Ta có : \(-\frac{4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-.....-\frac{4}{\left(n+4\right)n}\)

\(=-\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+......+\frac{4}{n\left(4+n\right)}\right)\)

\(=-\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+......+\frac{1}{n}-\frac{1}{n+4}\right)\)

\(=-\left(1-\frac{1}{n+4}\right)\)

\(=-\left(\frac{n+4}{n+4}-\frac{1}{n+4}\right)\)

\(=-\frac{n+3}{n+4}\)

\(S=\frac{-4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-...-\frac{4}{\left(n-4\right).n}\)

\(=-\left(\frac{1}{1}-\frac{1}{5}\right)-\left(\frac{1}{5}-\frac{1}{9}\right)-\left(\frac{1}{9}-\frac{1}{13}\right)-...-\left(\frac{1}{n-4}-\frac{1}{n}\right)\)

\(=-\frac{1}{1}+\frac{1}{5}-\frac{1}{5}+\frac{1}{9}-\frac{1}{9}+\frac{1}{13}-...-\frac{1}{n-4}+\frac{1}{n}\)

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

M=\(\frac{4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-...-\frac{4}{\left(n-4\right).n}\)

\(M=1-\frac{1}{5}-\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{n-4}-\frac{1}{n}\right)\)

\(M=1-\frac{1}{5}-\frac{1}{5}+\frac{1}{n}\)

\(M=\frac{3}{5}+\frac{1}{n}\)

Mình chỉ giải đến đây thôi vì chẳng biết n bằng mấy cả

= - (1-1/5 +1/5 -1/9 +1/9 -1/13 +1/n + 1/n+4)

=-(1-1/n+4)

=-1+1/n+4

Phạm Thị Mai Anh ko có dấu( - )đằng trước đâu ạ

Ta có : 

\(M=-\frac{4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-...-\frac{4}{\left(n+4\right)n}\)

\(\Leftrightarrow\)\(M=-\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{n+4}-\frac{1}{n}\right)\)

\(\Leftrightarrow\)\(M=-\left(1-\frac{1}{n}\right)\)

\(\Leftrightarrow\)\(M=-\frac{n}{n}+\frac{1}{n}\)

\(\Leftrightarrow\)\(M=\frac{-n+1}{n}\)

Vậy \(M=\frac{-n+1}{n}\)

\(\text{Đề bài sai : }\frac{4}{\left(n-4\right)^n}->\frac{4}{\left(n-4\right)^n}\)

\(\text{Ta có :}\)

                                               \(S=\frac{4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-...-\frac{4}{\left(n-4\right)n}\)

                                                  \(=\left(\frac{1}{1}-\frac{1}{5}\right)-\left(\frac{1}{5}-\frac{1}{9}\right)-...-\left(\frac{1}{n-4}-\frac{1}{n}\right)\)

                                                  \(=\frac{1}{1}-\frac{1}{5}-\frac{1}{5}+\frac{1}{9}-...-\frac{1}{n-4}+\frac{1}{n}\)

                                                  \(=\frac{1}{1}-\frac{1}{5}-\frac{1}{5}+\frac{1}{n}\)

                                                  \(=\frac{3}{5}+\frac{1}{n}\)

                                                  \(=\frac{3}{5}+\frac{1}{n}\)

                                                  \(=\frac{3n+5}{5n}\)

\(\text{Vậy ...}\)

M = - \(\frac{4}{1.5}\) - \(\frac{4}{5.9}\) - ... - \(\frac{4}{n\left(n+4\right)}\)

    =  - (\(\frac{4}{1.5}\) + \(\frac{4}{5.9}\) + ... + \(\frac{4}{n\left(n+4\right)}\)

    = - ( 1 - \(\frac{1}{5}\) + \(\frac{1}{5}\) - \(\frac{1}{9}\) + ... + \(\)\(\frac{1}{n}\) - \(\frac{1}{n+4}\)

    = - ( 1 - \(\frac{1}{n+4}\))

    =  - \(\frac{n+3}{n+4}\)

1 Tính : 

a) \(A=\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-...-\frac{1}{\left(n-1\right).n}\)

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

\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)

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

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

\(=\frac{1}{n}\)

b) \(B=\frac{4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-...-\frac{4}{\left(n-4\right).n}\)

\(=\frac{4}{1.5}-\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{\left(n-4\right).n}\right)\)

\(=\frac{4}{5}-\left(\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{\left(n-4\right).n}\right)\)

\(=\frac{4}{5}-\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{n-4}-\frac{1}{n}\right)\)

\(=\frac{4}{5}-\left(\frac{1}{5}-\frac{1}{n}\right)\)

\(=\frac{4}{5}-\frac{1}{5}+\frac{1}{n}\)

\(=\frac{3}{5}+\frac{1}{n}\)

c) \(C=1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{10}}\)

\(=1-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)

Đặt \(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)

\(\Rightarrow C=1-B\left(1\right)\)

\(\Rightarrow2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)

Lấy 2B trừ B ta có : 

\(2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)

\(B=1-\frac{1}{2^{10}}\left(2\right)\)

Thay (2) vào (1) ta có :

\(C=1-\left(1-\frac{1}{10}\right)\)

\(=1-1+\frac{1}{10}\)

\(=\frac{1}{10}\)

Vậy \(C=\frac{1}{10}\)

Trả lời :

\(E=-\left(\frac{4}{1\times5}+\frac{4}{5\times9}+\frac{4}{9\times13}+...+\frac{4}{n\left(n+4\right)}\right)\)

\(\Rightarrow E=-\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{n}-\frac{1}{n+4}\right)\)

\(\Rightarrow E=-\left(1-\frac{1}{n+4}\right)\)

\(\Rightarrow E=1+\frac{1}{n+4}\)

P/s : Sai thì thông cảm nha chị. Dạng này lâu chưa làm nên không nhớ rõ.

\(E=-\frac{4}{1.5}-\frac{4}{5.9}-\frac{4}{9.11}-...-\frac{4}{\left(n-4\right)n}\)

\(\Rightarrow E=-\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.11}+...+\frac{4}{\left(n-4\right)n}\right)\)

\(\Rightarrow E=-\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{n-4}-\frac{1}{n}\right)\)

\(\Rightarrow E=-\left(1-\frac{1}{n}\right)\)

\(\Rightarrow E=-1+\frac{1}{n}\)

cảm ơn các bạn rất nhìu