1 : 3 + 1:15 + 1: 35 + .... + 1:9999
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30 tháng 6 2016
\(=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+.......+\frac{1}{99\cdot101}=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{101}=1-\frac{1}{101}=\frac{100}{101}\)
30 tháng 6 2016
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+.....+\frac{1}{9999}=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+.....+\frac{1}{99.}\)\(\frac{1}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
6 tháng 4 2016
1/3+1/15+1/35+1/63+1/99+……+1/9999
=1/(1×3)+1/(3×5)+1/(5×7)+1/(7×9)+1/(9×11)+……+1/(99×101)
=1/2(1-1/3)+1/2(1/3-1/5)+1/2(1/5-1/7)+1/2(1/7-1/9)+1/2(1/9-1/11)+……+1/2(1/99-1/101)
=1/2(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+……+1/99-1/101)
=1/2(1-1/101)
=1/2×(100/101)
=50/101
6 tháng 4 2016
1/3+1/15+1/35+1/63+1/99+……+1/9999
=1/(1×3)+1/(3×5)+1/(5×7)+1/(7×9)+1/(9×11)+……+1/(99×101)
=1/2(1-1/3)+1/2(1/3-1/5)+1/2(1/5-1/7)+1/2(1/7-1/9)+1/2(1/9-1/11)+……+1/2(1/99-1/101)
=1/2(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+……+1/99-1/101)
=1/2(1-1/101)
=1/2×(100/101)
=50/101
KK
1
XO
21 tháng 7 2019
\(\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{15}\right)+...+\left(1-\frac{1}{9999}\right)\)
= \(\left(1-\frac{1}{1.3}\right)+\left(1-\frac{1}{3.5}\right)+...+\left(1-\frac{1}{99.101}\right)\)(50 cặp)
= \(\left(1+1+1+...+1\right)-\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\right)\)(50 số hạng 1)
= \(1.50-\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
= \(50-\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
= \(50-\frac{1}{2}.\left(1-\frac{1}{101}\right)\)
= \(50-\frac{1}{2}.\frac{100}{101}\)
= \(50-\frac{50}{101}\)
= \(\frac{5000}{101}\)
15 tháng 6 2017
\(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{15}-\dfrac{1}{35}-\dfrac{1}{63}-...-\dfrac{1}{9999}\)
\(=\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+...+\dfrac{1}{9999}\right)\)
\(=\dfrac{1}{2}-\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{99.101}\right)\)
\(=\dfrac{1}{2}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{1}{2}-\dfrac{1}{2}\left(1-\dfrac{1}{101}\right)\)
\(=\dfrac{1}{2}-\dfrac{1}{2}.\dfrac{100}{101}\)
\(=\dfrac{1}{2}-\dfrac{50}{101}\)
\(=\dfrac{1}{202}.\)
T
26 tháng 9 2023
\(S=\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{9999}\)
\(=\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{99\cdot101}\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{99\cdot101}\right)\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{3}-\dfrac{1}{101}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{98}{303}\)
\(=\dfrac{49}{303}\)
Vậy \(S=\dfrac{49}{303}\)
#\(Toru\)
AT
9
1 tháng 1 2017
1/3 + 1/15 + 1/35 + 1/63 + 1/99 + 9999
= 1/3 + ( 1/5 + 1/35 + 1/63 ) + 1/99 = 9999
= 1/3 + 1111/9999 + 1/99
= 3333/9999 + 1111/9999 +101/9999
= 4545/9999
XO
10 tháng 6 2020
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{9999}=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{101}\right)=\frac{1}{2}.\frac{100}{101}=\frac{50}{101}\)
\(S=1:3+1:15+1:35+...+1:9999\)
\(S=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{9999}\)
\(S=2\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)
\(2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(2S=1-\frac{1}{101}\)
\(2S=\frac{100}{101}\)
\(S=\frac{100}{101}:2\)
\(S=\frac{50}{101}\)