Chứng minh rằng:
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2n+1\right)\left(2n+3\right)}=\dfrac{n+1}{2n+3}\)
Giúp mik vs các bn
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TG
25 tháng 6 2021
a)
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}=\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)=\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)< \dfrac{1}{2}\)
P/s: Cj chỉ biết làm ý a thôi nhé! Có j ko hiểu cmt nhé!
18 tháng 1
\(u_{n+1}=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{\left(2n-1\right)\cdot\left(2n+1\right)}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2n+1-1}{2n+1}=\dfrac{n}{2n+1}\)
=>\(u_{50}=u_{49+1}=\dfrac{49}{2\cdot49+1}=\dfrac{49}{99}\)
ND
27 tháng 2 2018
\(S=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)
\(\Rightarrow2S=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\)
\(\Rightarrow2S=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\) \(\Rightarrow2S=1-\dfrac{1}{2n+1}\)
\(\Rightarrow S=\dfrac{n}{2n+1}\)
27 tháng 2 2018
Ta có : \(\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}=\dfrac{1}{2}\left(\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)
ta được \(\dfrac{1}{1.3}=\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}\right);\dfrac{1}{3.5}=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}\right);\dfrac{1}{5.7}=\dfrac{1}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\)
\(\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}=\dfrac{1}{2}\left(\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\) vậy \(S=\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)=\dfrac{n}{2n+1}\)
DV
8 tháng 12 2017
\(2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\)
\(=1-\dfrac{1}{2n+1}\Rightarrow A=\left(1-\dfrac{1}{2n+1}\right)\cdot\dfrac{1}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2n+1}< \dfrac{1}{2}\)
Vậy A < \(\dfrac{1}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 1 2021
\(=\lim\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
\(=\lim\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)
\(=\lim\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)=\dfrac{1}{2}\)
5 tháng 6 2022
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: \(=\dfrac{1}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{\left(4n-3\right)\left(4n+1\right)}\right)\)
\(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{4n-3}-\dfrac{1}{4n+1}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{4n}{4n+1}=\dfrac{n}{4n+1}\)
7 tháng 6 2022
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: \(=\dfrac{1}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{\left(4n-3\right)\left(4n+1\right)}\right)\)
\(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{4n-3}-\dfrac{1}{4n+1}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{4n}{4n+1}=\dfrac{n}{4n+1}\)
\(P=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{\left(2n+1\right)\left(2n+3\right)}\\ 2P=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{\left(2n+1\right)\left(2n+3\right)}\\ =\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n+1}-\dfrac{1}{2n+3}\\ =1-\dfrac{1}{2n+3}\\ =\dfrac{2\left(n+1\right)}{2n+3}\\ P=\dfrac{2\left(n+1\right)}{2n+3}:2\\ =\dfrac{n+1}{2n+3}\)
thanks nha