Tìm n để \(\frac{1}{1.3}\)+ \(\frac{1}{3.5}\)+ \(\frac{1}{5.7}\)+ . . . + \(\frac{1}{n\left(n+2\right)}\)< \(\frac{2003}{2004}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
DP
2 tháng 8 2017
Sửa đề . \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{n\left(n+2\right)}=\frac{71}{216}\)
\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+2}\right)=\frac{71}{216}\)
\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{n+2}\right)=\frac{71}{216}\)
\(\Leftrightarrow\frac{1}{n+2}=1-\frac{71}{216}\div\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{n+2}=\frac{37}{108}\)
\(\Leftrightarrow x=\frac{34}{37}\Rightarrow\text{(đề sai) }\)
29 tháng 12 2016
A\(A=\frac{1}{1.3}+..+\frac{1}{x\left(x+1\right)}\)
\(2A=\frac{1}{1}-\frac{1}{\left(x+1\right)}\)
\(A=\frac{x}{2.\left(x+1\right)}=\frac{8}{17}=\frac{16}{2.17}\)
X=16
1 tháng 1 2016
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{n\cdot\left(n+2\right)}<\frac{2003}{2004}\)
\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+2}<\frac{2003}{2004}\)
\(\Rightarrow1-\frac{1}{n+2}<\frac{2003}{2004}\)
\(\Rightarrow\frac{1}{n+2}>\frac{1}{2004}\)
\(\Rightarrow n+2<2004\)
\(\Rightarrow n=2002\)
29 tháng 4 2018
\(\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left\{\left(2x+1\right).\left(2x+3\right)\right\}}\right)=\frac{49}{99}\)
\(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}\right)=\frac{49}{99}\)
\(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{2x+3}\right)=\frac{49}{99}\)
\(\frac{1}{2}.\cdot\left(\frac{2x+3}{2x+3}-\frac{1}{2x+3}\right)=\frac{49}{99}\)
\(\frac{1}{2}.\frac{2x+2}{2x+3}=\frac{49}{99}\)
\(\frac{2x+2}{2x+3}=\frac{49}{99}:\frac{1}{2}\)
\(\frac{2x+2}{2x+3}=\frac{98}{99}\)
=) \(2x+2=98\)và \(2x+3=99\)
TH1 : \(2x+2=98\)
\(2x=98-2\)
\(2x=96\)
\(x=96:2\)
\(x=48\)( THỎa mãn )
TH2 :
\(2x+3=99\)
\(2x=99-3\)
\(2x=96\)
\(x=96:2\)
\(x=48\)( THỎa mãn )
Vậy x = 48
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{n\left(n+2\right)}< \frac{2003}{2004}\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{n}+\frac{1}{n+2}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{n+2}\right)\)
\(=\frac{1}{2}\left(\frac{n+2}{n+2}-\frac{1}{n+2}\right)\)
\(=\frac{1}{2}.\frac{n+1}{n+2}\)
\(=\frac{n+1}{2\left(n+2\right)}< \frac{2003}{2004}\)
\(\Leftrightarrow\hept{\begin{cases}n+1< 2003\\2\left(n+2\right)< 2004\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}n< 2002\\\left(n+2\right)< 1002\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}n< 2002\\n< 1000\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}n+1=2002\\2\left(n+2\right)=1000\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}n=2001\\n=498\end{cases}}\)