64.9 = 1.4.2 + 61.64.67

9(1.4 + 4.7 + 7.10+ ...+ 61.64) = 261576

  1.4 + 4.7 + 7.10 +...+ 61.64 = 261576 : 9

1.4 + 4.7 + 7.10 + ... + 61.64 = 29064 

\(\text{Đặt: S= biểu thức cần tính}\)

\(\Rightarrow9S=1.4.7+4.7.9+......+19.22.9+4.2\)

\(\Rightarrow9S=1.4.7+4.7\left(10-1\right)+...+19.22\left(25-16\right)+8\)

\(\Rightarrow9S=19.22.25+8\Rightarrow S=1162\)

sai rồi 9S = 1.4.9 mà 

\(B=\frac{5}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{2014}-\frac{1}{2017}\right)\)

\(B=\frac{5}{3}.\left(1-\frac{1}{2017}\right)\)

\(B=\frac{5}{3}.\frac{2016}{2017}=\frac{10080}{6051}\)

\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{2014.2017}\)

\(3M=5\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{2014.2017}\right)\)

\(3M=5\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{2014}-\frac{1}{2017}\right)\)

\(3M=5\left(1-\frac{1}{2017}\right)\)

\(3M=5.\frac{2016}{2017}\)

\(3M=\frac{10080}{2017}\)

\(\Rightarrow M=\frac{3360}{2017}\)

Ta có: 

\(B=\frac{5}{1.4}+\frac{5}{4.7}+.....+\frac{5}{2014.2017}\)

\(\Rightarrow B.\frac{3}{5}=\frac{3}{1.4}+\frac{3}{4.7}+.....+\frac{3}{2014.2017}\)

\(=\left(1-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{7}\right)+......+\left(\frac{1}{2014}-\frac{1}{2017}\right)\)

\(=1-\frac{1}{2017}\)

\(=\frac{2016}{2017}\)

\(\Rightarrow B=\frac{2016}{2017}:\frac{3}{5}\)

\(=\frac{3360}{2017}\)

Vậy B \(=\frac{3360}{2017}\)~~~~~