Tính
S= 1/1.3 + 1/5.9+ ....+ 1/2017.2021
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\(M=\frac{16}{1.5}+\frac{16}{5.9}+........+\frac{16}{2017.2021}\)
\(M=4.\left(\frac{4}{1.5}+\frac{4}{5.9}+.......+\frac{4}{2017.2021}\right)\)
\(M=4.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+.........+\frac{1}{2017}-\frac{1}{2021}\right)\)
\(M=4.\left(1-\frac{1}{2021}\right)\)
\(M=4.\frac{2020}{2021}\)
\(M=\frac{8080}{2021}\)
\(N=\frac{1}{1.7}+\frac{1}{7.13}+.......+\frac{1}{2007.2013}\)
\(N=\frac{1}{6}.\left(\frac{6}{1.7}+\frac{6}{7.13}+........+\frac{6}{2007.2013}\right)\)
\(N=\frac{1}{6}.\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+......+\frac{1}{2007}-\frac{1}{2013}\right)\)
\(N=\frac{1}{6}.\left(1-\frac{1}{2013}\right)\)
\(N=\frac{1}{6}.\frac{2012}{2013}\)
\(N=\frac{1006}{6039}\)
\(N=\frac{1}{1.7}+\frac{1}{7.13}+...+\frac{1}{2007.2013}\)
\(N=\frac{1}{1}-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+...+\frac{1}{2007}-\frac{1}{2013}\)
\(N=1-\frac{1}{2013}\)
\(N=\frac{2012}{2013}\)
\(4.B=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{93.97}\)
\(4.B=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{93}-\frac{1}{97}\)
\(4.B=1-\frac{1}{97}\)
\(4.B=\frac{96}{97}\)
\(B=\frac{96}{97}:4\)
\(B=\frac{24}{97}\)
a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2020.2021}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2020}-\frac{1}{2021}\)
\(=1-\frac{1}{2021}=\frac{2020}{2021}\)
b) \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{21.23}=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{21.23}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{21}-\frac{1}{23}\right)=\frac{1}{2}\left(1-\frac{1}{23}\right)=\frac{1}{2}.\frac{22}{23}=\frac{11}{23}\)
c) \(\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{2.1}=\frac{1}{99}-\left(\frac{1}{98.99}+\frac{1}{97.98}+...+\frac{1}{1.2}\right)\)
\(=\frac{1}{99}-\left(\frac{1}{98}-\frac{1}{99}+\frac{1}{97}-\frac{1}{98}+...+1-\frac{1}{2}\right)=\frac{1}{99}-\left(-\frac{1}{99}+1\right)=\frac{1}{99}-\frac{98}{99}\)
\(=-\frac{97}{99}\)
d) bạn xem lại đề
a)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2020}-\frac{1}{2021}\)
\(=\frac{1}{1}-\frac{1}{2021}\)
\(=\frac{2020}{2021}\)
b)
\(=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{21\cdot23}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{21}-\frac{1}{23}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{1}{1}-\frac{1}{23}\right)\)
\(=\frac{1}{2}\cdot\frac{22}{23}\)
\(=\frac{11}{23}\)
c)
\(=\frac{1}{99}-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}\right)\)
\(=\frac{1}{99}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}\right)\)
\(=\frac{1}{99}-\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{99}-\frac{98}{99}\)
\(=\frac{-97}{99}\)
d)
đề sai hay sao á mong bạn xem ljai ạ
Bài 1: Bạn ơi số 2004 không thuộc dãy A
A có số số hạng là: (2005 - 5) : 4 + 1 = 501 (số hạng)
A = (2005 + 5) x 501 : 2 = 503505
Bài 2:
a) B = 4 . 1 + 4 . 5 + 4 . 52 + 4 . 53 + ... + 4 . 51000
=> B = 4 . ( 1 + 5 + 52 + 53 + .... + 51000)
b) 5 + 52 + 53 + .... + 51000 có tận cùng là 0 (Do các lũy thừa với cơ số là 5 thì có tận cùng là 5 [25] mà ở đây có số số hạng là chẵn)
=> 1 + 5 + 52 + 53 + .... + 51000 có tận cùng là số 1
=> 4 . ( 1 + 5 + 52 + 53 + .... + 51000) có tận cùng là 4.
Vậy B có tận cùng là 4.
Bài 3:
1. A = 1.3 + 3.5 + 5.7 + ......... + 97.99
=> A = 1.(1 + 2) + 3.(3 + 2) + 5.(5 + 2) + .... + 97.(97 + 2)
=> A = 12 + 1.2 + 32 + 3.2 + 52 + 5.2 + .... + 972 + 97.2
=> A = (12 + 32 + 52 + .... + 972) + (1.2 + 3.2 + 5.2 + .... + 97.2)
=> A = (12 + 32 + 52 + .... + 972) + 2(1 + 3 + 5 + .... + 97)
=> A = (12 + 32 + 52 + .... + 972) + 2 { (97 + 1) . [(97 - 1) : 2 + 1] : 2 }
=> A = (12 + 32 + 52 + .... + 972) + 24802
Đặt B = (12 + 32 + 52 + .... + 972)
=> B = 1.1 + 3.3 + 5.5 + .... + 97.97
=> B = 1.(0 + 1) + 3.(1 + 2) + 5.(4 + 1) + ..... + 97.(96 + 1)
=> B = 0 + 1.1 + 3 + 2.3 + 5 + 4.5 + .... + 97 + 96.97
=> B = (0 + 3 + 5 + .... + 97) + (1.1 + 2.3 + 4.5 + .... + 96.97)
=> B = 2400 + \(\frac{\left(97-1\right).97\left(97+1\right)}{6}\)
=> B = 2400 + 152096 = 154496
=> A = 154496 + 4802 = 159298
(Làm tương tự ở câu 2 nha)
Ta có:\(\text{ -1 - 2 - 3 - 4 - 5 - ... - 1999 - 2000 - 2001 - 2002}\)
\(\text{= -(1 + 2 + 3 + 4 + 5 + 6 + ... + 2001 + 2002) }\)
\(=-8020012\)
{Số số hạng của dãy trong ngoặc là:\(\text{ (2002 - 1) : 1 + 1 = 2002}\)
\(\Rightarrow\)Tổng là: \(\left(1+2002\right)\cdot2002:2=8020012\)}
a: \(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2n+1-1}{2n+1}=\dfrac{1}{2}\cdot\dfrac{2n}{2n+1}=\dfrac{n}{2n+1}\)
b: \(=\dfrac{1}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{\left(4n-3\right)\left(4n+1\right)}\right)\)
\(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{4n-3}-\dfrac{1}{4n+1}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{4n}{4n+1}=\dfrac{n}{4n+1}\)
a: \(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2n+1-1}{2n+1}\)
\(=\dfrac{n}{2n+1}\)
b: \(=\dfrac{1}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{\left(4n-3\right)\left(4n+1\right)}\right)\)
\(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{4n-3}-\dfrac{1}{4n+1}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{4n}{4n+1}=\dfrac{n}{4n+1}\)