Đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+\frac{1}{4.5.6}+\frac{1}{5.6.7}+\frac{1}{6.7.8}+\frac{1}{7.8.9}+\frac{1}{8.9.10}\)

\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+\frac{2}{4.5.6}+\frac{2}{5.6.7}+\frac{2}{6.7.8}+\frac{2}{7.8.9}+\frac{2}{8.9.10}\)

\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{4.5}-\frac{1}{5.6}+...+\frac{1}{8.9}-\frac{1}{9.10}\)

\(2A=\frac{1}{1.2}-\frac{1}{9.10}=\frac{22}{45}\)

\(A=\frac{22}{45}:2=\frac{11}{45}\)
 

1/ 1.2.3 + 1/ 2.3.4 + 1/ 3.4.5+1/4.5.6+1/5.6.7+1/6.7.8+1/7.8.9+1/8.9.10

= 1 - 1/2 - 1/3 + 1/2 - 1/3 - 1/4 + 1/3 - 1/4 - 1/5 + 1/5 - 1/6 - 1/7 + 1/6 - 1/7 - 1/8 + 1/7 - 1/8 - 1/9 + 1/8 - 1/9 - 1/10

= 1 - 1/10 

= 9/10  

Lời giải:
\(2A=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+....+\frac{38-36}{36.37.38}+.\frac{39-37}{37.38.39}\)

\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{36.37}-\frac{1}{37.38}+\frac{1}{37.38}-\frac{1}{38.39}\)

\(=\frac{1}{1.2}-\frac{1}{38.39}=\frac{370}{741}\)

 S = 1.2.3 + 2.3.4 +..+ (n-1).n.(n+1) 

4S = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 +..+ (n-1)n(n+1).4 

ghi dọc cho dễ nhìn: 
(k-1)k(k+1).4 = (k-1)k(k+1)[(k+2) - (k-2)] = (k-1)k(k+1)(k+2) - (k-2)(k-1)k(k+1) 
ad cho k chạy từ 2 đến n ta có: 
1.2.3.4 = 1.2.3.4 
2.3.4.4 = 2.3.4.5 - 1.2.3.4 
3.4.5.4 = 3.4.5.6 - 2.3.4.5 

(n-2)(n-1)n.4 = (n-2)(n-1)n(n+1) - (n-3)(n-2)(n-1)n 
(n-1)n(n+1).4 = (n-1)n(n+1)(n+2) - (n-2)(n-1)n(n+1) 
+ + cộng lại vế theo vế + + (chú ý cơ chế rút gọn) 
4S = (n-1)n(n+1)(n+2) 

=> S = (n-1)n(n+1)(n+2)/4 

Tick nha Hoàng Thái

B= 1/ 1.2.3 + 1/ 2.3 4 + 1/ 3.4.5 + .... + 1/ 48.49.50

Mà ta có:

1/ 1.2 - 1/ 2.3 = 2/ 1.2.3

1/ 2.3 - 1/3.4 = 2/ 2.3.4

Từ đó=> B = 1/2 . ( 2/ 1.2.3 + 2/ 2,3.4 + ... + 2/ 18. 19. 20 )

= 1/2 .( 1/ 1.2 – 1/ 2.3 + 1/ 2.3 - .....- 1/19.20)

= 1/2. ( 1/ 1.2 – 1/ 19.20 ) = 1/ 2 . 189/380 = 189/760

A=[n.(n+1).(n+2).(n+3)-0.1.2.3]:4

Chắc chắn đúng. 

mik nhé

THANKS bạn!

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{1999.2000}-\frac{1}{2000.2001}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2000.2001}\right)=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4002000}\right)=\frac{1}{2}\left(\frac{2000999}{4002000}\right)=\frac{2000999}{8004000}\)

A = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ... + 1/1999.2000.2001

A = 1/2.(2/1.2.3 + 2/2.3.4 + 2/3.4.5 + 2/3.4.5 + ... + 2/1999.2000.2001)

A = 1/2.(1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + 1/3.4 - 1/4.5 + ... + 1/1999.2000 - 1/2000.2001)

A = 1/2.(1/1.2 - 1/2000.2001)

A = 1/2.(1/2 - 1/4002000)

Đến đây số to wa, bn tự lm típ

Chú ý: tính hiệu giữa: 1/1.2 - 1/2.3 = 3/1.2.3 - 1/1.2.3 = 2/1.2.3, nhân thêm 2 vào tử

Ủng hộ mk nha ^_-

Coi \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)

\(\Rightarrow2A=2x\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\right)\)

\(=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\)

\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\)

\(=\frac{1}{1.2}-\frac{1}{99.100}\)

\(=\frac{4950}{9900}-\frac{1}{9900}\)

\(=\frac{4949}{9900}\)

các bn làm đúng rồi

tk mk nha

thnak

Ta có:

B=1.2.3+2.3.4+3.4.5+...+20.21.22

=>4B=1.2.3(4-0)+2.3.4(5-1)+3.4.5(6-2)+...+20.21.22(23-19)

<=>4B=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+...+20.21.22.23-19.20.21.22

<=>4B=20.21.22.23

<=>B=5.21.22.23=53130

53130