Đặt S= 1.2 + 2.3 + 3.4 + ...+ 99.100
 3S = 1.2.3+2.3.3+3.4.3+...+98.99.3+99.100.3
3S= 1.2.3+2.3(4-1)+3.4(5-2)+...+98.99(100-97)+99.100(101-98)
3S= 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...-97.98.99+99.100.101-98.99.100
3S = 99.100.101  3S = 3.33.100.101 
 S=33.100.101= 333300

Bạn rút gọn chéo đi 2 với 2 ,3 với 3 cứ như thế còn mỗi 1/100. k nhé

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

=> 3C=3 (1.2 + 2.3 + 3.4 + ... + 99.100)

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

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

=> 3C=99.100.101

=> C=99.100.101/3=333300

Nha bạn      

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=1-\frac{1}{50}\)

\(A=\frac{49}{50}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=1-\frac{1}{50}\)

\(A=\frac{49}{50}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=\left(1-\frac{1}{50}\right)+\left(\frac{-1}{2}+\frac{1}{2}+\frac{-1}{3}+\frac{1}{3}+...+\frac{1}{49}\right)\)

\(A=\frac{49}{50}+0\)

\(A=\frac{49}{50}\)

A=1.2+2.3+3.4+........+98.99 
3A=1.2.3+2.3.3+3.4.3+........+98.99.3 
3A=1.2.3+2.3.(4 -1) +3.4.(5 -2)+........+98.99.(100 -97) 
3A=1.2.3+2.3.4 -1.2.3 +3.4.5 -2.3.4 +........+98.99.100 -97.98.99 
3A=98.99.100 
===>A=(98.99.100)/3

#Japhkiel#

A=1.2+2.3+3.4+........+98.99 
3A=1.2.3+2.3.3+3.4.3+........+98.99.3 
3A=1.2.3+2.3.(4 -1) +3.4.(5 -2)+........+98.99.(100 -97) 
3A=1.2.3+2.3.4 -1.2.3 +3.4.5 -2.3.4 +........+98.99.100 -97.98.99 
3A=98.99.100 
A=\(\frac{98.99.100}{3}=\frac{970200}{3}=323400\)

Đặt A\(=1\cdot2+2\cdot3+3\cdot4+......+98\cdot99\), ta có

\(3A=1\cdot2\cdot3+2\cdot3\cdot3+.....+98\cdot99\cdot3\)

\(=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+....+98\cdot99\cdot\left(100-97\right)\)

\(=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+....+98\cdot99\cdot100-97\cdot98\cdot99\)

\(=98\cdot99\cdot100\)

\(\Rightarrow A=\frac{98\cdot99\cdot100}{3}=323400\)

S=1.2+2.3+...+39.40

3S=1.2.(3 - 0)+2.3.(4 - 1)+...+39.40.(41 - 38)

3S=1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 +...+ 39.40.41 - 38.39.40

3S=39.40.41

S=13.40.41

S=21320

euhdewjs

sai đề 

sai đề là cái chắc

sai đề hả bạn????

A=\(x = {n(n+1)(n+2){} \over 3}\)

S=1.2+2.3+3.4+.............+n(n+1)

=1(1+1) + 2(2+1) + 3(3+1) +...+n(n+1)
=(1^2 + 2^2 + 3^2 +...+ n^2) + (1 + 2 + 3 + ...+ n)

Ta có các công thức:

1^2 + 2^2 + 3^2 +...+ n^2 = n(n+1)(2n+1)/6

1 + 2 + 3 + ...+ n = n(n+1)/2

Thay vào ta có:

S = n(n+1)(2n+1)/6 + n(n+1)/2

=n(n+1)/2[(2n+1)/3 + 1]

=n(n+1)(n+2)/3

(870 – 1.2).(870 – 2.3).(870 – 3.4) … (870 – 99.100)

Ta có: 870 = 29.30

Nên suy ra: 870 – 29.30 = 29.30 – 29.30 = 0

G = 0.

k cho mik nha, cô mik giảng vậy 

k cho mik nha 

Đặt A= 1.2+2.3 +.......+99.100

3A= 1.2.3+2.3.4+3.4.3 +......+ 99.100.3

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

3A = (1.2.3 + 2.3.4 + 3.4.5 +...... + 99.100.101) - (0.1.2 + 1.2.3 + 2.3.4 +.......+ 98.99.100)

3A = 99.100.101 - 0.1.2

3A = 999900 - 0

3A= 999900

A= 999900 : 3

A = 333300 

A=1.2+2.3+3.4+…+99.100

3A = 1.2.3 + 2.3.3 + ... + 99.100.3

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

3A = 1.2.3 + 2.3.4 - 1.2.3 + ... + 99.100.101 - 98.99.100

3A = 99.100.101

=> A = \(\frac{99.100.101}{3}\)= 333 300

3A = 1 × 2 × 3 + 2 × 3 × ( 4 - 1 ) + ... + 99 × 100 × ( 101 - 98 )

3A = 1 × 2 × 3 + 2 × 3 × 4 - 1 × 2 × 3 + ... + 99 × 100 × 101 - 98 × 99 × 100

3A = 99 × 100 × 101 = 999900

A = 999900 ÷ 3 = 333300

Tích mình cái nha