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\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+....+\dfrac{1}{99\cdot100}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=1+\left(-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}\right)-\dfrac{1}{100}\)
\(A=1+0-\dfrac{1}{100}\)
\(A=1-\dfrac{1}{100}< 1\)
\(\Rightarrow A< 1\)
= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100
= 1 - 1/100
= 99/100
1/1.2 +1/2.3 +1/3.4 +...+1/98.99 +1/99.100
=1-1/2+1/2-1/3+1/3-1/4+...+1/98-1/99+1/99-1/100
=1-1/100=100/100-1/100=99/100
Ta có: \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow1-\frac{1}{100}=\frac{99}{100}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{100.101}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-....-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{101}{101}-\frac{1}{101}=\frac{100}{101}\)
A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{100}\)\(\frac{1}{100}\)
A = \(1-\frac{1}{100}\)
A = \(\frac{100}{100}-\frac{1}{100}\)
A = \(\frac{99}{100}\)
Đặt A = 1/1x2 + 1/2x3 + 1/3x4 + .... + 1/99x100
=> A = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + .... + 1/99 - 1/100
=> A = 1 - 1/100
=> A = 99/100
\(M=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{99\times100}\)
\(M=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{99}-\frac{1}{100}\right)\)
\(M=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(M=\frac{1}{1}-\frac{1}{100}\)
\(M=\frac{100}{100}-\frac{1}{100}\)
\(M=\frac{99}{100}\)
\(M=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{99\times100}\)
\(M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{100}-\frac{1}{100}\)
\(M=1-\frac{1}{100}\)
\(M=\frac{99}{100}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
=\(1-\frac{1}{100}\)
=\(\frac{99}{100}\)