\(\frac{49}{50}nha\)

F= \(\frac{1}{1.2.3}\)- \(\frac{1}{2.3.4}\)+ \(\frac{1}{2.3.4}\)- \(\frac{1}{3.4.5}\)+....+\(\frac{1}{47.48.49}\)- \(\frac{1}{48.49.50}\)

F=\(\frac{1}{1.2.3}\)- \(\frac{1}{48.49.50}\)

F=\(\frac{6533}{39200}\)

\(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{47.48.49.50}\)

\(=\frac{1}{3}\cdot\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{47.48.49}-\frac{1}{48.49.50}\right)\)

\(=\frac{1}{3}\cdot\left(\frac{1}{1.2.3}-\frac{1}{48.49.50}\right)\)

\(=\frac{1}{3}\cdot\frac{6533}{39200}=\frac{6533}{117600}\)

\(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{97.98.99.100}=\frac{1}{3}.\left(\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+...+\frac{3}{97.98.99.100}\right)=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{97.98.99}-\frac{1}{98.99.100}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{98.99.100}\right)=\frac{1}{3}.\left(\frac{1}{6}-\frac{1}{970200}\right)=\frac{1}{18}-\frac{1}{6.970200}\)

        \(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{97.98.99.100}\)

\(=\frac{1}{3}.\left(\frac{3}{1.2.3.4}+ \frac{3}{2.3.4.5}+...+\frac{3}{97.98.99.100}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{97.98.99}-\frac{1}{98.99.100}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{98.99.100}\right)\)

\(=\frac{1}{3}.\frac{161699}{970200}=\frac{161699}{299106000}\)

hai bạn trước đó gửi sai hết rùi. đúng theo sách NÂNG CAO PHÁT TRIỂN TOÁN 6 TẬP 2 thì bài này có đáp án thì bằng 1353/8120 nhé

Lại phải giải hết 
Gọi dãy số trên là A
\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+.....+\frac{1}{200.201.202.203}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-.....+\frac{1}{200.201.202}-\frac{1}{201.202.203}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{201.202.203}\)(chỗ này lm hơi tắt tí )
\(3A=\frac{1}{6}-\frac{1}{8242206}=\frac{1373701}{8242206}-\frac{1}{8242206}=\frac{1373700}{8242206}\)
\(A=\frac{1373700}{8242206}:3=\frac{457900}{8242206}\)