đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}\left(\frac{1}{18.19}-\frac{1}{19.20}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{380}\right)=\frac{189}{760}\)

Đặt \(B=\frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{19.20}=\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{3}+...+\frac{3}{19}-\frac{3}{20}\)

\(=3-\frac{3}{20}=\frac{57}{20}\)

\(D=A-B=\frac{189}{760}-\frac{57}{20}=-\frac{1977}{760}\)

Gọi \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)là A

\(\frac{3}{1.2}-\frac{3}{2.3}-...-\frac{3}{19.20}\)là B

\(A=\left[\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{18.19}-\frac{1}{19.20}\right)\right]\)

\(A=\left[\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\right]\)

\(A=\left[\frac{1}{2}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\right)\right]\)

\(A=\left[\frac{1}{2}.\left(1-\frac{1}{20}\right)\right]\)

\(A=\frac{1}{2}.\frac{19}{20}\)

\(A=\frac{19}{40}\)

\(B=\frac{3}{1.2}-\frac{3}{2.3}-...-\frac{3}{19.20}\)

\(B=\left(\frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{19.20}\right)\)

\(B=\left[3.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{19.20}\right)\right]\)

\(B=\left[3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{2}{3}+...+\frac{1}{19}-\frac{1}{20}\right)\right]\)

\(B=\left[3.\left(\frac{19}{20}\right)\right]\)

\(B=\frac{57}{20}\)

Vậy A - B = \(\frac{19}{40}-\frac{57}{20}\)

\(=-\frac{95}{40}=-\frac{19}{8}\)

Nếu đúng thì k nha

a/

\(b=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)

\(2b=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{99-97}{97.99}=\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}=\)

\(=1-\dfrac{1}{99}=\dfrac{98}{99}\Rightarrow b=\dfrac{98}{2.99}=\dfrac{49}{99}\)

b/

\(c=\dfrac{3-1}{1.2.3}+\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}=\)

\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+\dfrac{1}{98.99}-\dfrac{1}{99.100}=\)

\(=\dfrac{1}{2}-\dfrac{1}{99.100}\)

c/

\(\dfrac{2}{5}.d=\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}+\dfrac{101-99}{99.100.101}=\)

\(=\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}+\dfrac{1}{99.100}-\dfrac{1}{100.101}=\)

\(=\dfrac{1}{2.3}-\dfrac{1}{100.101}\Rightarrow d=\left(\dfrac{1}{2.3}-\dfrac{1}{100.101}\right):\dfrac{2}{5}\)

Ta có : \(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)

\(\Leftrightarrow2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{18.19.20}\)

\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{18.19}-\frac{1}{19.20}\)

\(=\frac{1}{2}-\frac{1}{19.20}=\frac{189}{380}\)

\(\Rightarrow B=\frac{189}{760}\)

\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}\left(\frac{1}{18.19}-\frac{1}{19.20}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{19.20}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{380}\right)\)

\(=\frac{1}{2}.\frac{189}{380}=\frac{189}{760}\)

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)

*\(1+2+3+...+\left(n-1\right)+n\)

Số số hạng là:

\(\left(n-1\right):1+1=n-1+1=n\)(số hạng)

Tổng của dãy số là: 

\(\left(n+1\right)\cdot\dfrac{n}{2}=\dfrac{n\left(n+1\right)}{2}\)

*\(1+3+5+...+\left(2n-1\right)\)

Số số hạng của dãy số là: 

\(\left(2n-1-1\right):2+1=\dfrac{\left(2n-2\right)}{2}+1=n-1+1=n\)(số hạng)

Tổng của dãy số là: 

\(\left(2n-1+1\right)\cdot\dfrac{n}{2}=\dfrac{2n^2}{2}=2n\)

Lời giải:

Gọi tổng trên là $A$

$A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}$

$2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{18.19.20}$

$=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+....+\frac{20-18}{18.19.20}$

$=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{18.19}-\frac{1}{19.20}$

$=\frac{1}{1.2}-\frac{1}{19.20}=\frac{189}{380}$

$\Rightarrow A=\frac{189}{760}$

ĐÚNG ĐÓ BẠN ƠI

D=1/2.[1/1.2-1/2.3+1/2.3-1/3.4+...+1/18.19-1/19.20]-3.[1-1/2+1/2-1/3+1/3-1/4+...+1/19-1/20]

  =1/2.[1/2-1/380]-3.[1-1/20]

  =1/2.[189/380]-3.[19/20]

  =189/760-57/20

  =189/760-2166/760

  =-1977/760

Nhớ nhak

a)\(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}.\frac{1}{n+1}.\left(\frac{1}{n}-\frac{1}{n+2}\right)\)=\(\frac{1}{2}.\frac{1}{n\left(n+1\right)}-\frac{1}{2}.\frac{1}{\left(n+1\right)\left(n+2\right)}\)= \(\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+1}\right)-\frac{1}{2}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)\)

=> a = \(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{2}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{3}\right)\)+\(\frac{1}{2}\left(\frac{1}{2}-\frac{1}{3}\right)-\frac{1}{2}\left(\frac{1}{3}-\frac{1}{4}\right)\)+....+\(\frac{1}{2}\left(\frac{1}{2018}-\frac{1}{2019}\right)-\frac{1}{2}\left(\frac{1}{2019}-\frac{1}{2020}\right)\)=\(\frac{1}{2}\left(1-\frac{1}{2}\right)-\frac{1}{2}\left(\frac{1}{2019}-\frac{1}{2020}\right)\)=\(\frac{1}{4}\left(1-\frac{1}{2019.1010}\right)\)=\(\frac{2019.1010-1}{2.2019.2020}\)

b) tương tự \(\frac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}=\left(\frac{1}{n}-\frac{1}{n+1}\right)\left(\frac{1}{n+2}-\frac{1}{n+3}\right)\)=\(\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)-\left(\frac{1}{n+1}-\frac{1}{n+2}\right)\)-\(\frac{1}{3}\left(\frac{1}{n}-\frac{1}{n+3}\right)+\frac{1}{2}\left(\frac{1}{n+1}-\frac{1}{n+3}\right)\)=\(\frac{1}{6}\left(\frac{1}{n}-\frac{1}{n+1}\right)-\frac{1}{3}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)\)+\(\frac{1}{6}\left(\frac{1}{n+2}-\frac{1}{n+3}\right)\)= M-P+N

Với n từ 1 đến 2017 thì

M= \(\frac{1}{6}\left(\frac{1}{1}-\frac{1}{2}\right)+\frac{1}{6}\left(\frac{1}{2}-\frac{1}{3}\right)+...\)+\(\frac{1}{6}\left(\frac{1}{2017}-\frac{1}{2018}\right)\)=\(\frac{1}{6}\left(1-\frac{1}{2018}\right)=\frac{2017}{6.2018}\)

N= \(\frac{1}{6}\left(\frac{1}{3}-\frac{1}{4}\right)+\frac{1}{6}\left(\frac{1}{4}-\frac{1}{5}\right)+...+\)\(\frac{1}{6}\left(\frac{1}{2019}-\frac{1}{2020}\right)=\)\(\frac{1}{6}\left(\frac{1}{3}-\frac{1}{2020}\right)=\frac{2017}{6.3.2020}\)

P= \(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{3}\right)+\frac{1}{3}\left(\frac{1}{3}-\frac{1}{4}\right)+...+\)\(\frac{1}{3}\left(\frac{1}{2018}-\frac{1}{2019}\right)\)= \(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{2019}\right)=\frac{2017}{3.2.2019}\)

M+N-P = \(\frac{2017}{6}\left(\frac{1}{2018}+\frac{1}{3.2020}-\frac{1}{2019}\right)\)=\(\frac{2017}{6}.\left(\frac{1}{2018.2019}+\frac{1}{3.2020}\right)\)

=  \(\frac{2017\left(1010+1009.673\right)}{3.2018.2019.2020}\)