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Bài 5:
a) Ta có: \(A=1\cdot2+2\cdot3+3\cdot4+...+9\cdot10\)
\(\Leftrightarrow3\cdot A=3\cdot\left(1\cdot2+2\cdot3+3\cdot4+...+9\cdot10\right)\)
\(\Leftrightarrow3A=1\cdot2\cdot\left(3-0\right)+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+9\cdot10\cdot\left(11-8\right)\)
\(\Leftrightarrow3A=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+3\cdot4\cdot5-2\cdot3\cdot4+...+8\cdot9\cdot10-8\cdot9\cdot10+9\cdot10\cdot11\)
\(\Leftrightarrow3\cdot A=9\cdot10\cdot11=90\cdot11=990\)
hay A=330
Vậy: A=330
- \(B=\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{93.97}\)
\(4.B=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{93.97}\)
\(4.B=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{93}-\frac{1}{97}\)
\(4.B=1-\frac{1}{97}\)
\(4.B=\frac{96}{97}\)
\(B=\frac{96}{97}:4\)
\(B=\frac{24}{97}\)
b)
\(\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{32}{99}\)
c)
\(\frac{7}{3.4}+\frac{7}{4.5}+.....+\frac{7}{60.61}\)
\(=7\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{60}-\frac{1}{61}\right)\)
\(=7\left(\frac{1}{3}-\frac{1}{61}\right)\)
\(=\frac{406}{183}\)
d)
\(\frac{6}{2.4}+\frac{6}{4.6}+....+\frac{1}{72.74}\)
\(=3\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+.....+\frac{1}{72}-\frac{1}{74}\right)\)
\(=3\left(\frac{1}{2}-\frac{1}{74}\right)\)
=57/37
\(B=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+..+\frac{1}{55}\)
\(B=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{110}\)
\(B=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{10.11}\)
\(B=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(B=2.\left(\frac{1}{2}-\frac{1}{11}\right)=2.\frac{9}{22}=\frac{9}{11}\)
\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{49.50}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)
5,có thể bn sai đề chỗ \(\frac{1920}{111.2013}\),phải là \(\frac{1902}{111.2013}\)
\(F=\frac{3}{15.18}+\frac{11}{18.29}+\frac{12}{29.41}+\frac{70}{41.111}+\frac{1902}{111.2013}\)
\(=\frac{18-15}{15.18}+\frac{29-18}{18.29}+\frac{41-29}{12}+\frac{111-41}{41.111}+\frac{2013-111}{1902}\)
\(=\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{29}+\frac{1}{29}-\frac{1}{41}+\frac{1}{41}-\frac{1}{111}+\frac{1}{111}-\frac{1}{2013}\)
\(=\frac{1}{15}-\frac{1}{2013}=\frac{222}{3355}\)