`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

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

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

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

3S=99.100.101

S=(99.100.101):3=333300

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=333300

333300

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

ta có \(3S=1\cdot2\cdot3+2\cdot3\cdot3+.....+99\cdot100\cdot3\)

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

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

\(3S=99.100.101\)

\(S=\frac{99\cdot100\cdot101}{3}\)

S=...

3S=1.2.3+2.3.3+3.4.3+4.5.3+...+99.100.3

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

3S=99.100.101

S=33.100.101

S=333300

Vậy S=333300

Dễ mà , cô giáo minh vừa dạy xong:

Nhận xét:Khoảng cách giữa 2 thừa số trong mỗi số hạng là 1. Ta nhân 2 vế của S với 3 lần khoảng cách này ,ta được:

 3S=3.(1.2+2.3+3.4+4.5+...+99.100)

3S=1.2.3+2.3.3+3.4.3+4.5.3+....+99.100.3

3S=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+....+99.100.(101-98)

3S=1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+....+99.100.101-98.99.100

3S=99.100.101

S=99.100.101 /3

S=99.100.101/3 nha bạn

Chúc các bạn học giỏi 

Tết vui vẻ nha

S = 1.2 + 2.3 + ... + 99.100

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

4S = 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 +...+ 99.100.101 - 98.99.100

4S = (1.2.3 + 2.3.4 +...+ 99.100.101) - (0.1.2 + 1.2.3 +...+ 98.99.100)

4S = 99.100.101 - 0.1.2

4S = 99.100.101

S = 99.25.101

S = 249975

\(S=1.2+2.3+3.4+4.5+5.6+...+99.100\)

\(3S=1.2.3+2.3.3+3.4.3+4.5.3+...+99.100.3\)

\(3S=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+4.5.\left(6-3\right)+...+99.100.\left(101-98\right)\)\(1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101+98.99.100\)

\(3S=\left(1.2.3-1.2.3\right)+\left(2.3.4-2.3.4\right)+...+\left(98.99.100-98.99.100\right)+99.100.101\)

\(3S=99.100.101=9999000\)

\(S=9999000:3=3333000\)

\(\Rightarrow S=3333000\)