bai nay mik lam sai roi nha

      \(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{37.38.39}\right).1428+185.8\)

\(=\frac{2}{2}.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\right).1428+185.8\)

\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{37.38.39}\right).1428+1480\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\right).1428+1480\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{38.39}\right).1428+1480\)

\(\frac{1}{2}.\frac{370}{741}.1428+1480\)

\(=\frac{185}{741}.1428+1480\)

\(=356,52+1480=1836,52\)

chỗ\(\frac{185}{741}.1428\)mk làm tròn số lun á nha

mk ko chắc tính đúng hay sai nha nhưng cách làm thì kiểu vậy

\(\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{37.38}-\frac{1}{38.39}\right).1428+185.8\)

\(=\left(\frac{1}{2}-\frac{1}{1482}\right).1482+1840\)

\(=\frac{740}{1482}.1482+1840\)

\(=740+1840\)

\(=2580\)

giúp mình vs ạ

a) \(A=1\cdot2\cdot3-2\cdot3\cdot4+3\cdot4\cdot5+...+98\cdot99\cdot100\)

\(\Rightarrow4A=4\cdot\left(1\cdot2\cdot3+2\cdot3\cdot4+...+98\cdot99\cdot100\right)\)

\(\Rightarrow4A=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot4+3\cdot4\cdot5\cdot4+...+98\cdot99\cdot100\cdot4\)

\(\Rightarrow4A=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot\left(5-1\right)+3\cdot4\cdot5\cdot\left(6-2\right)+....+98\cdot99\cdot100\cdot\left(101-97\right)\)

\(\Rightarrow4A=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot5-1\cdot2\cdot3\cdot4+3\cdot4\cdot5\cdot6-....-97\cdot98\cdot99\cdot100\)

\(\Rightarrow4A=\left(1\cdot2\cdot3\cdot4-1\cdot2\cdot3\cdot4\right)+\left(2\cdot3\cdot4\cdot5-2\cdot3\cdot4\cdot5\right)+...+98\cdot99\cdot100\cdot101\)

\(\Rightarrow4A=0+0+0+...+98\cdot99\cdot100\cdot101\)

\(\Rightarrow4A=98\cdot99\cdot100\cdot101\)

\(\Rightarrow A=\dfrac{98\cdot99\cdot100\cdot101}{4}\)

b) \(B=1+3^2+3^3+3^4+...+3^{99}\)

\(\Rightarrow3B=3\cdot\left(1+3^2+3^3+...+3^{99}\right)\)

\(\Rightarrow3B=3+3^3+3^4+3^5+...+3^{99}+3^{100}\)

\(\Rightarrow3B-B=\left(3+3^3+3^4+...+3^{100}\right)-\left(1+3^2+3^3+...+3^{99}\right)\)

\(\Rightarrow2B=\left(3^3-3^3\right)+\left(3^4-3^4\right)+...+\left(3-1\right)+\left(3^{100}-3^2\right)\)

\(\Rightarrow2B=2+3^{100}-3^2\)

\(\Rightarrow B=\dfrac{2+3^{100}-3^2}{2}\)

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/

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

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

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

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

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

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 ^_-

Đặt A = 1.2.3 + 2.3.4 + 3.4.5 + ... + 28.29.30

4A = 1.2.3.(4-0) + 2.3.4.(5-1) + 3.4.5.(6-2) + ... + 28.29.30.(31-27)

4A = 1.2.3.4 - 0.1.2.3. + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + ... + 28.29.30.31 - 27.28.29.30

4A = 28.29.30.31 - 0.1.2.3

4A = 28.29.30.31

\(A=\frac{28.29.30.31}{4}=7.29.30.31=188790\)

Theo cách tính trên ta dễ dàng tính được:

1.2.3 + 2.3.4 + 3.4.5 + ... + (n - 1).n.(n + 1) = \(\frac{\left(n-1\right).n.\left(n+1\right).\left(n+2\right)}{4}\)

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

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

      \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{10.11.12}\)

\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{1}{10.11.12}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{10.11}-\frac{1}{11.12}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{11.12}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{132}\right)\)

\(=\frac{1}{2}.\frac{65}{132}=\frac{65}{264}\)