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\(1+\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=2-\frac{1}{100}\)
\(=\frac{199}{100}\)
Gọi biểu thức là A
A=1+1/2+1/2.3+1/3.4+...+1/98.99+1/99.100
A-1=1/2+1/2.3+1/3.4+...+1/98.99+1/99.100
A-1=1-1/2+1/2-1/3+1/3-1/4+...+/198-1/99+1/99-1/100
A-1=1-1/100
A-1=99/100
A=99/100+1
A=199/100
\(A=1+\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}+\frac{1}{100}\)
\(\Rightarrow A=1+\frac{1}{2}+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)+\frac{1}{100}\)
\(\Rightarrow A=1+\frac{1}{2}+\frac{1}{2}-\frac{1}{100}+\frac{1}{100}\)
\(\Rightarrow A=1+1\)
\(\Rightarrow A=2\)
Vậy A = 2
\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(B=1-\frac{1}{100}\)
\(B=\frac{99}{100}\)
Ta thấy: \(1-\frac{2}{n.\left(n+1\right)}=\frac{n.\left(n+1\right)-2}{n.\left(n+1\right)}=\frac{n^2+n-1-1}{n.\left(n+1\right)}=\frac{\left(n^2-1\right)+\left(n-1\right)}{n.\left(n+1\right)}=\frac{\left(n-1\right).\left(n+1\right)+\left(n-1\right)}{n.\left(n+1\right)}=\frac{\left(n-1\right).\left(n+2\right)}{n.\left(n+1\right)}\)
Lại có: \(\left(1-\frac{2}{2.3}\right).\left(1-\frac{2}{3.4}\right).\left(1-\frac{2}{4.5}\right).....\left(1-\frac{2}{99.100}\right)\)
\(=\left(1-\frac{2}{2.\left(2+1\right)}\right).\left(1-\frac{2}{3.\left(3+1\right)}\right).\left(1-\frac{2}{4.\left(4+1\right)}\right).....\left(1-\frac{2}{99.\left(99+1\right)}\right)\)
\(=\frac{\left(2-1\right).\left(2+2\right)}{2.\left(2+1\right)}.\frac{\left(3-1\right).\left(3+2\right)}{3.\left(3+1\right)}.\frac{\left(4-1\right).\left(4+2\right)}{4.\left(4+1\right)}.....\frac{\left(99-1\right).\left(99+2\right)}{99.\left(99+1\right)}\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.....\frac{98.101}{99.100}\)
\(=\frac{1.4.2.5.3.6.....98.101}{2.3.3.4.4.5.....99.100}\)
\(=\frac{\left(1.2.3.....98\right).\left(4.5.6.....101\right)}{\left(2.3.4.....99\right).\left(3.4.5.....100\right)}\)
\(=\frac{1.101}{99.3}\)
\(=\frac{101}{297}\)
\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)-\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{99.100.101}\right)\)
\(=\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)
\(=\left(1-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)\)
\(=\frac{99}{100}-\frac{1}{2}\cdot\frac{5049}{10100}=\frac{99}{100}-\frac{5049}{20200}=\frac{14949}{20200}\)
đặt A = 1/ 2.3 +.....................................+1
ta có A = ................(viết lại biểu thức )
A = 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ......+ 1/999 - 1/ 1000 + 1
= 1/2 - 1/1000 + 1
= 499/ 1000 + 1
= 1499/ 1000
1/2.3 +1/3.4+...+1/999.100 +1
=1/2-1/3+1/3-1/4+...+1/999-1/100 +1
=1/2-1/100+1
=49/100+1
=149/100
\(\frac{2}{1.2}+\frac{2}{2.3}+..........+\frac{2}{x\left(x+1\right)}=1\frac{2013}{2015}\)
\(\Rightarrow2\left(\frac{1}{1.2}+\frac{1}{2.3}+........+\frac{1}{x\left(x+1\right)}\right)=\frac{4028}{2015}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..........+\frac{1}{x}-\frac{1}{x+1}=\frac{4028}{2015}:2\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{2014}{2015}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{2014}{2015}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2015}\)
\(\Rightarrow x+1=2015\Rightarrow x=2014\)
\(\frac{2}{1\times2}+\frac{2}{2\times3}+\frac{2}{3\times4}+...+\frac{2}{x\left(x+1\right)}=1\frac{2013}{2015}\)
\(2\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{x\times\left(x+1\right)}\right)=1\frac{2013}{2015}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=1\frac{2013}{2015}\div2\)
\(1-\frac{1}{x+1}=\frac{2014}{2015}\)
\(\frac{1}{x+1}=1-\frac{2014}{2015}\)
\(\frac{1}{x+1}=\frac{1}{2015}\)
\(x+1=2015\)
\(x=2015-1\)
\(x=2014\)
\(A-1=\frac{1}{1.2}+\frac{1}{2.3}..+\frac{1}{99.100}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}\)\(=\frac{99}{100}\)
\(A=1+\frac{99}{100}=\frac{199}{100}\)
=1+1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/98-1/99+1/99-1/100
=1+1/2+1/2-1/100
=199/100