4\3.5 .6\5.7......2016\2015.2017
Cm tổng trên >1\6
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AL
0
30 tháng 1 2022
Câu 2:
\(D=\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{100}{101}=\dfrac{150}{101}\)
Câu 3:
\(E=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{205}-\dfrac{1}{207}\right)\)
\(=2\cdot\left(1-\dfrac{1}{207}\right)=2\cdot\dfrac{206}{207}=\dfrac{412}{207}\)
Câu 5:
\(G=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{17}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{16}{17}=\dfrac{4}{17}\)
SB
1
Z
6 tháng 4 2017
\(\sum\limits^{2016}_{x=1}\left(\dfrac{x^2}{\left(2x-1\right)\left(2x+1\right)}\right)\)
H
5 tháng 4 2016
Mk bik câu B nè!
2B = 2/3.5 + 2/5.7 + 2/7.9 +.......+2/97.99
2B = 1/3 - 1/5 + 1/5 - 1/7 +.......+ 1/97 - 1/99
2B = 1/3 - 1/99
2B = 32/99
=> B = 16/99
S
2 giờ trước (15:44)
\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2025\cdot2016}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2025}-\dfrac{1}{2026}\\ =\dfrac{1}{1}-\dfrac{1}{2026}=\dfrac{2025}{2026}\\ B=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{99\cdot101}\\ =\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\\ =\dfrac{1}{1}-\dfrac{1}{101}=\dfrac{100}{101}\\ C=\dfrac{5}{1\cdot3}+\dfrac{5}{3\cdot5}+\dfrac{5}{5\cdot7}+...+\dfrac{5}{99\cdot101}\\ =\dfrac{5}{2}\cdot\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\ =\dfrac{5}{2}\cdot\left(\dfrac{1}{1}-\dfrac{1}{101}\right)=\dfrac{5}{2}\cdot\dfrac{100}{101}=\dfrac{250}{101}\)
\(D=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{9900}\\ =\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{1}-\dfrac{1}{100}=\dfrac{99}{100}\)
TT
2
10 tháng 8 2016
\(\frac{2016}{1.3}+\frac{2016}{3.5}+\frac{2016}{5.7}+....+\frac{2016}{2015.2017}\)
\(=1008.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2015.2017}\right)\)
\(=1008.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(=1008.\left(1-\frac{1}{2017}\right)\)
\(=1008.\frac{2016}{2017}\)
NA
0