\(\left(1-\frac{1}{1\cdot2}\right)+\left(1-\frac{1}{2\cdot3}\right)+...+\left(1-\frac{1}{1995\cdot1996}\right)\)

\(=\left(1+1+1+...+1\right)-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{1995\cdot1996}\right)\)

\(=\left(1995\cdot1\right)-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1995}-\frac{1}{1996}\right)\)

\(=1995-\left(1-\frac{1}{1996}\right)\)

\(=1995-\frac{1995}{1996}\)

\(C=\left(1-\frac{1}{1.2}\right)+\left(1-\frac{1}{2.3}\right)+...+\left(1-\frac{1}{1995.1996}\right)\)

\(C=\left(1-1-\frac{1}{2}\right)+\left(1-\frac{1}{2}-\frac{1}{3}\right)+...+\left(1-\frac{1}{1995}-\frac{1}{1996}\right)\)

\(C=1\times998+1-\frac{1}{1996}\)

\(C=998+\frac{1995}{1996}\)

a)

`1/1-1/2`

`=2/2-1/2`

`=1/2`

b)

`1/(1*2)+1/(2*3)`

`=1/1-1/2+1/2-1/3`

`=1/1-1/3`

`=3/3-1/3`

`=2/3`

c)

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{1}-\dfrac{1}{100}\\ =\dfrac{99}{100}\)

d) 

\(\dfrac{3}{1\cdot2}+\dfrac{3}{2\cdot3}+...+\dfrac{3}{99\cdot100}\) đề phải như thế này chứ nhỉ?

\(=\dfrac{1\cdot3}{1\cdot2}+\dfrac{1\cdot3}{2\cdot3}+...+\dfrac{1\cdot3}{99\cdot100}\\ =3\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\\ =3\cdot\dfrac{99}{100}\\ =\dfrac{297}{100}\)

Lời giải :

Đặt S=1.2+2.3+3.4+4.5+…+99.100+100.101

3S=1.2.3+2.3.3+3.4.3+4.5.3+…+99.100.3+100.101.3

=1.2(3−0)+2.3(4−1)+3.4(5−2)+4.5(6−3)+…+99.100(101−98)+100.101(102−99)

=0.1.2-1.2.3+1.2.3-2.3.4+...+99.100.101-100.101.102

=100.101.102

S=100.101.34=343400

1.Tính 

a) Ta có: 

  A=(1-1/2²).(1-1/3²)...(1-1/100²)

=>A=3/2².8/3².....9999/100²

=>A=(1.3/2.2).(2.4/3.3).....(99.101/100.100)

=>A=(1.2.3.....99/2.3.4.....100).(3.4.5.....101/2.3.4.....100)

=>A=1/100.101/2

=>A=101/200

b) Ta có: 

  B=-1/1.2-1/2.3-1/3.4-...-1/100.101

=>B=-(1/1.2+1/2.3+1/3.4+...+1/100.101)

=>B=-(1-1/2+1/2-1/3+1/3-1/4+...+1/100-1/101)

=>B=-(1-1/101)

=>B=-100/101

 c) Ta có:

 C=1.2+2.3+3.4+...+100.101

       =>3C=1.2.3+2.3.3+3.4.3+...+100.101.3

       =>3C=1.2.3+2.3.(4-1)+3.4.(5-2)+...+100.101.(102-99)

       =>3C=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-3.4.5+...+100.101.102

       =>3C=100.101.102

       =>3C=1030200

       =>C=343400

Chúc bạn hok tốt nhé >:)!!!!!

v

Answer:

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}\)

\(=\dfrac{100}{100}-\dfrac{1}{100}\)

\(=\dfrac{99}{100}\)

A = 1 - \(\dfrac{1}{1.2}\) - \(\dfrac{1}{2.3}-\dfrac{1}{3.4}-\dfrac{1}{4.5}...-\dfrac{1}{97.98}\)

A= 1-\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{97.98}\right)\)

A= 1- \(\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}...+\dfrac{1}{97}-\dfrac{1}{98}\right)\)

A= 1- \(\left(\dfrac{1}{1}-\dfrac{1}{98}\right)\)

A=1-  1 + \(\dfrac{1}{98}\)

A= \(\dfrac{1}{98}\)

Lời giải:

$1-A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{97.98}$

$1-A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{98-97}{97.98}$

$1-A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{97}-\frac{1}{98}$

$=1-\frac{1}{98}$

$\Rightarrow A=\frac{1}{98}$

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2013.2014}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2013}-\dfrac{1}{2014}\\ =1-\dfrac{1}{2014}\\ =\dfrac{2013}{2014}\)

\(=\frac{0}{1.2}+\frac{0}{2.3}+\frac{0}{3.4}+...+\frac{0}{2015.2016}\)

\(=0+0+0+...+0=0\)