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

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

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

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

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

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

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

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

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 )
\(3A=\frac{1}{6}-\frac{1}{8242206}=\frac{1373701}{8242206}-\frac{1}{8242206}=\frac{1373700}{8242206}\)
\(A=\frac{1373700}{8242206}:3=\frac{457900}{8242206}\)

Đặt A là biểu thức của đề bài.

Ta có: 3/ 1.2.3.4 = 1/ 1.2.3 -1/ 2.3.4

          3/ 2.3.4.5 = 1/ 2.3.4 -1/ 3.4.5

          3/ n(n+1)(n+2)(n+3) = 1/ n(n+1)(n+2) -1/ (n+1)(n+2)(n+3)

Do đó: 3A = 1/ 1.2.3 -1/ 2.3.4 + 1/ 2.3.4 - 1/ 3.4.5 +...+ 1/ n(n+1)(n+2) - 1/ (n+1)(n+2)(n+3)

3A = 1/ 1.2.3 - 1/ (n+1)(n+2)(n+3)

3A = 1/6 - 1/ (n+1)(n+2)(n+3)

A = 1/18 - 1/ 3(n+1)(n+2)(n+3)

Đó là kết quả rút gọn. Chúc bạn học tốt.

Đặt \(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{n.\left(n+1\right).\left(n+2\right).\left(n+3\right)}\)

\(\Rightarrow3A=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+\frac{3}{3.4.5.6}+...+\frac{3}{n.\left(n+1\right).\left(n+2\right).\left(n+3\right)}\)

\(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{n.\left(n+1\right).\left(n+2\right)}-\frac{1}{\left(n+1\right).\left(n+2\right).\left(n+3\right)}\)

\(3A=\frac{1}{1.2.3}-\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)

\(A=\frac{\frac{1}{1.2.3}-\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}}{3}\)

B tự làm nốt nhé

Bài này áp dụng công thức:

 \(\frac{a}{b.c.d.e}=\frac{1}{b.c.d}-\frac{1}{c.d.e}\)( đk: \(e-b=a\))

hãy k cho tui

tui ko k lại đâu

mại dô!!!!!

THANKS

=1/1.2-1/3.4+1/2.3-1/3.4+...+1/116.117-1/118.119

=1-1/2-1/3+1/4+1/2-1/3-1/3-1/4+...+1/116-1/117-1/118+1/119

=1+1/119=120/119(ko nhầm thì z)

\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)

\(3A=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+...+\frac{3}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)

\(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)

\(3A=\frac{1}{6}-\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)

\(3A=\frac{\left(n+1\right)\left(n+2\right)\left(n+3\right)-6}{6\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)

=>\(A=\frac{\left(n+1\right)\left(n+2\right)\left(n+3\right)-6}{18\left(n+1\right)\left(n+2\right)\left(n+3\right)}=\frac{n^3+3n^2+3n^2+9n+6-6}{18\left(n+1\right)\left(n+2\right)\left(n+3\right)}=\frac{n^3+6n^2+9n}{18\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)

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