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\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{14\cdot15\cdot16}\)
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KT
31 tháng 7 2018
\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+....+\frac{1}{47.48.49.50}\)
\(=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{47.48.49}-\frac{1}{48.49.50}\right)\)
\(=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{48.49.50}\right)\)
\(=\frac{1}{3}.\frac{6533}{39200}=\frac{6533}{117600}\)
7 tháng 2 2017
\(A=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{7-2}{2.7}+\frac{11-7}{7.11}+\frac{14-11}{11.4}+\frac{15-14}{14.15}+\frac{28-15}{15.28}\)
\(\frac{A}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)
\(A=7.\frac{13}{28}\)
\(A=\frac{13}{4}\)
7 tháng 7 2017
\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{27.28.29.30}\)
\(A=\frac{1}{4.6}+\frac{1}{10.12}+\frac{1}{18.20}+...+\frac{1}{810.812}\)
.......
~ Chúc học tốt ~
Ai ngang qua xin để lại 1 L - I - K - E
N
7 tháng 7 2017
\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+.....+\frac{1}{27.28.29.30}\)
\(3A=3.\left(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+......+\frac{1}{27.28.29.30}\right)\)
\(3A=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+..........+\frac{3}{27.28.29.30}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+........+\frac{1}{27.28.29}-\frac{1}{28.29.30}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{28.29.30}\)
\(3A=\frac{1}{6}-\frac{1}{24360}\)
\(3A=\frac{1353}{8120}\)
\(A=\frac{1353}{8120}:3\)
\(A=\frac{451}{8120}\)
19 tháng 9 2016
\(M=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{37.38.39}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+..+\frac{1}{37.38}-\frac{1}{38.39}\)
\(=\frac{1}{1.2}-\frac{1}{38.39}=\frac{1}{1}-\frac{1}{2}-\frac{1}{38}+\frac{1}{39}=\frac{370}{741}\)
26 tháng 2 2018
Ta có 1/1.2-1/2.3=2/1.2.3;1/2.3-1/3.4=2/2.3.4 .....1/98.99-1/99.100=2/98.99.100 2A=2/1.2.3+2/2.3.4+....+2/98.99.100 = 1/1.2-1/2.3+1/2.3-1/3.4+...+1/98.99-1/99.100 = 1/2-1/99.100 = 4949/9900 A =4949/19800
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{14.15.16}\)
\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{14.15.16}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{14.15}-\frac{1}{15.16}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{15.16}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{240}\right)\)
\(=\frac{1}{2}.\frac{119}{240}\)
\(=\frac{119}{480}\)
Bài làm:
Ta có:\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{14.15.16}\)
\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{14.15.16}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{14.15}-\frac{1}{15.16}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{15.16}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{240}\right)\)
\(=\frac{1}{2}.\frac{119}{240}=\frac{119}{480}\)