Đặt \(A=1\cdot2\cdot4+2\cdot3\cdot5+3\cdot4\cdot6+\cdots+100\cdot101\cdot103\)

\(=1\cdot2\cdot\left(3+1\right)+2\cdot3\cdot\left(4+1\right)+\cdots+100\cdot101\cdot\left(102+1\right)\)

\(=\left(1\cdot2\cdot3+2\cdot3\cdot4+\cdots+100\cdot101\cdot102\right)+\left(1\cdot2+2\cdot3+\cdots+100\cdot101\right)\)

Đặt \(B=1\cdot2\cdot3+2\cdot3\cdot4+\cdots+100\cdot101\cdot102\)

\(=\left(2-1\right)\cdot2\cdot\left(2+1\right)+\left(3-1\right)\cdot3\cdot\left(3+1\right)+\cdots+\left(101-1\right)\cdot101\cdot\left(101+1\right)\)

\(=2\left(2^2-1\right)+3\left(3^2-1\right)+\cdots+101\left(101^2-1\right)\)

\(=\left(2^3+3^3+\cdots+101^3\right)-\left(2+3+\cdots+101\right)\)

\(=\left(1^3+2^3+3^3+\cdots+101^3\right)-1-\left(2+3+\cdots+101\right)\)

\(=\left(1^3+2^3+\cdots+101^3\right)-\left(1+2+3+\cdots+101\right)\)

\(=\left(1+2+3+\cdots+101\right)^2-\left(1+2+3+\cdots+101\right)\)

\(=\left\lbrack101\cdot\frac{102}{2}\right\rbrack^2-101\cdot\frac{102}{2}=\left(101\cdot51\right)^2-101\cdot51\)

Đặt \(C=1\cdot2+2\cdot3+\cdots+100\cdot101\)

\(=1\left(1+1\right)+2\left(2+1\right)+\cdots+100\left(100+1\right)\)

\(=\left(1^2+2^2+\cdots+100^2\right)+\left(1+2+\cdots+100\right)\)

\(=\frac{100\left(100+1\right)\left(2\cdot100+1\right)}{6}+\frac{100\cdot101}{2}=\frac{100\cdot101\cdot201}{6}+50\cdot101\)

\(=50\cdot101\cdot67+50\cdot101=50\cdot101\cdot68\)

Ta có: A\(=\left(1\cdot2\cdot3+2\cdot3\cdot4+\cdots+100\cdot101\cdot102\right)+\left(1\cdot2+2\cdot3+\cdots+100\cdot101\right)\)

=B+C

\(=\left(101\cdot51\right)^2-101\cdot51+50\cdot101\cdot68\)

\(=101^2\cdot51^2-101\cdot51+50\cdot101\cdot68=101\left(101\cdot51^2-51+50\cdot68\right)=101\cdot266050\)

Đặt \(D=1\cdot2^2+2\cdot3^2+\cdots+100\cdot101^2\)

\(=2^2\left(2-1\right)+3^2\left(3-1\right)+\cdots+101^2\left(101-1\right)\)

\(=\left(2^3+3^3+\cdots+101^3\right)-\left(2^2+3^2+\cdots+101^2\right)\)

\(=\left(1^3+2^3+\cdots+101^3\right)-\left(1^2+2^2+\cdots+101^2\right)\)
\(=\left(1+2+\cdots+101\right)^2-101\cdot\frac{\left(101+1\right)\left(2\cdot101+1\right)}{6}\)

\(=\left(101\cdot\frac{102}{2}\right)^2-101\cdot17\cdot2023=101^2\cdot51^2-101\cdot17\cdot2023\)

\(=101\cdot17\left(101\cdot17\cdot3^2-2023\right)=101\cdot17\cdot13430\)

Ta có: \(\frac{1\cdot2\cdot4+2\cdot3\cdot5+3\cdot4\cdot6+\cdots+100\cdot101\cdot103}{1\cdot2^2+2\cdot3^2+\cdots+100\cdot101^2}\)

\(=\frac{101\cdot266050}{101\cdot17\cdot13430}=\frac{1565}{1343}\)

\(\frac{B}{2}=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{100\cdot101}\)

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

\(\frac{B}{2}=\frac{100}{101}\)

\(B=\frac{200}{101}\)

B = \(2\left(\frac{1}{1x2}+\frac{1}{2x3}+....+\frac{1}{100x101}\right)\)

B = \(2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}...+\frac{1}{101}\right)\)

B = \(2\left(1-\frac{1}{101}\right)\)

B = \(2x\frac{100}{101}\)

B = \(\frac{200}{101}\)

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

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

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

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

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

có công thức nè bạn 1²+2²+3³+...+n²= n(n+1)(2n+1):6

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

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

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

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

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

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

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

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

201/101 nhé bạn 

CÁC BẠN LƯỚT QUA NHỚ TICK CHO MÌNH NHÉ 
AI TICK MÌNH MÌNH TICK LẠI CHO

TICK MÌNH ĐƯỢC MAY MẮN CẢ NĂM