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S
4
8 tháng 8 2021
A = 1/1.3 + 1/3.5 + 1/5.7 + ... + 1/2017.2019
A = 1/2 (1 - 1/3 + 1/3 - 1/5 + 1/5 - ... - 1/2019)
A = 1/2 (1 - 1/2019)
A = 1/2 . 2018/2019
A = 1009/2019
@Cỏ
8 tháng 8 2021
\(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{2017\cdot2019}\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{2019}\right)=\frac{1}{2}\cdot\frac{2018}{2019}\)
\(=\frac{1009}{2019}\)
18 tháng 12 2019
\(M=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2017.2019}\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{2019}\right)\)
\(=\frac{1}{2}.\frac{2018}{2019}\)
\(=\frac{2018}{4038}\)
\(\Rightarrow\frac{2018}{4038}< \frac{1}{2}\)( lấy máy tính )
HN
18 tháng 12 2019
\(M=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+.....+\frac{1}{2017.2019}\)
\(\Rightarrow M=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-......-\frac{1}{2017}+\frac{1}{2017}-\frac{1}{2019}\)
\(\Rightarrow M=1-\frac{1}{2019}\)
\(\Rightarrow M=\frac{2019}{2019}-\frac{1}{2019}\)
\(\Rightarrow M=\frac{2018}{2019}\)
Có \(\frac{2018}{2019}=\frac{2018.2}{2019.2}=\frac{4036}{4038}\)
\(\frac{1}{2}=\frac{1.2019}{2.2019}=\frac{2019}{4038}\)
Mà \(\frac{4036}{4038}< \frac{2019}{4038}\Rightarrow M< \frac{1}{2}\)
Vậy M < \(\frac{1}{2}\)
AL
0
DN
0
BC
5 tháng 3 2019
bn lên ngạng hoặc và xem câu hỏi tương tự nha!
Nhớ k mk đấy nha!
thanks nhìu!
OK..OK..OK
5 tháng 3 2019
\(C=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2n-1\right)\left(2n+1\right)}\)
\(2C=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}\)
Ta có :
\(\frac{2}{1.3}=1-\frac{1}{3}\)
\(\frac{2}{3.5}=\frac{1}{3}-\frac{1}{5}\)
...............................
\(\frac{2}{\left(2n-1\right)\left(2n+1\right)}=\frac{1}{2n-1}-\frac{1}{2n+1}\)
\(\Rightarrow2C=1-\frac{1}{2n+1}=\frac{2n}{2n+1}\)
\(\Rightarrow C=\frac{n}{2n+1}\)
DT
Tính S = 1.3/3.5 + 2.4/5.7 + 3.5/7.9 + ... + ( n-1)( n+1) / (2n-1)(2n+1) + ... + 1002.1004/2005.2007
2
KN
13 tháng 8 2020
\(S=\frac{1.3}{3.5}+\frac{2.4}{5.7}+\frac{3.5}{7.9}+...+\frac{\left(n-1\right)\left(n+1\right)}{\left(2n-1\right)\left(2n+1\right)}+...+\frac{1002.1004}{2005.2007}\)
\(\Rightarrow S=\frac{\left(2-1\right)\left(2+1\right)}{\left(2.2-1\right)\left(2.2+1\right)}+\frac{\left(3-1\right)\left(3+1\right)}{\left(3.2-1\right)\left(3.2+1\right)}+...+\frac{\left(n-1\right)\left(n+1\right)}{\left(2n-1\right)\left(2n+1\right)}\)
\(+..+\frac{\left(1003-1\right)\left(1003+1\right)}{\left(1003.2-1\right)\left(1003.2+1\right)}\)
\(\Rightarrow S=\frac{1}{4}-\frac{3}{8}\left(\frac{1}{2.2-1}-\frac{1}{2.2+1}\right)+\frac{1}{4}-\frac{3}{8}\left(\frac{1}{3.2-1}-\frac{1}{3.2+1}\right)+...\)
\(+\frac{1}{4}-\frac{3}{8}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)+...+\frac{1}{4}-\frac{3}{8}\left(\frac{1}{1003.2-1}-\frac{1}{1003.2+1}\right)\)
\(\Rightarrow S=1002.\frac{1}{4}-1002.\frac{3}{8}\left(\frac{1}{2.2-1}-\frac{1}{2.2+1}+\frac{1}{3.2-1}-...-\frac{1}{1003.2+1}\right)\)
\(\Rightarrow S=\frac{501}{2}-\frac{1503}{4}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2005}-\frac{1}{2007}\right)\)
\(\Rightarrow S=\frac{501}{2}-\frac{1503}{4}\left(\frac{1}{3}-\frac{1}{2007}\right)\)
\(\Rightarrow S=\frac{501}{2}-\frac{1503}{4}.\frac{668}{2007}\)
\(\Rightarrow S=\frac{501}{2}-\frac{27889}{223}\)
\(\Rightarrow S=125,4372197\)
\(\)
17 tháng 9 2017
Cách làm :
Áp dụng công thức : \(\dfrac{n}{a\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\)
\(C=\dfrac{1}{1.2}+\dfrac{1}{2.3}+..........+\dfrac{1}{999.1000}\)
\(\Leftrightarrow C=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+..........+\dfrac{1}{999}-\dfrac{1}{1000}\)
\(\Leftrightarrow C=1-\dfrac{1}{1000}\)
\(\Leftrightarrow C=\dfrac{999}{1000}\)
\(F=\dfrac{1}{1.3}+\dfrac{1}{3.5}+.........+\dfrac{1}{99.101}\)
\(\Leftrightarrow2F=\dfrac{2}{1.3}+\dfrac{2}{3.5}+............+\dfrac{2}{99.101}\)
\(\Leftrightarrow2F=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+........+\dfrac{1}{99}-\dfrac{1}{101}\)
\(\Leftrightarrow2F=1-\dfrac{1}{101}\)
\(\Leftrightarrow2F=\dfrac{100}{101}\)
\(\Leftrightarrow F=\dfrac{50}{101}\)
G
17 tháng 9 2017
Giải:
\(C=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{999.1000}\)
\(\Leftrightarrow C=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{999}-\dfrac{1}{1000}\)
\(\Leftrightarrow C=\dfrac{1}{1}-\dfrac{1}{1000}\)
\(\Leftrightarrow C=\dfrac{999}{1000}\)
Sửa đề:
\(F=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{999.1001}\)
\(\Leftrightarrow F=\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{999}-\dfrac{1}{1001}\right)\)
\(\Leftrightarrow F=\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{1001}\right)\)
\(\Leftrightarrow F=\dfrac{1}{2}.\dfrac{1000}{1001}\)
\(\Leftrightarrow F=\dfrac{500}{1001}\)
Chúc bạn học tốt!
\(C=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{2017+2019}\)
\(\Rightarrow2C=2\cdot\left(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{2017\cdot2019}\right)\)
\(2C=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{2017\cdot2019}\)
Ta co: \(\frac{2}{3\cdot5}=\frac{1}{3}-\frac{1}{5}\)
\(\frac{2}{5\cdot7}=\frac{1}{5}-\frac{1}{7}\)
.............................
\(\frac{2}{2017\cdot2019}=\frac{1}{2017}-\frac{1}{2019}\)
\(\Rightarrow2C=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\)
\(2C=\frac{1}{3}-\frac{1}{2019}\)
\(2C=\frac{673}{2019}-\frac{1}{2019}\)
\(2C=\frac{672}{2019}\)
\(C=\frac{672}{2019}\div2\)
\(C=\frac{112}{673}\)
Vay \(C=\frac{112}{673}\).