Question

[Dương Hải Đường]

a)1/1.2.3+1/2.3.4+1/3.4.5+...+1/2018.2019.2020

b?)1/1.2.3.4+1/2.3.4.5+1/3.4.5.6+...+1/2017.2018.2019.2020

MN GIÚP MÌNH VỚI Ạ! MÌNH CẢM ƠN CC NHÌU NHÌU!!!>.<

Answer 1

[Vũ Tiến Manh]

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

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