\(\text{S}\)= 1 - \(\frac{1}{2}\)+ \(\frac{1}{2}\)-\(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{4}\)+ .... + \(\frac{1}{99}\)- \(\frac{1}{100}\)

\(S\)= ( 1 - \(\frac{1}{100}\)) : 2

\(S\)= \(\frac{99}{100}\): 2 

\(S\)= \(\frac{99}{200}\)

tick nhé Lê Thiên Hương

99/200 dạng chuỗi mà bạn

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2010}-\frac{1}{2011}\)

\(=1-\frac{1}{2011}\)

\(=\frac{2010}{2011}\)

=1/1-1/2+1/2-1/3+1/3-1/4+...+1/2010-1/2011

= 1 - 1/2011

= 2010/ 2011

Đáp số: 2010/2011

Chúy ý công thức: \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)

= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2010 - 1/2011

= 1 - 1/2011

= 2010/2011

Đáp sô: 2010/2011

Chú ý công thưc:              \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)

1+1/1 - 1/1000 = tự tính nhé

\(\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{999x1000}+1\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}-\frac{1}{1000}+1\)

\(=2-\frac{1}{1000}=\frac{1999}{1000}\)

Ta có: 1/1x2 + 1/2x3 + 1/3x4 +...+ 1/X x (X + 1) = 499/500

=> 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +...+ 1/X - 1/(X + 1) = 499/500

=> 1 - 1/(X + 1) = 499/500

=>      1/(X + 1) = 1 - 499/500

=>      1/(X + 1) = 1/500

=>          X + 1 = 500

=>          X       = 500 - 1

=>          X       = 499 

Đáp số: X = 499

\(G=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)

\(G=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\)

\(3G=3+1+\frac{1}{3}+...+\frac{1}{3^4}\)

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

\(2G=3-\frac{1}{3^5}\)

\(2G=3-\frac{1}{243}\)

\(2G=\frac{729}{243}-\frac{1}{243}\)

\(G=\frac{728}{243}:2\)

\(G=\frac{364}{243}\)

\(\frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{x.\left(x+1\right)}=\frac{6042}{2015}\)

\(3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{6042}{2015}\)

\(1-\frac{1}{x+1}=\frac{6042}{2015}:3\)

\(1-\frac{1}{x-1}=\frac{2014}{2015}\)

\(\frac{1}{x-1}=1-\frac{2014}{2015}\)

\(\frac{1}{x-1}=\frac{1}{2015}\)

\(\Rightarrow x-1=2015\)

\(\Rightarrow x=2016\)

\(S=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\)

Áp dụng công thức : \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)

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

\(S=1-\frac{1}{100}=\frac{99}{100}\)

Dap an la 99/100.nho k cho minh.bai giai se gui sau

= 1/9900

đúng đấy

\(A=\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{7.8}\)

\(\Rightarrow5A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}\)

\(\Rightarrow5A=1.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{8}\right)\)

\(\Rightarrow5A=1-\frac{1}{8}\)

\(\Rightarrow A=\left(1-\frac{1}{8}\right).\frac{1}{5}=\frac{7}{40}\)

\(A=\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{7.8}\)

\(A=5\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{5}{7.8}\right)\)

\(A=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}\right)\)

\(A=5\left(1-\frac{1}{8}\right)\)

\(A=5.\frac{7}{8}\)

\(A=\frac{38}{8}\)

nhầm rồi :)

\(A=5.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-....-\frac{1}{8}\right)\)

\(\Rightarrow A=\frac{5.7}{8}\)

\(\Rightarrow A=\frac{35}{8}\)

S=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+...+\(\frac{1}{2009}\)-\(\frac{1}{2010}\)

S=1-\(\frac{1}{2010}\)

S=\(\frac{2009}{2010}\)

k nha bn

\(S=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{2008\times2009}+\frac{1}{2009\times2010}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2008}-\frac{1}{2009}+\frac{1}{2009}-\frac{1}{2010}\)

\(=1-\frac{1}{2010}\)

\(=\frac{2009}{2010}\)

Vậy \(S=\frac{2009}{2010}\)

Học tốt #

s = 1- 1/2+ 1/2- 1/3+ 1/3- 1/4 ......-1/2008- 1/2009+ 1/2009- 1/2010

s =1- 1/2010

s = 2009/2010

\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}\)\(=\frac{24}{50}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x.1}\)=\(\frac{24}{50}\)

=\(\frac{1}{2}-\frac{1}{x.1}=\frac{24}{50}\)

=\(\frac{1}{x.1}=\frac{1}{2}-\frac{24}{50}\)

=\(\frac{1}{x.1}=\frac{1}{50}\)

\(\Rightarrow\)\(x.1=50\)

\(\Rightarrow x=50\)