M = 1 - 1/ 999 = 998/999

=> 2M = 2/1*3 + 2/3*5 + ... + 2/995*997  + 2/ 997*999  =  1-1/3 + 1/3 - 1/5 +... + 1/ 995 - 1/997 + 1/997 - 1 / 999 = 1- 1/999 = 998/999 

=> m = 998/999   /  2   =  499/999  (bn tính lại xem nha mk ko có máy tính nên sợ sai)

 Vậy M = ..... 

\(M=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{995.997}+\frac{1}{997.999}\)

\(M=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{995}-\frac{1}{997}+\frac{1}{997}-\frac{1}{999}\)

\(M=1-\frac{1}{999}\)

\(M=\frac{998}{999}\)

\(M=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2017.2019}\)

\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{2019}\right)\)

\(=\frac{1}{2}.\frac{2018}{2019}\)

\(=\frac{2018}{4038}\)

\(\Rightarrow\frac{2018}{4038}< \frac{1}{2}\)( lấy máy tính ) 

\(M=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+.....+\frac{1}{2017.2019}\)

\(\Rightarrow M=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-......-\frac{1}{2017}+\frac{1}{2017}-\frac{1}{2019}\)

\(\Rightarrow M=1-\frac{1}{2019}\)

\(\Rightarrow M=\frac{2019}{2019}-\frac{1}{2019}\)

\(\Rightarrow M=\frac{2018}{2019}\)

Có \(\frac{2018}{2019}=\frac{2018.2}{2019.2}=\frac{4036}{4038}\)

\(\frac{1}{2}=\frac{1.2019}{2.2019}=\frac{2019}{4038}\)

Mà \(\frac{4036}{4038}< \frac{2019}{4038}\Rightarrow M< \frac{1}{2}\)

Vậy M < \(\frac{1}{2}\)

tớ làm câu b thôi, câu a nhân 1/2 lên là đc 

\(A=\frac{1}{2}.\left[\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left(2n-1\right).\left(2n+1\right)}\right)\right]\)

\(A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2.n-1}-\frac{1}{2n+1}\right)\)

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

p/s: lưu ý không có dấu "=" đâu nhé vì \(\frac{1}{2.\left(2n+1\right)}>0\left(n\text{ thuộc }N\right)\)

a) 1/1 - 1/3 +1/3 - 1/5 +........+1/49 - 1/51

=1/1 - 1/51 (các số liền kề nhau cộng lại bằng 0)

=50/51

còn câu b bạn tự giải

nhớ thank mik nha!!!!!

b,khoảng cách của nó là 3 mà tử của nó bằng 3 chứng  tỏ nó là dạng đủ 

1/1-1/4+1/4-1/7+...+1/97-1/100

1-1/100=99/100

\(\frac{1}{2}-\frac{1}{1.3}-\frac{1}{3.5}-\frac{1}{5.7}-\frac{1}{7.9}-\frac{1}{9.11}=\frac{4}{5}-x\)

<=> \(2.\frac{1}{2}-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)=\frac{8}{5}-2x\)

<=> \(1-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right)=\frac{8}{5}-2x\)

<=> \(1-\left(1-\frac{1}{11}\right)-\frac{8}{5}=-2x\)

<=> \(-\frac{83}{55}=-2x\)

<=> \(x=\frac{83}{110}\)

Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\), ta có:

\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2015.2017}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)

\(=\frac{1}{2}.\frac{2016}{2017}=\frac{1008}{2017}\)

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2015.2017}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2015}-\frac{1}{2017}+\frac{1}{2017}\)

\(=1-\frac{1}{2017}\)

\(=\frac{2016}{2017}\)

mk đầu tiên đấy

1008/2017