កំណែ​លំហាត់​សិស្ស​ពូកែ​គណិតវិទ្យា​ភ្នំពេញ​២០១០ (10)
On 07.22.10, In Education, Mathematics, by វិចិត្រ

គេឱ្យ x_0=2010 និង

\displaystyle x_{n+1}=\frac{1+x_n}{1-x_n}, n\ge 0

គណនា x_{2012} ។

ដំណោះស្រាយ

យើងមាន

\displaystyle x_{n+1}=\frac{1+x_n}{1-x_n} \implies x_{n+2}=\frac{1+x_{n+1}}{1-x_{n+1}}=\frac{1+\frac{1+x_n}{1-x_n}}{1-\frac{1+x_n}{1-x_n}}=\frac{2}{-2x_n}=-\frac{1}{x_n}
\implies x_{n+4}=-\frac{1}{x_{n+2}}= x_n

ដូច្នេះ x_0=x_4=x_8=x_{12}=\dotsi=x_{4\times 503}=x_{2012}=2010 ។ x_{2012}=2010 ។

Share and Enjoy:
  • Digg
  • del.icio.us
  • Facebook
  • Mixx
  • Google Bookmarks
  • MySpace
  • StumbleUpon
  • Twitter

Related Posts

វិចិត្រ

Visit Authors Website  |  All Articles From This Author

Leave Your Response

* Name, Email, Comment are Required