Question

\(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{2n\left(2n+2\right)}=\dfrac{1009}{4038}\) Tìm n?

Answer 1

Ta có \(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{2n\left(2n+2\right)}=\dfrac{1009}{4038}\)

\(\Leftrightarrow\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2n\left(2n+2\right)}=\dfrac{1009}{2019}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2n}-\dfrac{1}{2n+2}=\dfrac{1009}{2019}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2n+2}=\dfrac{1009}{2019}\)

\(\Leftrightarrow\dfrac{n}{2n+2}=\dfrac{1009}{2019}\)

\(\Leftrightarrow2019n=1009\left(2n+2\right)\)

\(\Leftrightarrow2019n=2018n+2018\)

\(\Leftrightarrow n=2018\)