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

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

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

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

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

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

\(S=\frac{3}{\left(1\times2\right)^2}+\frac{5}{\left(2\times3\right)^2}+...+\frac{201}{\left(100\times101\right)^2}\)

\(=\frac{2^2-1^2}{\left(1\times2\right)^2}+\frac{3^2-2^2}{\left(2\times3\right)^2}+...+\frac{101^2-100^2}{\left(100\times101\right)^2}\)

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

\(=1-\frac{1}{101^2}\)

\(=\frac{10200}{10201}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{100.101}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-....-\frac{1}{101}\)

\(=1-\frac{1}{101}=\frac{101}{101}-\frac{1}{101}=\frac{100}{101}\)

201/101 nhé bạn 

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

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

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

=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

(6x + 35 ) = 330 : 6

6x+35=55

6x=55-35

6x=20

x=20:6

x=10/3