1/1.3 + 1/3.5 + 1/5.7 +...+ 1/2019.2021
Cho \(P=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2019.2021}\)
chứng tỏ rằng P<1
Ta có : \(P=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2019}-\dfrac{1}{2020}=1-\dfrac{1}{2020}=\dfrac{2019}{2020}\)
mà \(2019< 2020\)nên P < 1 ( đpcm )
\(P=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2019.2021}\)
\(P=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2019}-\dfrac{1}{2021}\)
\(P=1-\dfrac{1}{2021}\)
\(P=\dfrac{2020}{2021}\)
Vì \(\dfrac{2020}{2021}< 1\) ⇒ \(P< 1\) ( điều phải chứng minh )
c, C = 2020/1.2 + 2020/2.3 + 2020/3.4 + ... + 2020/2019.2020
d, D = 2020/1.3 + 2020/3.5 + 2020/5.7 + ... + 2020/2019.2021
e, E = 2023/ 1.3 + 2023/3.5 + 2023/5.7 + ... + 2023/2019.2020
f, F = 1/15 + 1/35 + 1/63 + ... + 1/657
giúp với mình cần gấp lắm
(1+1/1.3) . (1+1/2.4) . (1+1/3.5) .... (1+1/2019.2021
Tính :
(1 + \(\dfrac{1}{1.3}\) ) . ( 1+\(\dfrac{1}{2.4}\) ) . (1+\(\dfrac{1}{3.5}\)) . ... . ( 1+\(\dfrac{1}{2019.2021}\))
Lời giải:
Gọi tích trên là $A$
Xét thừa số tổng quát: $1+\frac{1}{n(n+2)}=\frac{n(n+2)+1}{n(n+2)}=\frac{(n+1)^2}{n(n+2)}$
Thay $n=1,2,3....,2019$ ta có:
$A=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}....\frac{2020^2}{2019.2021}$
$=\frac{2^2.3^2...2020^2}{(1.3)(2.4)(3.5)...(2019.2021)}$
$=\frac{(2.3....2020)(2.3...2020)}{(1.2.3...2019)(3.4...2021)}$
$=2020.\frac{2}{2021}=\frac{4040}{2021}$
a) 1/1.3+1/3.5+1/5.7
b) 1/1.3+1/3.5+1/5.7+...+1/2007.2009+1/2009.2011
a)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{7}\right)\)
\(=\frac{1}{2}.\frac{6}{7}\)
\(=\frac{3}{7}\)
b)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009.2011}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)
\(=\frac{1}{2}.\frac{2010}{2011}\)
\(=\frac{1005}{2011}\)
Thực hiện phép tính (1+1/1.3)(1+1/2/4)(1+1/3.5).....(1+1/2019.2021)
Giúp mình với ngày mai mình phải nộp rồi!
Thực hiện phép tính:
\(\left(1+\dfrac{1}{1.3}\right)+\left(1+\dfrac{1}{2.4}\right)+\left(1+\dfrac{1}{3.5}\right)+...+\left(1+\dfrac{1}{2019.2021}\right)\)
Sửa đề: A=(1+1/1*3)(1+1/2*4)*...*(1+1/2019*2021)
\(=\dfrac{2^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(3-1\right)\left(3+1\right)}\cdot...\cdot\dfrac{2020^2}{\left(2020-1\right)\left(2020+1\right)}\)
\(=\dfrac{2}{1}\cdot\dfrac{3}{2}\cdot...\cdot\dfrac{2020}{2019}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2020}{2021}=2020\cdot\dfrac{2}{2021}=\dfrac{4040}{2021}\)
Thực hiện phép tính:
\(\left(1+\dfrac{1}{1.3}\right)+\left(1+\dfrac{1}{2.4}\right)+\left(1+\dfrac{1}{3.5}\right)+...+\left(1+\dfrac{1}{2019.2021}\right)\)
1/1.3+1/3.5+1/5.7+...+1/2009.2011
\(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{2009\cdot2011}\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{2009\cdot2011}\right)\)
\(=\dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\)
\(=\dfrac{1}{2}\cdot\left(1-\dfrac{1}{2011}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2010}{2011}=\dfrac{1005}{2011}\)
= 1/2 . (1/1 - 1/3 + 1/3 - 1/5 +... + 1/2009 - 1/2011)
= 1/2 . (1/1 - 1/2011)
= 1/2 . 2010 / 2011
= 1005/2011