Tính nhanh có diễn giải.
B=2/3+2/15+2/35+2/63+...+2/9999.
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12 tháng 7 2016
2/3+2/15+2/35+2/63+...+2/9999
=2/1.3+2/3.5+2/5.7+...+2/99x101
=1-1/3+1/3-1/5+...+1/99-1/101
=1-1/101=100/101
7 tháng 7 2016
\(=2\times\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\right)\)
\(=2\times\left(\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{99\times101}\right)\)
\(=2\times\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=2\times\left(\frac{1}{1}-\frac{1}{101}\right)\)
\(=2\times\frac{100}{101}\)
\(=\frac{200}{101}\)
DP
12 tháng 7 2017
\(\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+...+\frac{9998}{9999}\)
\(=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{15}\right)+\left(1-\frac{1}{35}\right)+\left(1-\frac{1}{63}\right)+...+\left(1-\frac{1}{9999}\right)\)
\(=\left(1-\frac{1}{1\cdot3}\right)+\left(1-\frac{1}{3\cdot5}\right)+\left(1-\frac{1}{5\cdot7}\right)+\left(1-\frac{1}{7\cdot9}\right)+...+\left(1-\frac{1}{99\cdot101}\right)\)
\(=\left(1+1+1+1+...+1\right)-\frac{1}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
Có tất cả : (101 - 3) : 2 + 1 = 50 chữ số 1 => (1 + 1 + 1 + .... + 1) = 1 x 50 = 50
\(\Rightarrow50-\frac{1}{2}\cdot\left(1-\frac{1}{101}\right)\)
\(=50-\frac{1}{2}\cdot\frac{100}{101}=50-\frac{100}{101}=\frac{4950}{101}\)
Vậy \(\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+...+\frac{9998}{9999}=\frac{4950}{101}\)
3 tháng 5 2015
\(M=1-\frac{1}{3}+1-\frac{1}{15}+1-\frac{1}{35}+1-\frac{1}{63}+...+1-\frac{1}{9999}\)
\(M=\left(1+1+1+...+1\right)-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\right)\)
\(M=\left(1+1+1+...+1\right)-\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\right)\)(Có (99 - 1): 2+ 1 = 50 số 1)
\(M=50-\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\right)\)
\(M=50-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(M=50-\left(1-\frac{1}{101}\right)=50-\frac{100}{101}=\frac{5050-100}{101}=\frac{4950}{101}\)
MT
31 tháng 8 2015
\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{9999}\)
\(=\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{99\times101}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{1}-\frac{1}{101}=\frac{101}{101}-\frac{1}{101}=\frac{100}{101}\)
NT
2
HT
23 tháng 7 2015
A = \(\frac{2}{3}+\frac{3}{15}+\frac{2}{35}+.....+\frac{2}{9999}\)
A = \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{99.101}\)
A = \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{99}-\frac{1}{101}\)
A = \(1-\frac{1}{101}\)
A = \(\frac{100}{101}\)
15 tháng 3 2017
biến đổi ra là đc
\(1-\left(\frac{2}{3}+\frac{2}{3.5}+\frac{2}{5.7}+..+\frac{2}{99.101}\right)\)
tới đây thôi bạn tự làm đi
NT
3
16 tháng 3 2017
\(-2\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{9999}\right)\)
\(=-2\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{99.101}\right)\)\(=-2\cdot\left(\frac{1}{3}-\frac{1}{101}\right)\)
=.....
mình quên đem máy tính nên k ghi đc đấp số
THÔNG CẢM
15 tháng 6 2017
\(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{15}-\dfrac{1}{35}-\dfrac{1}{63}-...-\dfrac{1}{9999}\)
\(=\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+...+\dfrac{1}{9999}\right)\)
\(=\dfrac{1}{2}-\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{99.101}\right)\)
\(=\dfrac{1}{2}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{1}{2}-\dfrac{1}{2}\left(1-\dfrac{1}{101}\right)\)
\(=\dfrac{1}{2}-\dfrac{1}{2}.\dfrac{100}{101}\)
\(=\dfrac{1}{2}-\dfrac{50}{101}\)
\(=\dfrac{1}{202}.\)
\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{9999}\)
\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{1}{5.7}+....+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}+0+0+...+0\)
\(=\frac{100}{101}\)