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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 \(\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+...+\frac{9998}{9999}=\frac{4950}{101}\)
HN
0
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)
\(M=50-\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\right)\)
\(M=50-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(M=50-\left(1-\frac{1}{101}\right)=50-\frac{100}{101}=\frac{5050-100}{101}=\frac{4950}{101}\)
M
9 tháng 4 2018
1, A=\(\left(1+1+1+1\right)\)-\(\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}\right)\)
=4-\(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)\)
= 4-\(\left(\frac{1}{1}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{9}\right)\)
=4-\(\left(1-\frac{1}{9}\right)\)
= 4-\(\frac{8}{9}\)
= \(\frac{7}{9}\)
NN
1
MV
3 tháng 8 2017
\(A=\dfrac{2}{3}+\dfrac{14}{15}+\dfrac{34}{35}+...+\dfrac{9998}{9999}\\ =\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{15}\right)+\left(1-\dfrac{1}{35}\right)+...+\left(1-\dfrac{1}{9999}\right)\\ =\left(1+1+1+...+1\right)\left(\text{có 50 số 1}\right)-\left(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{9999}\right)\\ =50\cdot1-\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{99\cdot101}\right)\\ =50-\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\ =50-\left(1-\dfrac{1}{101}\right)\\ =50-1+\dfrac{1}{101}\\ =49+\dfrac{1}{101}\\ =\dfrac{4949+1}{101}\\ =\dfrac{4950}{101}\)
MP
2 tháng 7 2020
\(A=\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+\frac{98}{99}+\frac{142}{143}\)
\(=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{15}\right)+\left(1-\frac{1}{35}\right)+\left(1-\frac{1}{63}\right)+\left(1-\frac{1}{99}\right)+\left(1-\frac{1}{143}\right)\)
\(=\left(1+1+1+1+1+1\right)-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\right)\)
\(=6-\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\right)\)
\(=6-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=6-\left(1-\frac{1}{13}\right)\)
\(=6-1+\frac{1}{13}\)
\(=5+\frac{1}{13}\)
\(=\frac{66}{13}\)
MP
2 tháng 7 2020
Mk sửa lại 1 tí nha dòng thứ 5 :
\(A=6-\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=6-\frac{1}{2}\left(1-\frac{1}{13}\right)\)
\(=6-\frac{1}{2}.\frac{12}{13}\)
\(=6-\frac{6}{13}=\frac{72}{13}\)
Mong bn bỏ qua nha
NQ
2
WR
8 tháng 6 2015
=2/1.3+2/3.5+2/5.7+....+2/99.101
=1-1/3+1/3-1/5+`/5-1/7+...+1/99-1/101
=1-1/101
=100/101
8 tháng 6 2015
\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{99.101}\)
\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
\(=3,571\)