Tìm STN n biết :
\(\frac{1}{3}\) + \(\frac{1}{6}\)+ \(\frac{1}{10}\)+...+\(\frac{2}{n\left(n+1\right)}\)=\(\frac{2003}{2004}\)
Giúp mình vs nha. Mình đang cần gấp. Nếu dc thì cố xog trước thứ 2 nhé các bạn. Mình sẽ like và nhờ bn mik like jùm
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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
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
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{3004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(\Rightarrow P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)
\(\Rightarrow P=\frac{1}{5}-\frac{2}{3}\)
\(\Rightarrow P=\frac{3}{15}-\frac{10}{15}\)
\(\Rightarrow P=\frac{-7}{15}\)
Vậy \(P=\frac{-7}{15}\)
\(\frac{1}{3}+\frac{1}{6}+...+\frac{2}{n\left(n+1\right)}=\frac{2003}{2004}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+...+\frac{2}{n\left(n+1\right)}=\frac{2003}{2004}\)
\(\Rightarrow\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{n\left(n+1\right)}=\frac{2003}{2004}\)
\(\Rightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\right)=\frac{2003}{2004}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{2003}{4008}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{n+1}=\frac{2003}{4008}\)\(\Rightarrow\frac{1}{n+1}=\frac{1}{4008}\)\(n+1=4008\Rightarrow n=4007\)
A) \(\frac{1}{2}\cdot\left(\frac{2}{9}+\frac{3}{7}-\frac{5}{27}\right)\)
\(=\frac{1}{2}\cdot\frac{1}{2}\)
\(=\frac{1}{4}\)
B) \(\left(\frac{-5}{28}+1.75+\frac{8}{35}\right):\left(-3\frac{9}{20}\right)\)
\(=\left(\frac{-5}{28}+\frac{7}{4}+\frac{8}{35}\right):\frac{-69}{20}\)
\(=\frac{14}{5}:\frac{-69}{20}\)
\(=\frac{-56}{69}\)
n=2015 . bạn nhân 2 vào lần lượt các p/s rồi đưa 2 ra ngoài
hoi kho day bn a hihihihihiiiiiiiiiiiiiiiiiiiiiiiiiii