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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(\Rightarrow x-1=2015\)

\(\Rightarrow x=2016\)

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

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{499}{500}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{499}{500}\)

\(1-\frac{1}{x+1}=\frac{499}{500}\)

\(\frac{1}{x+1}=1-\frac{499}{500}=\frac{1}{500}\)

=> x + 1 = 500

=> x = 500 - 1

=> x = 499

Vậy x = 499

1/1.2 + 1/2.3 + 1/3.4 +...+ 1/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

=>x/(x+1)=499/500

=>x=499

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{15.16}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{15}-\frac{1}{16}\)

\(=1-\frac{1}{16}=\frac{15}{16}\)

\(\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{15x16}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{15}-\frac{1}{16}\)

\(=1-\frac{1}{16}\)

\(=\frac{15}{16}\)

Cảm ơn nhé

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

Đặt \(M=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(M=1-\frac{1}{100}\)

\(M=\frac{99}{100}\)

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+....+\frac{1}{99\times100}\)

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

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

\(=\frac{99}{100}\)

\(\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{n\times\left(n+1\right)}=\frac{49}{100}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{49}{100}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{n+1}=\frac{49}{100}\)

\(\Rightarrow\frac{n+1-2}{2\left(n+1\right)}=\frac{49}{100}\)

\(\Rightarrow\frac{n-1}{2n+2}=\frac{49}{100}\)

\(\Rightarrow100\left(n-1\right)=49\left(2n+2\right)\)

\(\Rightarrow100n-100=98n+98\)

\(\Rightarrow2n=198\)

=> n = 99

Vậy n =  99

\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+....+\(\frac{1}{n}\)-\(\frac{1}{n+1}\)=\(\frac{49}{100}\)

\(\frac{1}{2}\)-\(\frac{1}{n+1}\)=\(\frac{49}{100}\)

         \(\frac{1}{n+1}\)=\(\frac{1}{2}\)-\(\frac{49}{100}\)

          \(\frac{1}{n+1}\)=\(\frac{1}{100}\)

=> n+1=100

        n=100-1

       n=99

\(\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{nx\left(n+1\right)}=\frac{49}{100}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{49}{100}\)

\(\frac{1}{2}-\frac{1}{n+1}=\frac{49}{100}\)

\(\frac{1}{n+1}=\frac{1}{2}-\frac{49}{100}\)

\(\frac{1}{n+1}=\frac{1}{100}\)

n + 1 = 100

n = 100 - 1

n = 99

Bn k cho mik nhé!

\(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)

\(=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\right)\)

\(=1+\left(1-\frac{1}{2018}\right)\)

\(=1+\left(\frac{2018}{2018}-\frac{1}{2018}\right)\)

\(=1+\left(\frac{2017}{2018}\right)\)

\(=\frac{2018}{2018}+\frac{2017}{2018}=\frac{4035}{2018}\)

\(1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}...+\frac{1}{2017\cdot2018}\)

\(=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}...+\frac{1}{2017}-\frac{1}{2018}\right)\)

\(=1+\left(1-\frac{1}{2018}\right)\)

\(=1+\frac{2017}{2018}\)

\(=1+\frac{2017}{2018}\)

\(=\frac{4035}{2018}\)

\(1+\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{2017x2018}\)

\(=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\right)\)

\(=1+\left(1-\frac{1}{2018}\right)\)

\(=1+\frac{2017}{2018}\)

\(=\frac{4035}{2018}\)

1/1x2+1/2x3+...+1/49x50

=1-1/2+1/2-1/3+.....+1/49-1/50

=1-1/50(1)

Ta co   1(2)

So sanh (1) voi (2) ta thay 1-1/50<1

=>1/1x2+...+1/49x50<1

(Phuong phap khu)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}\)

=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}\)

=\(\frac{1}{1}-\frac{1}{50}=\frac{50}{50}-\frac{1}{50}=\frac{49}{50}<1\)

Vậy \(\frac{49}{50}<1\)