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\(M=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(\Rightarrow M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow M=1-\frac{1}{100}\)
\(\Rightarrow M=\frac{100}{100}-\frac{1}{100}=\frac{99}{100}\)
\(b,N=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(\Rightarrow N=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
\(\Rightarrow N=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{97}-\frac{1}{99}\right)\)
\(\Rightarrow N=\frac{1}{2}.\left(1-\frac{1}{99}\right)=\frac{1}{2}.\frac{98}{99}\)
\(\Rightarrow N=\frac{1.98}{2.99}=\frac{49.2}{2.99}=\frac{49}{99}\)
\(a,M=1-\frac{1}{100}=\frac{99}{100}\)
\(b=2N=\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+...+\frac{2}{97x99}\)
\(=1-\frac{1}{99}=\frac{98}{99}\)
=>\(N=\frac{98}{99}:2=\frac{49}{99}\)
\(S=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\)
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
\(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\)
\(=\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+\frac{5-4}{4\times5}+\frac{6-5}{5\times6}\)
\(=\frac{3}{2\times3}-\frac{2}{2\times3}+\frac{4}{3\times4}-\frac{3}{3\times4}+\frac{5}{4\times5}-\frac{4}{4\times5}+\frac{6}{5\times6}-\frac{5}{5\times6}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=\frac{1}{2}-\frac{1}{6}\)
\(=\frac{1}{3}\)
2)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+...+\left(\frac{1}{98}-\frac{1}{99}\right)+\left(\frac{1}{99}-\frac{1}{100}\right)\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}+....+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}=\frac{1.50}{100}-\frac{1}{100}=\frac{50-1}{100}=\frac{49}{100}\)
\(\frac{2}{1x3}+\)\(\frac{2}{3x5}+\)\(\frac{2}{5x7}+\)\(\frac{2}{7x9}+\frac{2}{9x11}+\frac{2}{11x13}\)
= \(\frac{3-1}{1x3}+\frac{5-3}{3x5}+\frac{7-5}{5x7}+\frac{9-7}{7x9}+\frac{11-9}{9x11}\)\(+\frac{13-11}{11x13}\)
= \(\frac{3}{1x3}-\frac{1}{1x3}+\frac{5}{3x5}-\frac{3}{3x5}+\frac{7}{5x7}-\frac{5}{5x7}+\frac{9}{7x9}-\frac{7}{7x9}+\frac{11}{9x11}\)\(-\frac{9}{9x11}\)\(+\frac{13}{11x13}-\frac{11}{11x13}\)
= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\)\(\frac{1}{13}\)
= \(1-\frac{1}{13}=\frac{12}{13}\)
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{y\times\left(y+1\right)}=\frac{996}{997}\)
\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{y}-\frac{1}{y+1}=\frac{996}{997}\)
\(\Leftrightarrow1-\frac{1}{y+1}=\frac{996}{997}\)
\(\Leftrightarrow\frac{1}{y+1}=1-\frac{996}{997}=\frac{1}{997}\)
\(\Leftrightarrow y+1=997\Leftrightarrow y=996\)
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. Đ/S: 99/100
2. Đ/S: 38/123