\(S=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{\left(n-1\right).n.\left(n+1\right)}+...+\frac{1}{23.24.25}\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}-\frac{1}{n.\left(n+1\right)}+...+\frac{1}{23.24}-\frac{1}{24.25}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{24.25}\right)=\frac{299}{1200}\)

\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{23.24.25}\right)=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{23.24}-\frac{1}{24.25}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{600}\right)=\frac{1}{2}.\frac{299}{600}=\frac{299}{1200}\)

\(S=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{23.24.25}\)

\(S=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{23.24}-\frac{1}{24.25}\right)\)

\(S=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{24.25}\right)\)

\(S=\frac{1}{4}-\frac{1}{24.50}\)

Dễ thấy với mọi số tự nhiên n > 1 , ta có :

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

Sử dụng  hệ thức trên cho từng số hạng trong tổng sau :

\(2S=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{\left(n-1\right).n.\left(n+1\right)}+\frac{2}{23.24.25}\)

     \(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{\left(n-1\right).n}-\frac{1}{n.\left(n+1\right)}+...+\frac{1}{23.24}-\frac{1}{24.25}\)

Để ý rằng trong vế phải của hệ thức trên , trừ 2 số hạng đầu và cuối , các số hạng còn lại tạo thành từng cặp đối nhau.

Do đó , có thể rút gọn : 

\(2S=\frac{1}{1.2}-\frac{2}{24.25}=\frac{299}{600}\)

Vậy , ta được \(S=\frac{299}{600}\)

A = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{99.100.101}\)

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

= \(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)\)

= \(\frac{1}{2}.\frac{5049}{10100}\)

= \(\frac{5049}{20200}\)

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{99.100.101}\)

\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{99.100.101}\)

Ta thấy:

\(\frac{2}{1.2.3}=\frac{1}{1.2}-\frac{1}{2.3};\frac{2}{2.3.4}=\frac{1}{2.3}-\frac{1}{3.4};...;\frac{2}{99.100.101}=\frac{1}{99.100}-\frac{1}{100.101}\)

\(\Rightarrow2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\)

\(\Rightarrow2A=\frac{1}{1.2}-\frac{1}{100.101}\)

\(\Rightarrow2A=\frac{1}{2}-\frac{1}{10100}\)

\(\Rightarrow2A=\frac{5050}{10100}-\frac{1}{10100}\)

\(\Rightarrow2A=\frac{5049}{10100}\Rightarrow A=\frac{5049}{10100}:2=\frac{5049}{20200}\)

x = 9/11

(x=9/11)có đúng không????

gọi A=1/1*2*3+1/2*3*4+...+1/49*50*51

      2A=2(1/1*2*3+1/2*3*4+...+1/49*50*51)

       2A=2/1*2*3+2/2*3*4+...+2/49*50*51

       2A=1/1*2-1/2*3+1/2*3-1/3*4+...+1/49*50-1/50*51

      2A=1/2-1/2550

      2A=637/1275

      A=637/1275:2

      A=637/2550

qua bài trên ta có công thức \(\frac{1}{n\cdot\left(n+1\right)\cdot\left(n+2\right)}\)= \(\frac{1}{n\cdot\left(n+1\right)}\)-\(\frac{1}{\left(n+1\right)\cdot\left(n+2\right)}\)

lộn công thức là 2/n*(n+1)*(n+2)=1/n*(n+1)-1/(n+1)*(n+2) cho tui xin lỗi

mà tick nhé

cảm ơn bạn rất nhiều

Đặt biểu thức là A

\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+.2018.2019\)

\(2A=\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\left(\frac{1}{2017.2018}-\frac{1}{2018.2019}\right)\)

\(2A=\frac{1}{2}-\frac{1}{2018.2019}\)

A= 1/4 - 1/(2018.2019)

Vậy A = ... (tự ghi)

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2017.2018.2019}\)

\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2017.2018.2019}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2017.2018}-\frac{1}{2018.2019}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{2018.2019}\right)\)

Tự làm nốt

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2017.2018.2019}\)

\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2017.2018.2019}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2017.2018}-\frac{1}{2018.2019}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2018.2019}\right)\)

Đến đây tự làm tiếp :))) 

\(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{8\cdot9\cdot10}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{8\cdot9\cdot10}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{x}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}-\frac{1}{9\cdot10}\right)=\frac{22}{45}\)

\(\Rightarrow\frac{x}{2}\left(\frac{1}{2}-\frac{1}{90}\right)=\frac{22}{45}\)

\(\Rightarrow\frac{x}{2}\cdot\frac{22}{45}=\frac{22}{45}\)

\(\Rightarrow\frac{x}{2}=1\)

\(\Rightarrow x=2\)