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3) Ta có : \(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
4)
A = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
A = \(\frac{1}{2}.\left(1-\frac{1}{3}\right)+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+...+\frac{1}{2}.\left(\frac{1}{99}-\frac{1}{101}\right)\)
A = \(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
A = \(\frac{1}{2}.\left(1-\frac{1}{101}\right)\)
\(A=\frac{1}{2}.\frac{100}{101}\)
A = \(\frac{50}{101}\)
2, đặt tên biểu thức trên là A. Ta có :
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{10100}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\)
\(A=1-\frac{1}{101}\)
\(A=\frac{100}{101}\)
1) \(\frac{1}{1}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\)
\(=1-\frac{1}{5}\)
\(=\frac{4}{5}\)
Nhận thấy \(\)\(\dfrac{1}{1.1!}=1\); \(\dfrac{1}{2.2!}=\dfrac{1}{4}\)
Đặt \(P=\dfrac{1}{3.3!}+...+\dfrac{1}{2013.2013!}\)
\(P=\dfrac{1}{3.1.2.3}+...+\dfrac{1}{2013.1.2...2013}\)
\(P< \dfrac{1}{1.2.3}+...+\dfrac{1}{2011.2012.2013}\)
\(P< \dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+...+\dfrac{1}{2011.2012}-\dfrac{1}{2012.2013}\right)\)
\(P< \dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2012.2013}\right)=\dfrac{1}{4}-\dfrac{1}{2.2012.2013}\)
\(P< \dfrac{1}{4}\)
\(A< \dfrac{1}{4}+\dfrac{1}{4}+1=\dfrac{3}{2}\left(đpcm\right)\)
chung to :C = \(\frac{1}{1.1!}+\frac{1}{2.2!}+\frac{1}{3.3!}+...+\frac{1}{2019.2019!}< \frac{3}{2}\)
Thấy : \(\frac{1}{1.1!}=\frac{1}{1}\)
\(\frac{1}{2.2!}=\frac{1}{4}\)
\(\frac{1}{3.3!}< \frac{1}{1.2.3}\)( Vì 3.3! > 1.2.3 )
...
\(\frac{1}{2019.2019!}< \frac{1}{2017.2018.2019}\)( vì 2019.2019! < 2017.2018.2019)
Cộng từng vế có :
\(\frac{1}{3.3!}+\frac{1}{4.4!}+...+\frac{1}{2019.2019!}< \frac{1}{1.2.3}+...+\frac{1}{2017.2018.2019}\)
\(\Rightarrow\frac{1}{1.1!}+\frac{1}{2.2!}+...+\frac{1}{2019.2019!}< \frac{1}{1}+\frac{1}{4}+\frac{1}{1.2.3}+...+\frac{1}{2017.2018.2019}\)
\(\Rightarrow C< \frac{1}{1}+\frac{1}{4}+\left(\frac{1}{1.2}-\frac{1}{2.3}+...+\frac{1}{2017.2018}-\frac{1}{2018.2019}\right):2\)
\(\Rightarrow C< \frac{1}{1}+\frac{1}{4}+\left(\frac{1}{2}-\frac{1}{2018.2019}\right):2\)
\(\Rightarrow C< \frac{3}{2}-\frac{1}{2.2018.2019}\)
Vì \(\frac{1}{2.2018.2019}>0\Rightarrow C< \frac{3}{2}\)
c) \(C=1.2+2.3+3.4+...+98.99\)
\(\Rightarrow3C=1.2\left(3-0\right)+2.3\left(4-1\right)+3.4\left(5-2\right)+...+98.99\left(100-97\right)\)
\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+....+98.99.100-97.98.99\)
\(=98.99.100\)
\(\Rightarrow C=\frac{98.99.100}{3}=323400\)
d) \(D=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
Bài 2:
100! = 1.2.3.4. ................... .100