Tim n để
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{n\left(n+1\right)}=\frac{2003}{2004}\)
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16 tháng 2 2015
đặt a=1/3+1/6+1/10+...........+2/n(n+1)
1/2a=1/6+1/12+...........+1/n(n+1)
1/2a=1/2.3+1/3.4+........+1/n(n+1)
1/2a=1/2-1/3+1/3-1/4+.......+1/n-1/n+1
1/2a=1/2-1/n+1
a=(1/2--1/n+1):1/2=2003/2004
1/2-1/n+1=2003/2004.1/2
1/2-1/n+1=2003/4008
1/n+1=1/2-2003/4008
1/n+1=1/4008
suy ra n+1=4008
n=4007
11 tháng 12 2016
\(\frac{1}{3}+\frac{1}{6}+...+\frac{2}{n\left(n+1\right)}=\frac{2003}{2004}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+...+\frac{2}{n\left(n+1\right)}=\frac{2003}{2004}\)
\(\Rightarrow\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{n\left(n+1\right)}=\frac{2003}{2004}\)
\(\Rightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\right)=\frac{2003}{2004}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{2003}{4008}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{n+1}=\frac{2003}{4008}\)\(\Rightarrow\frac{1}{n+1}=\frac{1}{4008}\)\(n+1=4008\Rightarrow n=4007\)
23 tháng 8 2018
\(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{2002}-1\right)\left(\frac{1}{2003}-1\right)\)
\(=\left(-\frac{1}{2}\right)\left(-\frac{2}{3}\right)...\left(-\frac{2001}{2002}\right)\left(-\frac{2002}{2003}\right)\)
\(=\frac{-1.\left(-2\right).....\left(-2001\right)\left(-2002\right)}{2.3....2002.2003}\)
\(=\frac{1}{2003}\)
20 tháng 8 2016
1 + 1/3 + 1/6 + 1/10 + ... + 2/x(x + 1) = 4007/2004
2/2 + 2/6 + 2/12 + 2/20 + ... + 2/x(x + 1) = 4007/2004
2 × (1/1×2 + 1/2×3 + 1/3×4 + 1/4×5 + ... + 1/x(x + 1)) = 4007/2004
1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/x - 1/x + 1 = 4007/2004 : 2
1 - 1/x + 1 = 4007/2004 × 1/2
x/x + 1 = 4007/4008
=> x = 4007
DL
8 tháng 1 2019
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{n\left(n+1\right)}=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{n\left(n+1\right)}\)
\(=\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+...+\frac{2}{n\left(n+1\right)}=1-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+\frac{2}{4}-\frac{2}{5}+...+\frac{2}{n}-\frac{2}{n+1}\)
Tới đây dễ rồi bạn rút gọn rồi tìm n
24 tháng 3 2019
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{n\left(n+2\right)}< \frac{2003}{2004}\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{n}+\frac{1}{n+2}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{n+2}\right)\)
\(=\frac{1}{2}\left(\frac{n+2}{n+2}-\frac{1}{n+2}\right)\)
\(=\frac{1}{2}.\frac{n+1}{n+2}\)
\(=\frac{n+1}{2\left(n+2\right)}< \frac{2003}{2004}\)
\(\Leftrightarrow\hept{\begin{cases}n+1< 2003\\2\left(n+2\right)< 2004\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}n< 2002\\\left(n+2\right)< 1002\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}n< 2002\\n< 1000\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}n+1=2002\\2\left(n+2\right)=1000\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}n=2001\\n=498\end{cases}}\)
