\(=\frac{2}{6}+\frac{2}{12}+...+\frac{2}{n.\left(n+1\right)}\)

\(=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{n.\left(n+1\right)}\)

còn lại tự làm

mình giải nhé:

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{n\left(n+1\right)}=\frac{2015}{2016}\)

\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{n\left(n+1\right)}=\frac{2015}{2016}\)

\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{n\left(n+1\right)}=\frac{2015}{2016}\)

\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{n\left(n+1\right)}\right)=\frac{2015}{2016}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{n}-\frac{1}{n+1}\right)=\frac{2015}{2016}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{n+1}\right)=\frac{2015}{2016}\)

\(\Leftrightarrow1-\frac{2}{n+1}=\frac{2015}{2016}\Leftrightarrow\frac{2}{n+1}=\frac{1}{2016}\)

\(\Leftrightarrow n+1=4032\Leftrightarrow n=4031\)

Xong rồi nhé ^ ^ chúc bạn học tốt!

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}...+\frac{1}{n.\left(n+1\right)}=\frac{2015}{2016}\)

\(\frac{1.2}{3.2}+\frac{1.2}{6.2}+\frac{1.2}{10.2}+...+\frac{1}{n\left(n+1\right)}=\frac{2015}{2016}\)

\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{1}{n.\left(n+1\right)}=\frac{2015}{2016}\)

\(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{1}{n\left(n+1\right)}=\frac{2015}{2016}\)

\(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{n\left(n+1\right)}\right)=\frac{2015}{2016}\)

\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\right)=\frac{2015}{2016}\)

\(2.\left(\frac{1}{2}-\frac{1}{n+1}\right)=\frac{2015}{2016}\)

\(\frac{1}{2}-\frac{1}{n+1}=\frac{2015}{2016}:2\)

\(\frac{1}{2}-\frac{1}{n+1}=\frac{2015}{4032}\)

\(\frac{1}{n+1}=\frac{1}{2}-\frac{2015}{4032}\)

\(\frac{1}{n+1}=\frac{1}{4032}\)

\(\Rightarrow n+1=4032\)

\(\Rightarrow n=4031\)

1/3 + 1/6 + 1/10 + ... + 2/n(n+1) = 2003/2004

\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{n.\left(n+1\right)}=\frac{2003}{4008}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{2003}{4008}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{n+1}=\frac{2003}{4008}\)

\(\Rightarrow\frac{1}{n+1}=\frac{1}{2}-\frac{2003}{4008}\)

\(\Rightarrow\frac{1}{n+1}=\frac{1}{4008}\)

\(\Rightarrow n+1=4008\)

\(\Rightarrow n=4008-1=4007\)

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{n\left(n+1\right)}=\frac{2003}{2004}\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{n\left(n+1\right)}\right)=\frac{1}{2}.\frac{2003}{2004}\)

\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{n\left(n+1\right)}=\frac{2003}{4008}\)

\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{n\left(n+1\right)}=\frac{2003}{4008}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{2003}{4008}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{n+1}=\frac{2003}{4008}\)

\(\Leftrightarrow\frac{1}{n+1}=\frac{1}{2}-\frac{2003}{4008}=\frac{1}{4008}\)

\(\Rightarrow n+1=4008\Rightarrow n=4007\)

Vậy \(n=4007\)

Câu hỏi của trần như - Toán lớp 7 - Học toán với OnlineMath

Bài 1 em tham khảo tại link trên nhé.