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\(S=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\)
Áp dụng công thức : \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(S=1-\frac{1}{100}=\frac{99}{100}\)
=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/98-1/99+1/99-1/100
=1/1-1/100
=100/100-1/100
=99/100
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
= \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
= \(\frac{1}{1}-\frac{1}{100}\)
= \(\frac{99}{100}\)
~~~
#Sunrise
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{y\times\left(y+1\right)}=\frac{996}{997}\)
\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{y}-\frac{1}{y+1}=\frac{996}{997}\)
\(\Leftrightarrow1-\frac{1}{y+1}=\frac{996}{997}\)
\(\Leftrightarrow\frac{1}{y+1}=1-\frac{996}{997}=\frac{1}{997}\)
\(\Leftrightarrow y+1=997\Leftrightarrow y=996\)
Vậy y = 996
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 #
\(=1\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=1\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
\(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\)
\(=\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+\frac{5-4}{4\times5}+\frac{6-5}{5\times6}\)
\(=\frac{3}{2\times3}-\frac{2}{2\times3}+\frac{4}{3\times4}-\frac{3}{3\times4}+\frac{5}{4\times5}-\frac{4}{4\times5}+\frac{6}{5\times6}-\frac{5}{5\times6}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=\frac{1}{2}-\frac{1}{6}\)
\(=\frac{1}{3}\)
\(\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{999x1000}+1\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}-\frac{1}{1000}+1\)
\(=2-\frac{1}{1000}=\frac{1999}{1000}\)
1/1x2+1/2x3+....+1/2010x2011
=1-1/2+1/2-1/3+.......+1/2010-1/2011
=1-1/2011=2010/2011
ung ho va chon nha
Ta có:
\(A=\left(1-\frac{1}{1.2}\right)+\left(1-\frac{1}{2.3}\right)+....+\left(1-\frac{1}{2016.2017}\right)\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2016.2017}\right)\)
\(=2016-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}\right)\)
\(=2016-\left(1-\frac{1}{2017}\right)\)
\(=2016-\frac{2016}{2017}=\frac{4064256}{2017}\)
Vậy giá trị biểu thức là \(\frac{4064256}{2017}\)