Ta có: 
A-B =1.2+2.3+3.4+...+100.101- 
(1^2+2^2+3^2+4^2+...+100^2) 
= 1.2+2.3+3.4+...+100.101- 

Đặt B = 1.2+2.3 +.......+99.100+100.101

3B= 1.2.3+2.3.4+3.4.3 +......+ 99.100.3 + 100.101.3

3B= 1.2. (3 - 0) + 2.3.(4 - 1) +3.4. (5 - 2)....... . 99.100. (101 - 98) . 100.101.(102 - 99)

3B = (1.2.3 + 2.3.4 + 3.4.5 +...... + 99.100.101 + 100.101.102) - (0.1.2 + 1.2.3 + 2.3.4 +.......+ 98.99.100.99.100.101)

3B = 100.101.102 - 0.1.2

3B = 1030200 - 0

3B= 1030200

B =  1030200 : 3

B =  343400

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

Nhớ **** cho mjk với nhak!!!!!

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

Nhớ **** cko mjk nhak!!

A = 1.2 + 2.3 + 3.4 + 4.5 + ... + 99.100 + 100.101

3.A = 1.2.3 + 2.3.3 +3.4.3 + ... + 100.101.3

3A= 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 100.101.(102 - 99)

3A = 1.2.3 + 2.3.4 - 1.2.3 + 2.3.4 -3.4.5 + ... +99.100.101 -100.101.102

3A = 99.100.101

A = 99.100.101 : 3

A = 33.100.101

Vậy A = 33. 100 .101 (Tự tính)

A=1.2+2.3+3.4+.....+99.100+100.101

đáp án:

bằng 343400

học tốt

Đặt A = 1.2 + 2.3 + 3.4 + 4.5 + ... + 99.100 + 100.101

3A = 1.2.3 + 2.3.3 + 3.4.3 + 4.5.3 + ... + 99.100.3 + 100.101.3

3A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 99.100.(101 - 98) + 100.101.(102 - 99)

3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100 + 100.101.102 - 99.100.101

3A = 100.101.102

3A = 1030200

A = 343400

Gọi biểu thức này là S , ta có 

S =1.2 + 2.3 + 3.4 + 4.5 + ...+ 100.101

3S= 1. 2 .3 + 2. 3 .3 + 3 . 4 .3 + 4 .5 .3 + ...........+ 100 .101 .3

3S= 1.2 (3 - 0) + 2 . 3 .(4 - 1) + 3 . 4. (5 - 2 ) +.......+ 100 . 101 . (102 - 99)

3S = 1 . 2 . 3 - 0 . 1 .2  + 2 . 3 . 4 - 1 . 2 .3 + ................+ 100 . 101 .102 - 99  100 . 101 

S   =   \(\frac{100.101.102}{3}=\frac{100.101.34}{1}\)

S   = 343400

Được rồi:Để ý nhé số hạng tổng quát của dãy có dạng:

\(\dfrac{a+b}{ab}=\dfrac{1}{a}+\dfrac{1}{b}\)

\(S=\dfrac{3}{1.2}-\dfrac{5}{2.3}+...-\dfrac{201}{100.101}\)

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

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

@Bình Dị bn giỏi thật

B=1.2+2.3+3.4+...+100.101

=>3B=1.2.3+2.3.3+3.4.4+....+100.101.3

=1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+....+100.101.102-99.100.101

=-0.1.2-100.101.102

=1030200

=>B=1030200:3=343400

B = 1.2+2.3 +.......+99.100+100.101

3B= 1.2.3+2.3.4+3.4.3 +......+ 99.100.3 + 100.101.3

3B= 1.2. (3 - 0) + 2.3.(4 - 1) +3.4. (5 - 2)....... . 99.100. (101 - 98) . 100.101.(102 - 99)

3B = (1.2.3 + 2.3.4 + 3.4.5 +...... + 99.100.101 + 100.101.102) - (0.1.2 + 1.2.3 + 2.3.4 +.......+ 98.99.100.99.100.101)

3B = 100.101.102 - 0.1.2

3B = 1030200 - 0

3B= 1030200

B =  1030200 : 3

B =  343400

=> 3.( \(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+...+\(\dfrac{1}{100.101}\))

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

=> 3.(\(\dfrac{1}{1}\)-\(\dfrac{1}{101}\))

=> 3. \(\dfrac{100}{101}\)

=> \(\dfrac{300}{101}\)

Tick cho mk nhé, chúc bạn học tốt

\(\dfrac{3}{1.2}+\dfrac{3}{2.3}+\dfrac{3}{3.4}+...+\dfrac{3}{100.101}\)

= \(3.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{100.101}\right)\)

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

= \(3.\left(1-\dfrac{1}{101}\right)\)= \(3.\dfrac{100}{101}=\dfrac{300}{101}\).