Tinh tổng:(1-1/3)+(1-1/15)+(1-1/35)+.........+(1-1/9999)
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VT
2
10 tháng 1 2016
A=1/3*5 + 1/5*7 + ....+ 1/99*101
A=1/2(2/3*5 + 2/5*7 + ...+ 2/99*101)
A=1/2[(1/3-1/5)+(1/5-1/7)+...+(1/99-1/101)]
A=1/2(1/3-1/101)
A=1/2 * 98/303
A=49/303
10 tháng 1 2016
\(A=\frac{1}{5}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{9999}\)
\(=\frac{1}{3X5}+\frac{1}{5X7}+\frac{1}{7X9}+\frac{1}{9X11}+...+\frac{1}{99X101}\)
\(2A=\frac{2}{3X5}+\frac{2}{5X7}+\frac{2}{7X9}+\frac{2}{9X11}+...+\frac{2}{99X101}\)
\(=\frac{5-3}{3X5}+\frac{7-5}{7X9}+\frac{9-7}{9X7}+\frac{11-9}{9X11}+...+\frac{101-99}{101}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{3}-\frac{1}{101}=\frac{98}{303}\)
Từ đó ta suy ra:\(A=\frac{98}{303}:2=\frac{49}{303}\)
17 tháng 1 2016
2A=2/3.5+2/5.7+2/7.9+...+2/99.101=>2A=1/3-1/5+1/5-1/7+...+1/99-1/100=>2A=1/3-1/100=>2A=97/300=>A=97/600
VT
1
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
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+.....+\frac{1}{9999}=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+.....+\frac{1}{99.}\)\(\frac{1}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
4 tháng 5 2016
\(S=1:3+1:15+1:35+...+1:9999\)
\(S=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{9999}\)
\(S=2\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)
\(2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(2S=1-\frac{1}{101}\)
\(2S=\frac{100}{101}\)
\(S=\frac{100}{101}:2\)
\(S=\frac{50}{101}\)
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
15 tháng 6 2017
\(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{15}-\dfrac{1}{35}-\dfrac{1}{63}-...-\dfrac{1}{9999}\)
\(=\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+...+\dfrac{1}{9999}\right)\)
\(=\dfrac{1}{2}-\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{99.101}\right)\)
\(=\dfrac{1}{2}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{1}{2}-\dfrac{1}{2}\left(1-\dfrac{1}{101}\right)\)
\(=\dfrac{1}{2}-\dfrac{1}{2}.\dfrac{100}{101}\)
\(=\dfrac{1}{2}-\dfrac{50}{101}\)
\(=\dfrac{1}{202}.\)
25 tháng 12 2016
siêu tốc
\(2000+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)=2000+\frac{50}{3.101}\)
LA
25 tháng 12 2016
Ta có: \(2000+\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\right)\)
=\(2000+\left(\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+...+\frac{1}{99x101}\right)\)
Đặt A=\(\left(\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+...+\frac{1}{99x101}\right)\)
=> 2xA =\(\left(\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+...+\frac{2}{99x101}\right)\)
2xA = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-...-\frac{1}{99}+\frac{1}{99}-\frac{1}{101}\)
2xA = \(\frac{1}{3}-\frac{1}{101}\)
2xA = \(\frac{98}{303}\)
A = \(\frac{98}{606}=\frac{49}{303}\)
=> \(2000+\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\right)=2000+\frac{49}{303}=\frac{606049}{303}\)
Bạn xem rút gọn được thì rút nhé
\(\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{15}\right)+...+\left(1-\frac{1}{9999}\right)\)
= \(\left(1-\frac{1}{1.3}\right)+\left(1-\frac{1}{3.5}\right)+...+\left(1-\frac{1}{99.101}\right)\)(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)
= \(1.50-\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
= \(50-\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
= \(50-\frac{1}{2}.\left(1-\frac{1}{101}\right)\)
= \(50-\frac{1}{2}.\frac{100}{101}\)
= \(50-\frac{50}{101}\)
= \(\frac{5000}{101}\)