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\(\frac{1}{50\cdot51}+\frac{1}{51\cdot52}+\frac{1}{52\cdot53}+...+\frac{1}{99\cdot100}\)tính kết quả
Đặt A = \(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+.....+\frac{3}{99.100}\)
\(\frac{1}{3}A\)\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)
\(\frac{1}{3}A\)\(=1-\frac{1}{100}\)
=> \(\frac{1}{3}A=\frac{99}{100}\)
=> A = \(\frac{99}{100}.3=\frac{297}{100}\)
\(\frac{3}{1.2}+\frac{3}{2.3}+..................+\frac{3}{99.100}\)
\(=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+..................+\frac{1}{99.100}\right)\)
\(=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.................+\frac{1}{99}-\frac{1}{100}\right)\)
\(=3.\left(1-\frac{1}{100}\right)\)
\(=3.\frac{99}{100}\)
\(=\frac{297}{100}\)
Đặt \(D=\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\)
=>\(2D=\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}\)
=>\(2D=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\)
=>\(2D=\frac{1}{2}-\frac{1}{100}\)
=>\(2D=\frac{49}{100}\)
=>\(D=\frac{49}{50}\)
a) \(C=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)
\(=7\left(\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\right)\)
\(=7\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\right)\)
\(=7\left(\frac{1}{2}-\frac{1}{28}\right)\)
\(=7.\frac{13}{28}=\frac{7.13}{28}=\frac{13}{4}\)
b) \(B=\frac{6}{3.5}+\frac{6}{5.7}+\frac{6}{7.9}+...+\frac{6}{97.99}\)
\(=3\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\right)\)
\(=3\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=3\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=3.\frac{32}{99}=\frac{3.32}{99}=\frac{32}{33}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2014.2015}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-...+\frac{1}{2014}-\frac{1}{2015}=1-\frac{1}{2015}=\frac{2014}{2015}\)
\(B=\frac{2}{11.13}+\frac{2}{13.15}+...+\frac{2}{97.99}=\frac{2}{11}-\frac{2}{13}+...+\frac{2}{97}-\frac{2}{99}=\frac{2}{11}-\frac{2}{99}=\frac{16}{99}\)
C=(1x2)+(2x3)+...+(99x100)
3xC=1x2x3+2x3x3+...+3x99x100
3xC=1x2x3-0x1x2+2x3x4-1x2x3+...+99x100x101-98x99x100
3xC=99x100x101
3xC=999900
C=333300
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}\)
\(=1-\frac{1}{6}\)
\(=\frac{5}{6}\)
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\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}=\)
\(\frac{1}{1}-\frac{1}{6}=\frac{5}{6}\)