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a: \(A=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^7\)
=>\(2\cdot A=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^6\)
=>\(2A-A=1-\left(\dfrac{1}{2}\right)^7=1-\dfrac{1}{128}=\dfrac{127}{128}\)
=>\(A=\dfrac{127}{128}\)
b: \(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{10\cdot11}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{10}-\dfrac{1}{11}\)
\(=1-\dfrac{1}{11}=\dfrac{10}{11}\)
1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56
= 1/2x3 + 1/3x4 + 1/4x5 + 1/5x6 + 1/6x7 + 1/7X8
=1/2 - 1/3 + 1/3 -1/4 + 1/4 - 1/5 + 1/5 -1/6 + 1/6 - 1/7 + 1/7 - 1/8
= 1/2 - 1/8
= 4/8 - 1/8
= 3/8
A=1/20+1/30+1/42+1/56+1/72+1/90
A=1/4x5+1/5x6+1/6x7+1/7x8+1/8x9+1/9x10
A=1/4-1/5+1/5-1/6+...+1/9-1/10
A=1/4-1/10
A=3/20
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{5.4}+\frac{6-5}{5.6}+\frac{7-6}{7.6}+\frac{8-7}{7.8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{2}-\frac{1}{8}=\frac{4}{8}-\frac{1}{8}=\frac{3}{8}\)
A=1/20+1/30+1/42+1/56+1/72+1/90
A=1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10
A=1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10
A=1/4-1/10
A=3/20
1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90
=1/(1.2)+1/(2.3)+1/(3.4)+1/(4.5)
+1/(5.6)+1/(6.7)+1/(7.8)
+1/(8.9)+1/(9.10)
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5.+1/5-1/6...
+1/9-1/10
=1-1/10
=9/10
1/6+1/12+1/20+1/30+1/42+1/56
=1/2x3+1/3x4+1/4x5+1/5x6+1/6x7+1/7x8
=1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8
=1/2-1/8
=3/8
\(=\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{11.12}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{2}-\frac{1}{12}\)
\(=\frac{5}{12}\)
bn sẽ tinh theo kieeuranhaan 2 nha xin lỗi mik làm bi này rùi nhưng mik quên mik có sacks xem lại
\(\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
\(=\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)
\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}=\dfrac{1}{5}-\dfrac{1}{8}=\dfrac{8-5}{40}=\dfrac{3}{40}\)