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

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

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

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

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

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

A= \(\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+\frac{1}{3.4.5}-\frac{1}{4.5.6}+...+\frac{1}{19.20.21}-\frac{1}{20.21.22}\right)\)

=\(\frac{1}{3}\left(\frac{1}{6}-\frac{1}{9240}\right)\)

=\(\frac{171}{3080}\)

A=1/1-1/2-1/3+1/2-1/3-1/4-1/5+1/3-1/4-1/5-1/6+...+1/19-1/20-1/21-1/22

A=1/1-1/22

A=21/22

  Vậy A=21/22

Biến đổi ở phân số dạng tổng quát :

\(\frac{1}{n(n+1)(n+2)(n+3)}=\frac{3}{3n(n+1)(n+2)(n+3)}=\frac{3+n-n}{3n(n+1)(n+2)(n+3)}\)

\(=\frac{1}{3}\left[\frac{n+3}{n(n+1)(n+2)(n+3)}-\frac{n}{n(n+1)(n+2)(n+3)}\right]\)

\(=\frac{1}{3}\left[\frac{1}{n(n+1)(n+2)}-\frac{1}{(n+1)(n+2)(n+3)}\right]\)

Áp dụng kết quả này vào bài được :

\(\frac{1}{1\cdot2\cdot3\cdot4}=\frac{1}{3}\left[\frac{1}{1\cdot2\cdot3}-\frac{1}{2\cdot3\cdot4}\right],\frac{1}{2\cdot3\cdot4\cdot5}=\frac{1}{3}\left[\frac{1}{2\cdot3\cdot4}-\frac{1}{3\cdot4\cdot5}\right],...\)

\(\frac{1}{n(n+1)(n+2)(n+3)}=\frac{1}{3}\left[\frac{1}{n(n+1)(n+2)}-\frac{1}{(n+1)(n+2)(n+3)}\right]\)

Cộng từng vế,ta được : \(S=\frac{1}{3}\left[\frac{1}{1\cdot2\cdot3}-\frac{1}{(n+1)(n+2)(n+3)}\right]\)

P/S : Xong

Ta có: S= \(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)

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

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

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

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

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

Ta có: \(S=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)

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

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

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

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

chỗ phân số thiếu tử thì điền tử bằng 1 nha

dùng sai phân cuối cùng ra:

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

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

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

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

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

P/S:  tham khảo nha

Đến đây bn thu gọn và tính tiếp nhé

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

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

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

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

Biến đổi phân số ở dạng tổng quát:

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

\(=\frac{1}{3}\left[\frac{n+3}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)\left(n+2\right)}\right]\)

=\(\frac{1}{3}\left[\frac{1}{n\left(n+1\right)\left(n+2\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}\right]\)

Áp dụng kết quả vào bài, ta được:

\(\frac{1}{1.2.3.4}=\frac{1}{3}\left[\frac{1}{1.2.3}-\frac{1}{2.3.4}\right],\frac{1}{2.3.4.5}=\frac{1}{3}\left[\frac{1}{2.3.4}-\frac{1}{3.4.5}\right]\),...

\(\frac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}=\frac{1}{3}\left[\frac{1}{n\left(n+1\right)\left(n+2\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}\right]\)

Cộng từng vế, ta được:

\(S=\frac{1}{3}\left[\frac{1}{1.2.3}-\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}\right].\)

\(\frac{150}{5.8}+\frac{150}{8.11}+\frac{150}{11.14}+.....+\frac{150}{47.50}\)

\(=50.\left(\frac{3}{5.8}+\frac{5}{8.11}+.....+\frac{3}{47.50}\right)\)

\(=50.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{47}-\frac{1}{50}\right)\)

\(=50.\left(\frac{1}{5}-\frac{1}{50}\right)\)

\(=50.\frac{9}{50}=9\)