\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+......+\frac{1}{48.49.50}\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{48.49}-\frac{1}{49.50}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{49.50}\right)\)

\(=\frac{1}{2}.\frac{612}{1225}=\frac{612}{2450}=\frac{306}{1225}\)

Do not ask why hay quá!

Đặt \(T=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\)

Ta xét:

\(\frac{1}{1.2}-\frac{1}{2.3}=\frac{1}{1.2.3}\);\(\frac{1}{2.3}-\frac{1}{3.4}=\frac{1}{2.3.4}\);. . . ; \(\frac{1}{48.49}-\frac{1}{49.50}=\frac{1}{48.49.50}\)

 Rút ra dạng tổng quát,ta có: (mình nói thêm nhé)

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

\(\Rightarrow2T=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\)

Ta nhận thấy: \(-\frac{1}{2.3}+\frac{1}{2.3}=0\);\(-\frac{1}{3.4}+\frac{1}{3.4}=0\);.....

\(\Rightarrow2T=\frac{1}{1.2}-\frac{1}{49.50}=\frac{612}{1225}\)

\(\Rightarrow T=\frac{612}{\frac{1225}{2}}=\frac{306}{1225}\)

Vậy .. . . 

\(A=\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+...+\frac{1}{36\times37\times38}+\frac{1}{37\times38\times39}\)

\(2A=\frac{2}{1\times2\times3}+\frac{2}{2\times3\times4}+\frac{2}{3\times4\times5}+...+\frac{2}{36\times37\times38}+\frac{2}{37\times38\times39}\)

\(2A=\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+...+\frac{1}{37\times38}-\frac{1}{38\times39}\)

\(2A=\frac{1}{1\times2}-\frac{1}{38\times39}\)

\(2A=\frac{741}{1482}-\frac{1}{1482}\)

\(2A=\frac{370}{741}\)

\(A=\frac{370}{741}:2=\frac{185}{741}\)

cái đuôi ak ko hiểu còn cái đầu thì dễ 

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+.......+\frac{1}{8.9.10}\)

\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+......+\frac{2}{8.9.10}\right)\)

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

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{90}\right)=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)

Sửa đề chút:

\(\frac{1}{1x2x3}+\frac{1}{2x3x4}+...+\frac{1}{98x99x100}\)

\(=\frac{1}{2}.\left(\frac{2}{1x2x3}+\frac{2}{2x3x4}+...+\frac{2}{98x99x100}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1x2}-\frac{1}{2x3}+\frac{1}{2x3}-\frac{1}{3x4}+...+\frac{1}{98x99}-\frac{1}{99x100}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{99.100}\right)\)

\(=\frac{1}{4}-\frac{1}{99.200}< 1\)

                         đpcm

thank you nha!!!

Bạn xem lại đề, là cộng mới đúng chứ ???

Mình làm được rồi này :

\(B=\frac{1}{1.2.3}-\left(\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{97.98.99}\right)\)

    \(=\frac{1}{6}-\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{97.98}-\frac{1}{98.99}\right)\)

    \(=\frac{1}{6}-\left(\frac{1}{2.3}-\frac{1}{98.99}\right)\)

     \(=\frac{1}{6}-\frac{1}{6}+\frac{1}{9702}\)

     \(=\frac{1}{9702}\)

Trả lời:

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2018.2019.2020}+\frac{1}{2.2019.2020}\)

\(A=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2018.2019.2020}+\frac{2}{2.2019.2020}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2018.2019}-\frac{1}{2019.2020}+\frac{1}{2019.2020}\right)\)

\(A=\frac{1}{2}.\frac{1}{1.2}\)

\(A=\frac{1}{4}\)

H = \(\frac{1}{1.2}-\frac{1}{1.2.3}+\frac{1}{2.3}-\frac{1}{2.3.4}+\frac{1}{3.4}-\frac{1}{3.4.5}+...+\frac{1}{99.100}-\frac{1}{99.100.101}\)

   \(=\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\right)-\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{99.100.101}\right)\)

Đặt G = \(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\right)\)

          = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)

          = \(1-\frac{1}{100}\)

           = \(\frac{99}{100}\)

Đặt K = \(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{99.100.101}\right)\)

=>2K = \(\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{99.100.101}\right)\)

          = \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\)

          = \(\frac{1}{1.2}-\frac{1}{100.101}\)

          = \(\frac{1}{2}-\frac{1}{10100}\)

          = \(\frac{5049}{10100}\)

=> K =\(\frac{5049}{10100}:2=\frac{5049}{10100}.\frac{1}{2}=\frac{5049}{20200}\)

Thay G,K vào H ta có :

H = \(\frac{99}{100}-\frac{5049}{20200}\)

Tự tính :)

\(H=\frac{1}{1.2}-\frac{1}{1.2.3}+\frac{1}{2.3}-\frac{1}{2.3.4}+...+\frac{1}{99.100}-\frac{1}{99.100.101}\)

\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)-\left(\frac{1}{1.2.3}+\frac{1}{2.34}+...+\frac{1}{99.100.101}\right)\)

\(=\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{99.100.101}\right)\)

\(=\left(1-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)

\(=\frac{99}{100}-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)=\frac{99}{100}-\frac{1}{2}.\frac{5049}{10100}=\frac{99}{100}-\frac{5049}{20200}=\frac{14949}{20200}\)

A = 2/1x2x3 + 2/2x3x4 + 2/3x4x5 + ... + 2/36x37x38 + 2/37x38x39

A = 1/1x2 - 1/2x3 + 1/2x3 - 1/3x4 + 1/3x4 - 1/4x5 + ...+ 1/36x37 - 1/37x38 + 1/37x38 - 1/38x39

A = 1/2 - 1/38x39

A = 370/741

Tớ ko chắc là đúng đâu

bài 

Nguyễn Hà Thương sai chưa chia 2