=0

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

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

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

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

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

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 >:)!!!!!

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2016.2017}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{2016}-\frac{1}{2017}\)

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

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

                           \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\)

                       \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)

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

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

                    Ủng hộ mk nha!!!

ta có 
1=1 
1/1.2 =1-1/2 
1/2.3 1/2-1/3 
1/3.4 =1/3-1/4 
....... 
1/(99.100) =1/99 -1/100 
cộng theo vế các đẳng thức trên được 
S =1+1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100 
S =2-1/100

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

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

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

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

Vậy A = \(\frac{99}{100}\)

A=1-(-1/2+1/2-1/3+1/3-1/4.....+1/99-1/100)

A=1-1/100

A=100/100-1/100

A=99/100

CHÚC BN HỌC GIỎI NHA

a) Đặt A = 1.2 + 2.3 + ........ + (n-1)n

3A = 1.2.3 + 2.3.(4-1) + .... + (n-1)n[(n+1)-(n-2)]

3A = 1.2.3 + 2.3.4 - 1.2.3 + .... + (n-1)n(n+1) - (n-2)(n-1)n

3A = (1.2.3 - 1.2..3) + ... + (n-1)n(n+1)

A = \(\frac{\left(n-1\right)n\left(n+1\right)}{3}\)

b) Đặt B = 1² + 2² + ..... + n²

B = 1(2 - 1) + 2(3 - 1) + ..... + n[(n + 1) - 1]

B = 1.2 + 2.3 + .......... + n(n + 1) - (1+2+3+....+n)

B = A -  \(\frac{n\left(n+1\right)}{2}\)