1/28+1/70+1/130+1/208+...+1/3190
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\(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+\frac{1}{208}\)
\(=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}\)
\(=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}\right)\)
\(=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\right)\)
\(=\frac{1}{3}.\left(1-\frac{1}{16}\right)\)
\(=\frac{1}{3}.\frac{15}{16}=\frac{5}{16}\)
\(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+\frac{1}{208}\)
\(=\frac{1}{1\times4}+\frac{1}{4\times7}+\frac{1}{7\times10}+\frac{1}{10\times13}+\frac{1}{13\times16}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\)
\(=1-\frac{1}{16}=\frac{15}{16}\)
Chúc bạn học tốt ^^
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1/2A=1/2(6/4+6/28+6/70+6/130+6/208)
= 3/4+3/28+3/70+3/130+3/208
= 1-1/4+1/4-1/7+..................-1/16
=1-1/16
=15/16 => A=15/8
\(A=\dfrac{2}{1x3}+\dfrac{2}{3x5}+\dfrac{2}{5x7}+...+\dfrac{2}{21x23}\)
\(A=2x\left(\dfrac{1}{1x3}+\dfrac{1}{3x5}+\dfrac{1}{5x7}+...+\dfrac{1}{21x23}\right)\)
\(A=2x\dfrac{1}{2}x\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{21}-\dfrac{1}{23}\right)\)
\(A=1-\dfrac{1}{23}\)
\(A=\dfrac{22}{23}\)
\(B=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)
\(B=\dfrac{1}{2x3}+\dfrac{1}{3x4}+\dfrac{1}{4x5}+\dfrac{1}{5x6}+\dfrac{1}{6x7}+\dfrac{1}{7x8}+\dfrac{1}{8x9}+\dfrac{1}{9x10}\)
\(B=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(B=\dfrac{1}{2}-\dfrac{1}{10}\)
\(B=\dfrac{5}{10}-\dfrac{1}{10}\)
\(B=\dfrac{4}{10}\)
\(B=\dfrac{2}{5}\)
\(3M=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\)
\(3M=\frac{4-1}{1.4}+\frac{7-4}{4.7}+...+\frac{100-97}{97.100}\)
\(3M=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\)
\(3M=1-\frac{1}{100}\)
\(3M=\frac{99}{100}\)
\(M=\frac{33}{100}\)
Gọi tổng là A
⇒ A = \(\dfrac{1}{28}+\dfrac{1}{70}+\dfrac{1}{130}+\dfrac{1}{208}+...+\dfrac{1}{3190}\)
⇒ 3A = \(3\left(\dfrac{1}{28}+\dfrac{1}{70}+\dfrac{1}{130}+\dfrac{1}{208}+...+\dfrac{1}{3190}\right)\)
⇒ 3A = \(\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+\dfrac{3}{13.16}+...+\dfrac{3}{55.58}\)
⇒ 3A = \(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}+...+\dfrac{1}{55}-\dfrac{1}{58}\)
⇒ 3A = \(\dfrac{1}{4}-\dfrac{1}{58}\) = \(\dfrac{29}{116}-\dfrac{2}{116}\) = \(\dfrac{27}{116}\)
⇒ A = \(\dfrac{27}{116}\): 3 = \(\dfrac{27}{116}\).\(\dfrac{1}{3}\) = \(\dfrac{9}{116}\)