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

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

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

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

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

Đặt \(A=\frac{1}{5}+\frac{1}{5^3}+...+\frac{1}{5^{101}}\)

\(\Rightarrow25A=5+\frac{1}{5}+\frac{1}{5^3}+...+\frac{1}{5^{99}}\)

\(\Rightarrow25A-A=\left(5+\frac{1}{5}+\frac{1}{5^3}+...+\frac{1}{5^{99}}\right)-\left(\frac{1}{5}+\frac{1}{5^3}+\frac{1}{5^5}+...+\frac{1}{5^{101}}\right)\)

hay \(24A=5-\frac{1}{5^{101}}\)

\(\Rightarrow A=\frac{5-\frac{1}{5^{101}}}{24}\)

\(\Rightarrow A:\left(1-\frac{1}{5^{102}}\right)=\frac{5-\frac{1}{5^{101}}}{24}.\frac{1}{1-\frac{1}{5^{102}}}\)

\(=\frac{5\left(1-\frac{1}{5^{102}}\right)}{24}.\frac{1}{1-\frac{1}{5^{102}}}=\frac{5}{24}\)

ai giúp mình với @@

Ta có : 

\(A=\frac{1}{2016.2015}+\frac{1}{2015.2014}+\frac{1}{2014.2013}+...+\frac{1}{2}\)

\(A=\frac{1}{1.2}+...+\frac{1}{2013.2014}+\frac{1}{2014.2015}+\frac{1}{2015.2016}\)

\(A=\frac{1}{1}-\frac{1}{2}+...+\frac{1}{2013}-\frac{1}{2014}+\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2015}-\frac{1}{2016}\)

\(A=1-\frac{1}{2016}\)

\(A=\frac{2015}{2016}\)

Vậy \(A=\frac{2015}{2016}\)

Chúc bạn học tốt ~ 

ai kb vs tôi ko

Thôi được rồi .

Giải:

\(P=\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{\left(4n-3\right)\left(4n+1\right)}\)

\(\Rightarrow4A=\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{\left(4n-3\right)\left(4n+1\right)}\)

            \(=\frac{5-1}{1.5}+\frac{9-5}{5.9}+...+\frac{\left(4n+1\right)-\left(4n-3\right)}{\left(4n-3\right)\left(4n+1\right)}\)

            \(=\left(\frac{1}{1}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{9}\right)+...+\frac{1}{4n-3}-\frac{1}{4n+1}\)

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

Vậy \(A=\frac{4n}{4n+1}\)

cảm ơn nha

\(P=\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+...+\frac{1}{\left(4n-3\right)\left(4n+1\right)}\)

\(4P=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+...+\frac{4}{\left(4n-3\right)\left(4n+1\right)}\)

\(4P=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{4n-3}-\frac{1}{4n+1}\)

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

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

\(4P=\frac{4n}{4n+1}\)

\(P=\frac{4n}{4n+1}\cdot\frac{1}{4}\)

\(P=\frac{n}{4n+1}\)

\(B=\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{41.45}\)

\(4B=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{41}-\frac{1}{45}\)

\(4B=\frac{44}{45}\)

\(B=\frac{11}{45}\)

\(B=\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{41.45}\)

\(=\frac{1}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{41.45}\right)\)

\(=\frac{1}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{41}-\frac{1}{45}\right)\)

\(=\frac{1}{4}.\left(1-\frac{1}{45}\right)\)

\(=\frac{1}{4}.\frac{44}{45}\)

\(=\frac{11}{45}\)

\(B=\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{41.45}\)

= \(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\)

= \(1-\frac{1}{45}\)

= \(\frac{45}{45}-\frac{1}{45}\)

= \(\frac{44}{45}\)

Chúc bạn học tốt !!!

A=8/1.5 + 8/5.9 + 8/9.13+ ... +8/25.29

A=2 . (2/1.5 +4/5.9 + 4/9.13 + ...... +4/25.29

A=2.(1-1/5+1/5-1/9+1/9-1/13+...+1/25-1/29

A=2.(1-1/29)

A=2. 28/29

A=56/29

mn giải chi tiết ra hộ mình nhé!

cứu mik với

\(A=\frac{3}{1}+\frac{3}{1+2}+\frac{3}{1+2+3}+...+\frac{3}{1+2+3+...+99+100}\)

\(=3+\frac{3}{\frac{\left(1+2\right).2}{2}}+\frac{3}{\frac{\left(1+3\right).3}{2}}+...+\frac{3}{\frac{\left(1+100\right).100}{2}}\)

\(=3+\frac{6}{2.3}+\frac{6}{3.4}+...+\frac{6}{100.101}=3+6.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\right)\)

\(=3+6.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)

\(=3+6.\left(\frac{1}{2}-\frac{1}{101}\right)=3+6.\frac{99}{202}=\frac{600}{101}\)

Tốt nhất bạn nên nói mấy bài đơn giản ik dạng nâng cao ko có cho thi đâu đừng lo

tham khảo câu hỏi của mik giống đó! có nguwoif trả lời dc hok 24 tik zui

\(C=\frac{2}{4.7}-\frac{3}{5.9}+\frac{2}{7.10}-\frac{3}{9.13}+...+\frac{2}{301.304}-\frac{3}{401.405}\)

\(C=\left(\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{301.304}\right)-\left(\frac{3}{5.9}+\frac{3}{9.13}+...+\frac{3}{401.405}\right)\)

\(C=\frac{2}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{301.304}\right)-\frac{3}{4}\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{401.405}\right)\)

\(C=\frac{2}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{301}-\frac{1}{304}\right)-\frac{3}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+..+\frac{1}{401}-\frac{1}{405}\right)\) \(C=\frac{2}{3}\left(\frac{1}{4}-\frac{1}{304}\right)-\frac{3}{4}\left(\frac{1}{5}-\frac{1}{405}\right)\)

\(C=\frac{25}{152}-\frac{4}{27}\)

\(C=\frac{67}{4104}\)

ra 67/1014

67/4104