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

3A = 1.2.3 + 2.3.3 + 3.4.3 + ....... + 99 . 100 . 3

3A = 1.2.3 + 2.3.(4-1) + 3.4.(5-2)  +.... + 99.100.(101-98)

3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ..... + 99 . 100 . 101 - 98 . 99 . 100

3A = (1.2.3 - 1.2.3) + (2.3.4-2.3.4) + ... + (98.99.100 - 98.99.100) + 99  . 100 . 101

3A = 99 . 100 . 101 = 999900

A = 999900 : 3 = 333300

A=1*2+2*3+3*4+...+99*100

A=100*101*102:3

A=343400(công thức)

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`S = 1.2 + 2.3 + 3.4 + 4.5 + ... + 99.100.`

`3S =  1.2.3 + 2.3.(4-1) + 3.4.(5-4) + 4.5.(6-3) + ... + 99.100.(101-98)`

`3S =  1.2.3 + 2.3.4-1.2.3 + 3.4.5-4.5.6 + 4.5.6-3.4.5 + ... + 99.100.101-98.99.100`

`3S =  99.100.101`

`S = 33.100.101`

`S = 333300`

3S=1.2(3-0)+2.3(4-1)+.....+99.100(101-98)

=1.2.3-0.1.2+2.3.4-1.2.3+4.5.6-2.3.4+....+99.100.101-98-99-100

=99.100.101

S=33.100.101

=333300

nhân 3 vào mỗi hạng tử ta được:

3*(1.2+2.3+3.4+...+99.100)

= 1.2.(3-0)+ 2.3.(4-1)+ 3.4.(5-2)+... + 99.100.(101-98)

=1.2.3 + 2.3.4 -1.2.3 + 3.4.5 -2.3.4 +... + 99.100.101 - 98.99.100

= 99.100.101

Vậy tổng ban đầu 99.100.101/3= 33.100.101

Vậy tổng trên chia hết cho 2;3;4;5;10

A = 1.2+2.3+3.4+......+99.100 
Gấp A lên 3 lần ta có: 
A . 3 = 1.2.3 + 2.3.3 + 3.4.3 + … + 99.100.3 
A . 3 = 1.2.3 + 2.3.(4 - 1) + 3.4.( 5 - 2) + … + 99.100. (101 - 98) 
A . 3 = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + … + 99.100.101 - 98.99.100 
A . 3 = 99.100.101 
A = 99.100.101 : 3 
A = 33.100.101 
A = 333 300

\(A=1\cdot2+2\cdot3+3\cdot4+4\cdot5+...+99\cdot100\)

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

\(3A=1\cdot2\cdot\left(3-0\right)+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+99\cdot100\cdot\left(101-98\right)\)

\(3A=1\cdot2\cdot3-0+2\cdot3\cdot4-1\cdot2\cdot3+3\cdot4\cdot5-2\cdot3\cdot4+...99\cdot100\cdot101-98\cdot99\cdot100\)

\(3A=98\cdot99\cdot100\Rightarrow A=\frac{98\cdot99\cdot100}{3}=...\)

Ta có: \(S=1.2+2.3+3.4+...+99.100\)

\(\Rightarrow3S=1.2.3+2.3.3+3.3.4+....+99.100.3\)

\(\Rightarrow3S=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)....99.100.\left(101-98\right)\)

\(\Rightarrow3S=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100\)

\(\Rightarrow3S=99.100.101\)

\(\Rightarrow S=\frac{99.100.101}{3}=\frac{999900}{3}=333300\)

S=  1.2 + 2.3 +... + 99.100

=>S= \(\frac{99.100.101}{3}\)=333300

   \(S=1.2+2.3+3.4+4.5+...+99.100\)

\(3S=1.2.3+2.3.3+3.4.3+...+99.100.3\)

\(3S=1.2.3+2.3.\left(4-1\right)+...+99.100.\left(101-98\right)\)

Ta có : S = 1.2 + 2.3 + 3.4 + ..... + 99.100

=> 3S = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + .... + 99.100.101

=> 3S = 99.100.101

=> S = \(\frac{99.100.101}{3}=333300\)

ta xét

\(S\left(n\right)=1.2+2.3+..+n\left(n-1\right)\)

\(\Rightarrow3S\left(n\right)=1.2.3+2.3.3+..+3.n.\left(n-1\right)\)

\(\Leftrightarrow3S\left(n\right)=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+..+n\left(n-1\right)\left(n+1-\left(n-2\right)\right)\)

\(\Leftrightarrow3S\left(n\right)=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+..+n\left(n-1\right)\left(n+1\right)-n\left(n-1\right)\left(n-2\right)\)

\(\Leftrightarrow3S\left(n\right)=n\left(n-1\right)\left(n+1\right)\Rightarrow S\left(n\right)=\frac{n\left(n-1\right)\left(n+1\right)}{3}\)

Áp dụng ta có \(S\left(100\right)=\frac{99.100.101}{3}=333300\)

 S=1.2+ 2.3+4,5.......+99.100 
Nhân cả 2 vế với 3, ta được: 
3S=1.2.3+ 2.3.3+ 3.4.3+ 4.5.3+...... 99.100.3 
= 1.2.3 + 2.3(4-1) + 3.4.(5-2) +...+ 99.100.(101-98) 
= 1.2.3 + 2.3.4 -1.2.3 + 3.4.5-2.3.4 +...+ 99.100.101-98.99.100 
= 99.100.101 
----> S = (99.100.101):3 
 S= 333300 
Vậy A=333300 

S = 1.2 + 2.3 + 3.4 + 4.5 +...+ 99.100

S = 1.100

S = 100