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\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{n\left(n+1\right)}=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{n\left(n+1\right)}\)
\(=\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+...+\frac{2}{n\left(n+1\right)}=1-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+\frac{2}{4}-\frac{2}{5}+...+\frac{2}{n}-\frac{2}{n+1}\)
Tới đây dễ rồi bạn rút gọn rồi tìm n
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{n\left(n+2\right)}< \frac{2003}{2004}\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{n}+\frac{1}{n+2}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{n+2}\right)\)
\(=\frac{1}{2}\left(\frac{n+2}{n+2}-\frac{1}{n+2}\right)\)
\(=\frac{1}{2}.\frac{n+1}{n+2}\)
\(=\frac{n+1}{2\left(n+2\right)}< \frac{2003}{2004}\)
\(\Leftrightarrow\hept{\begin{cases}n+1< 2003\\2\left(n+2\right)< 2004\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}n< 2002\\\left(n+2\right)< 1002\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}n< 2002\\n< 1000\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}n+1=2002\\2\left(n+2\right)=1000\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}n=2001\\n=498\end{cases}}\)
Bạn tham khảo câu trả lời tương tự ở đây nhé:
Câu hỏi của Nguyễn Hải - Toán lớp 7 - Học toán với OnlineMath
\(\frac{1}{3}\)+\(\frac{1}{6}\)+\(\frac{1}{10}\)+...+\(\frac{2}{n\left(n+1\right)}\)=\(\frac{2017}{2019}\)
\(\frac{2}{6}\)+\(\frac{2}{12}\)+\(\frac{2}{20}\)+...+\(\frac{2}{n\left(n+1\right)}\)=\(\frac{2017}{2019}\)
2\(\times\)\((\)\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+\(\frac{1}{4.5}\)+...+\(\frac{1}{n.\left(n+1\right)}\)\()\)=\(\frac{2017}{2019}\)
2\(\times\)\((\)\(\frac{1}{2}\)_\(\frac{1}{3}\)+\(\frac{1}{3}\)_\(\frac{1}{4}\)+\(\frac{1}{4}\)_\(\frac{1}{5}\)+...+\(\frac{1}{n}\)_\(\frac{1}{n+1}\)\()\)=\(\frac{2017}{2019}\)
2\(\times\)\((\)\(\frac{1}{2}\)_\(\frac{1}{n+1}\)\()\)=\(\frac{2017}{2019}\)
\(\frac{1}{2}\)_\(\frac{1}{n+1}\)=\(\frac{2017}{4038}\)
\(\frac{1}{n+1}\)=\(\frac{1}{2}\)_\(\frac{2017}{4038}\)
\(\frac{1}{n+1}\)=\(\frac{1}{2019}\)
\(\Rightarrow\)n+1=2019
\(\Rightarrow\)n=2018\(\in\)Z
Vậy n=2018
n=2015 . bạn nhân 2 vào lần lượt các p/s rồi đưa 2 ra ngoài
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{2}{x\left(x+1\right)}=\frac{2003}{2005}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{x\left(x+1\right)}\right)=\frac{1}{2}.\frac{2003}{2005}\)
\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.....+\frac{1}{x\left(x+1\right)}=\frac{2003}{4010}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\left(x+1\right)}=\frac{2003}{4010}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2003}{4010}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2003}{4010}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2003}{4010}=\frac{1}{2005}\)
\(\Rightarrow x+1=2005\Rightarrow x=2004\)
a) Ta có: \(\frac{n}{n-3}\)có tử số lớn hơn mẫu số. \(\Rightarrow\frac{n}{n-3}>1\)
Ta lại có: \(\frac{\left(n+1\right)}{n+2}< 1\)( vì \(\frac{\left(n+1\right)}{n+2}\) có tử bé hơn mẫu)
\(\Rightarrow\frac{n}{n-3}>\frac{\left(n+1\right)}{n+2}\)
b)
Mà: \(\frac{2003.2004-1}{2003.2004}=1\)( Loại hai số giống nhau ở cả tử và mẫu: 2003 , 2004)
Còn: \(\frac{2004.2005-1}{2004.2005}=1\)
\(\Rightarrow\frac{2003.2004-1}{2003.2004}=\frac{2004.2005-1}{2004.2005}\)
P/s: Mình không chắc câu b) Nhé
Ta thấy : n > n - 3
=> \(\frac{n}{n-1}>1\)
Có : n + 1 < n + 2
=> \(\frac{n+1}{n+2}< 1\)
=> \(\frac{n}{n-3}>\frac{n+1}{n+2}\)
1.
a) \(A=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{97\cdot100}\\ A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\\ A=1-\frac{1}{100}=\frac{99}{100}\)
b) Sửa đề: B = 1/2.5 + 1/5.8 + 1/8.11 + ...
\(B=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{92\cdot95}+\frac{1}{95\cdot98}\\ B=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{92\cdot95}+\frac{3}{95\cdot98}\right)\\ B=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}\right)\\ B=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{98}\right)\\ B=\frac{1}{6}-\frac{1}{294}\\ B=\frac{49}{294}-\frac{1}{294}=\frac{48}{294}=\frac{8}{49}\)
2.
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{n\left(n+1\right)}=\frac{1999}{2000}\\ \frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{n\left(n+1\right)}=\frac{1999}{2000}\\ 2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{n\left(n+1\right)}\right)=\frac{1999}{2000}\\ 2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{n\left(n+1\right)}\right)=\frac{1999}{2000}\\ 2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{n}-\frac{1}{n+1}\right)=\frac{1999}{2000}\\ 2\left(\frac{1}{2}-\frac{1}{n+1}\right)=\frac{1999}{2000}\\ \frac{1}{2}-\frac{1}{n+1}=\frac{1999}{2000}:2\\ \frac{1}{2}-\frac{1}{n+1}=\frac{1999}{4000}\\ \frac{1}{2}-\frac{1999}{4000}=\frac{1}{n+1}\\ \frac{1}{n+1}=\frac{1}{4000}\\ \Rightarrow n+1=4000\\ \Rightarrow n=3999\)
Vậy n = 3999
\(\frac{1}{3}+\frac{1}{6}+...+\frac{2}{n\left(n+1\right)}=\frac{2003}{2004}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+...+\frac{2}{n\left(n+1\right)}=\frac{2003}{2004}\)
\(\Rightarrow\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{n\left(n+1\right)}=\frac{2003}{2004}\)
\(\Rightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\right)=\frac{2003}{2004}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{2003}{4008}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{n+1}=\frac{2003}{4008}\)\(\Rightarrow\frac{1}{n+1}=\frac{1}{4008}\)\(n+1=4008\Rightarrow n=4007\)
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