Ta có:

\(\frac{1}{20.21}+\frac{1}{21.22}+\frac{1}{22.23}+...+\frac{1}{60.61}\)

\(=\frac{1}{20}-\frac{1}{21}+\frac{1}{21}-\frac{1}{22}+\frac{1}{22}-\frac{1}{23}+...+\frac{1}{60}-\frac{1}{61}\)

\(=\frac{1}{2}-\frac{1}{61}=\frac{59}{122}\)

b) \(\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{45.49}\)

\(=\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+...+\frac{1}{45.49}\)

\(=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{45}-\frac{1}{49}\)

\(=\frac{1}{5}-\frac{1}{49}=\frac{44}{245}\)

Bn Tấn sai rùi

phần a , câu cuối là \(\frac{1}{20}\)chứ đâu phải \(\frac{1}{2}\)

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

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

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

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

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

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

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

\(S=\frac{5-1}{1.5}+\frac{9-5}{5.9}+\frac{13-9}{9.13}+..+\frac{2005-2001}{2001.2005}\)

\(=\left(1-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{9}\right)+\left(\frac{1}{9}-\frac{1}{13}\right)+...+\left(\frac{1}{2001}-\frac{1}{2005}\right)\)

\(=1+\left(-\frac{1}{5}+\frac{1}{5}\right)+\left(-\frac{1}{9}+\frac{1}{9}\right)+...+\left(-\frac{1}{2001}+\frac{1}{2001}\right)-\frac{1}{2005}\)

\(=1-\frac{1}{2005}\)

\(=\frac{2004}{2005}\)

a) \(=\frac{9}{1.4}+\frac{9}{4.7}+\frac{9}{7.10}+...+\frac{9}{61.64}\)

\(=3\left(\frac{1}{1}-\frac{1}{64}\right)\)

\(=\frac{189}{64}\)

b) \(=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{21}-\frac{1}{25}\)

\(=\frac{1}{1}-\frac{1}{25}\)

\(=\frac{24}{25}\)

c) Chưa học tới

b)1/1.5+1/5.9+1/9.13+...+1/21.25

=1/4.(4/1.5+4/5.9+4/9.13+4/21.25)

=1/4.(4-4/5+4/5-4/9+4/9-4/13+...+4/21-4/25)

=1/4.(4-4/25)

=1/4.(100/25-4/25)

=1/4.96/25

=24/25

M = - ( 4/1.5 + 4/5.9 + ..................+ 4/(n-4).n

M = - (1-1/5 + 1/5 - 1/9 +..............+1/(n-4) - 1/n

M = -(1-1/n)

M = -1 + 1/n 

M = -n + 1

xie xie bn nha

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 \(C=\frac{1}{10}\)

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

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

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

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