tìm n để 2/1.3 + 2/3.5 +.....+ 2/n(n+2) < 2003/2004
tìm n để 2/1.3+2/3.5+2/5.7+2/7.9+...+2/n.(n+2)<2003/2004
2/1.3+2/3.5+2/5.7+...+2/n.(n+2)=1-1/3+1/3-1/5+1/5-1/7+...+1/n-1/n+2. =1-1/n+2<2003/2004. =>1/n+2>1-2003/2004=1/2004. =>n+2<2004.=>n<2002. Vậy 1<n<2002.
TÌM N ĐỂ
2/1.3+2/3.5+2/5.7+....+2/n(n+2) < 2003/2004
tìm n biết: 2/1.3+2/3.5+...+2/n.(n+2)<2003/2004
\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{n\left(n+2\right)}\)
\(=\frac{3-1}{1.3}+\frac{5-3}{3.5}+...+\frac{n+2-n}{n\left(n+2\right)}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{n}-\frac{1}{n+2}\)
\(=1-\frac{1}{n+2}< \frac{2003}{2004}\)
\(\Leftrightarrow\frac{1}{n+2}>\frac{1}{2004}\)
\(\Leftrightarrow0< n+2< 2004\)
Tìm n để
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{n.\left(n+2\right)}<\frac{2003}{2004}\)
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{n\cdot\left(n+2\right)}<\frac{2003}{2004}\)
\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+2}<\frac{2003}{2004}\)
\(\Rightarrow1-\frac{1}{n+2}<\frac{2003}{2004}\)
\(\Rightarrow\frac{1}{n+2}>\frac{1}{2004}\)
\(\Rightarrow n+2<2004\)
\(\Rightarrow n=2002\)
nhầm bước cuối
\(\Rightarrow n<2002\)
Tìm số tự nhiên n để : 1/1.3+1/3.5+1/5.7+...+1/n.(n+2)<2003/2004
GIÚP MIK VỚI MIK ĐANG CẦN GẤP
\(1/1.3+1/3.5+1/5.7+...+1/n.(n+2)<2003/2004\)
Ta có :=2/2.(1/1.3+1/3.5+1/5.7+...+1/n.(n+2)
=1/2.(2/1.3+2/3.5+2/5.7+...+2/n.(n+2)
=1/2.(1-1/3+1/3-1/5+1/5-1/7+...+1/n-1/n+2)
=1/2.(1-1/n+2)
=1/2.(n+2/n+2-1/n+2)
=1/2.(n+2-1/n+2)
=1/2.n+1/n+2
=n+1/(n+2).2
Vì: n+1/(n+2).2<2003/2004
Suy ra:n+1/(n+2).2=x/2004
Suy ra:(n+2).2=2004
n+2 =1002
n =1000
Vậy n bằng 1000
tìm n để \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{n.\left(n+2\right)}< \frac{2003}{2004}\)
bạn nào ĐÚNG,NHANH,CÓ LỜI GIẢI mình sẽ tick
Đặt A = \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{n.\left(n+2\right)}\)
A=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{n}-\frac{1}{n+2}\)
A = \(1-\frac{1}{n+2}\)
A= \(\frac{n+1}{n+2}\)=> Để A<2003/2004 thì \(\left(n+1\right).2004< \left(n+2\right).2003\)
\(\Leftrightarrow2004n+2004< 2003n+4006\)
\(\Leftrightarrow n< 2002\)
1/1-1/3+1/3-1/5+1/5-1/7+....+1/n-1/(n+2)
=1-1/(n+2)=(n+1)/(n+2)
Suy ra n =2001
CMR 2/1.3 + 2/3.5 + 2/5.7 +...+ 2/n(n+2) < 2003/2004 (. là dấu nha)
Sửa đề \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{n\left(n+2\right)}< \frac{2014}{2014}=1\)
Ta có :
\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{n\left(n+2\right)}\)
\(=\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+...+\left(\frac{1}{n}-\frac{1}{n+1}\right)\)
\(=\left(1-\frac{1}{n+2}\right)+\left[\left(\frac{1}{3}+\frac{1}{5}+...+\frac{1}{n}\right)-\left(\frac{1}{3}+\frac{1}{5}+...+\frac{1}{n}\right)\right]\)
\(=1-\frac{1}{n+2}+0\)
=\(=1-\frac{1}{n+2}\)
Vì \(1-\frac{1}{n+2}< 1\) nên\(\frac{2}{1.3}+\frac{1}{3.5}+...+\frac{1}{n\left(n+2\right)}< 1\left(đpcm\right)\)
Tặng like !!! Tìm n để:
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{n\left(n+2\right)}<\frac{2003}{2004}\)
Chứng tỏ rằng
S = 1/2 + 1/3 + 1/4 + ...+ 1/63 > 2
Cho S = 1/11 + 1/12 + 1/13 + 1/14 + ...+ 1/20 . Hãy so sánh S và 1/2
Tìm n để : 2/1.3 + 2/3.5 + 2/5.7+ ...+ 2/n(n+2) < 2003/2004