\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{100\cdot101}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-...+\dfrac{1}{100}-\dfrac{1}{101}\)

\(=1-\dfrac{1}{101}\)

\(=\dfrac{100}{101}\)

\(1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}+\frac{1}{100\cdot101}\)

\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}\)

\(=1+1-\frac{1}{101}=2-\frac{1}{101}=1\frac{100}{101}=\frac{201}{101}\)

=1+1/1-1/2+1/2-1/3+1/3-1/+1/4-1/5+...+1/99-1/100+1/100-1/101

=1+1-1/101

=201/101

Đây là câu trả lời của mình :

 Đặt A = 1 x 2 + 2 x 3 + 3 x 4 + ........ + 99 x 100 + 100 x 101

 3A =  1 x 2 x3 + 2 x 3 x 3 + 3 x 4 x 3 + ........... + 99 x 100 x 3 + 100 x 101 x 3

 3A =  1 x 2 x 3 + 2 x 3 x ( 4 - 1 ) + 3 x 4 x ( 5 - 2 ) + ........... + 99 x 100 x ( 101 - 98 ) + 100 x 101 x ( 102 - 99 )

 3A =  1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + 3 x 4 x 5 - 2 x 3 x 4 + .......... + 99 x 100 x 101 - 98 x 99 x 100 + 100 x 101 x 102 - 99 x 100 x 101

 3A =  100 x 101 x 102

 3A =  1030200

 A = 1030200 : 3

 A = 343400

Vậy : 1 x 2 + 2 x 3 + 3 x 4 + ......... + 99 x 100 + 100 x 101 = 343400

343400

343400 

Học thày chẳng bằng học bn 

Các bn nhỉ

A=1x2+2x3+3x4+4x5+......+99x100+100x101

3A=1x2x(3-0)+2x3x(4-1)+3x4x(5-2)+4x5x(6-3)+...+99x100x(101-98)+100x101x(102-99)

3A=1x2x3-0x1x2+2x3x4-1x2x3+3x4x5-2x3x4+4x5x6-3x4x5+...+99x100x101-98x99x100+100x101x102-99x100x101

3A=(1x2x3+2x3x4+3x4x5+4x5x6+...+99x100x101+100x101x102)-(0x1x2+1x2x3+2x3x4+3x4x5+...+98x99x100+99x100x101)

3A=100x101x102

A=100x101x102:3

A=343400

A = 1x2 + 2x3 + 3x4 + 4x5 + ... + 99x100 + 100x101

3A = 1x2x(3-0) + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98) + 100x101x(102-99)

3A = 1x2x3 - 0x1x2 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100 + 100x101x102 - 99x100x101

3A = 100x101x102 - 0x1x2

3A = 100x101x102

A = 100x101x34

A = 343400

=1-1/2+1/2-1/3+...+1/99-1/100

=1-1/100=99/100

= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100

= 1 - 1/100

= 99/100

= 99/100

Ta có:

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

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

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

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

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\)

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

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

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

Vậy.....

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}.\)
\(=\frac{1}{1}-\frac{1}{100}=\frac{99}{100}\)

\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+....+\dfrac{1}{99\cdot100}\)

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

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

\(A=1+0-\dfrac{1}{100}\)

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

\(\Rightarrow A< 1\)

A=1⋅21​+2⋅31​+3⋅41​+....+99⋅1001​

�=1−12+12−13+13−14+...+199−1100A=1−21​+21​−31​+31​−41​+...+991​−1001​

�=1+(−12+12−13+13−14+...+199)−1100A=1+(−21​+21​−31​+31​−41​+...+991​)−1001​

�=1+0−1100A=1+0−1001​

�=1−1100<1A=1−1001​<1

$\Rightarrow A<1$

A<1

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

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

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

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

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

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

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

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

A = 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/99.100

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

A = 1 - 1/100

A = 99/100