1/3.5+1/5.7+1/7.9+........+1/99.101
giup minh vói!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
K
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KT
19 tháng 1 2018
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\frac{98}{303}\)
\(=\frac{49}{303}\)
17 tháng 4 2016
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
\(=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
=\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\frac{98}{303}\)
\(=\frac{49}{303}\)
17 tháng 4 2016
\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{99.101}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.......+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\frac{98}{101}=\frac{49}{101}\)
NN
1
TM
25 tháng 4 2016
\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
\(\Rightarrow2A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}...+\frac{2}{99.101}\)
\(\Rightarrow2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)
\(\Rightarrow2A=\frac{1}{3}-\frac{1}{101}\)
\(\Rightarrow2A=\frac{101}{303}-\frac{3}{303}\)
\(\Rightarrow2A=\frac{98}{303}\)
\(\Rightarrow A=\frac{98}{303}:2=\frac{98}{303.2}=\frac{98}{606}=\frac{49}{303}\)
lên 820 điểm hỏi đáp nha
25 tháng 1 2023
\(B=-\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\)
\(=\dfrac{-1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{100}{101}=-\dfrac{50}{101}\)
K
3
31 tháng 8 2017
Đặt :
\(A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+.........+\dfrac{1}{99.101}\)
\(\Leftrightarrow2A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+............+\dfrac{2}{99.101}\)
\(\Leftrightarrow2A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+............+\dfrac{1}{99}-\dfrac{1}{101}\)
\(\Leftrightarrow2A=\dfrac{1}{3}-\dfrac{1}{101}\)
\(\Leftrightarrow2A=\dfrac{98}{303}\)
\(\Leftrightarrow A=\dfrac{49}{303}\)
TV
6
18 tháng 6 2017
\(A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+............+\dfrac{1}{99.101}\)
\(A=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+.....+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(A=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{101}\right)=\dfrac{1}{2}.\dfrac{98}{303}=\dfrac{49}{303}\)
19 tháng 6 2017
\(A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+.................+\dfrac{1}{99.101}\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{101}\right)\)
\(=\dfrac{1}{2}.\dfrac{98}{303}\)
\(=\dfrac{49}{303}\)
OY
22 tháng 7 2021
Đặt A=\(\dfrac{2}{3.5}.\dfrac{2}{7.9}.....\dfrac{2}{99.101}\)
A=\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
A=\(\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}\)
22 tháng 7 2021
Ta có: \(P=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{1}{15}\)
\(=\dfrac{4}{15}\)
XO
12 tháng 5 2020
Ta có : S = 1.3 + 3.5 + 5.7 + .... + 97.99 + 99.101
=> 6S = 1.3.6 + 3.5.6 + 5.7.6 +...+ 97.99.6 + 99.101.6
= 1.3.(5 + 1) + 3.5.(7 - 1) + 5.7.(9 - 3) + .... + 97.99.(101 - 95) + 99.101.(103 - 97)
= 3 + 1.3.5 + 3.5.7 - 1.3.5 + 5.7.9 - 3.5.7 + ... + 97.99.101 - 95.97.99 + 99.101.103 - 97.99.101
= 3 + 99.101.103
= 1029900
=> 6S = 1029900
=> S = 171650
T
12 tháng 5 2020
Ta có: A = 1.3 + 3.5 + 5.7 +…+ 97.99 + 99.101
A = 1.(1 + 2) + 3.(3 + 2) + 5.(5 + 2) + … + 97.(97 + 2) + 99.(99 + 2)
A = (1^2 + 3^2 + 5^2 + … + 97^2 + 99^2) + 2.(1 + 3 + 5 + … + 97 + 99).
Đặt B = 1^2 + 3^2 + 5^2 + … + 99^2
=> B = (1^2 + 2^2 + 3^2 + 4^2 + … + 100^2) – 2^2.(1^2 + 2^2 + 3^2 + 4^2 + … + 50^2)
Tính dãy tổng quát C = 1^2 + 2^2 + 3^2 + … + n^2
C = 1.(0 + 1) + 2.(1 + 1) + 3.(2 + 1) + … + n.[(n – 1) + 1]
C = [1.2 + 2.3 + … + (n – 1).n] + (1 + 2 + 3 + … + n)
C = = n.(n + 1).[(n – 1) : 3 + 1 : 2] = n.(n + 1).(2n + 1) : 6
Áp dụng vào B ta được:
B = 100.101.201 : 6 – 4.50.51.101 : 6 = 166650
=> A = 166650 + 2.(1 + 99).50 : 2
=> A = 166650 + 5000 = 172650.
Đ/s: A = 172650.
DP
8 tháng 6 2017
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
Đặt A = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)
\(2A=\frac{1}{1}-\frac{1}{101}\)
\(2A=\frac{100}{101}\)
\(\Rightarrow A=\frac{100}{101}\div2\)
\(\Rightarrow A=\frac{50}{101}\)
Ta có: \(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{99\cdot101}\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{101}\right)=\frac{1}{2}\cdot\frac{98}{303}=\frac{49}{303}\)
đặt A=1/3.5+1/5.7+1/7.9+.....+1/99.101
2A=2/3.5+2/5.7+2/7.9+....+2/99.101
=>2A=1/3-1/5+1/5-1/7+1/7-1/9+...+1/99-1/101(chỗ này nếu ko hiểu thì bạn dùng máy tính mà thử)
=>2A=1/3-1/101
=>A=(1/3-1/101).1/2(chỗ này ko cần tính chi tiết chỉ cần để ở dạng rút gọn thôi)